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8.4: Trigonometric Substitutions - Mathematics


So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. This seems like a "reverse'' substitution, but it is really no different in principle than ordinary substitution.

Example (PageIndex{1})

Evaluate

[int sqrt{1-x^2},dx.]

Solution

Let (x=sin u) so (dx=cos u,du). Then

[ int sqrt{1-x^2},dx=intsqrt{1-sin^2 u}cos u,du= intsqrt{cos^2 u}cos u,du. ]

We would like to replace ( sqrt{cos^2 u}) by (cos u), but this is valid only if (cos u ) is positive, since ( sqrt{cos^2 u}) is positive. Consider again the substitution (x=sin u). We could just as well think of this as (u=arcsin x). If we do, then by the definition of the arcsine, (-pi/2le ulepi/2), so (cos uge0). Then we continue:

[eqalign{ intsqrt{cos^2 u}cos u,du&=intcos^2u,du=int {1+cos 2uover2},du = {uover 2}+{sin 2uover4}+Ccr &={arcsin xover2}+{sin(2arcsin x)over4}+C.cr }]

This is a perfectly good answer, though the term (sin(2arcsin x)) is a bit unpleasant. It is possible to simplify this. Using the identity (sin 2x=2sin xcos x), we can write

[sin 2u=2sin ucos u=2sin(arcsin x)sqrt{1-sin^2 u}= 2xsqrt{1-sin^2(arcsin x)}=2xsqrt{1-x^2}.]

Then the full antiderivative is

[ {arcsin xover2}+{2xsqrt{1-x^2}over4}= {arcsin xover2}+{xsqrt{1-x^2}over2}+C. ]

This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity (sin^2x+cos^2x=1) in one of three forms:

[ cos^2 x=1-sin^2x,]

[ sec^2x=1+ an^2x,]

or

[ an^2x=sec^2x-1.]

If your function contains (1-x^2), as in the example above, try (x=sin u); if it contains (1+x^2) try (x= an u); and if it contains ( x^2-1), try (x=sec u). Sometimes you will need to try something a bit different to handle constants other than one.

Example (PageIndex{2})

Evaluate [intsqrt{4-9x^2},dx.]

Solution

We start by rewriting this so that it looks more like the previous example:

[ intsqrt{4-9x^2},dx=intsqrt{4(1-(3x/2)^2)},dx =int 2sqrt{1-(3x/2)^2},dx. ]

Now let (3x/2=sin u) so ((3/2),dx=cos u ,du) or (dx=(2/3)cos u,du). Then

[eqalign{ int 2sqrt{1-(3x/2)^2},dx&=int 2sqrt{1-sin^2u},(2/3)cos u,du ={4over3}int cos^2u,ducr &={4uover 6}+{4sin 2uover12}+Ccr &={2arcsin(3x/2)over3}+{2sin u cos uover3}+Ccr &={2arcsin(3x/2)over3}+{2sin(arcsin(3x/2))cos(arcsin(3x/2))over3}+Ccr &={2arcsin(3x/2)over3}+{2(3x/2)sqrt{1-(3x/2)^2}over3}+Ccr &={2arcsin(3x/2)over3}+{xsqrt{4-9x^2}over2}+C,cr }]

using some of the work fromExample(PageIndex{1}),

Example (PageIndex{3})

Evaluate [intsqrt{1+x^2},dx.]

Solution

Let (x= an u), (dx=sec^2 u,du), so

$$ intsqrt{1+x^2},dx=int sqrt{1+ an^2 u}sec^2u,du= intsqrt{sec^2u}sec^2u,du. ]

Since (u=arctan(x)), (-pi/2le ulepi/2) and (sec uge0), so (sqrt{sec^2u}=sec u). Then $$intsqrt{sec^2u}sec^2u,du=int sec^3 u ,du.$$ In problems of this type, two integrals come up frequently: ( intsec^3u,du) and (intsec u,du). Both have relatively nice expressions but they are a bit tricky to discover.

First we do (intsec u,du), which we will need to compute (intsec^3u,du$):

$$eqalign{ intsec u,du&=intsec u,{sec u + an uover sec u + an u},ducr &=int{sec^2 u +sec u an uover sec u + an u},du.cr }]

Now let (w=sec u + an u), (dw=sec u an u + sec^2u,du), exactly the numerator of the function we are integrating. Thus

$$eqalign{ intsec u,du=int{sec^2 u +sec u an uover sec u + an u},du&= int{1over w},dw=ln |w|+Ccr &=ln|sec u + an u|+C.cr }]

Now for (intsec^3 u,du):

$$eqalign{ sec^3u&={sec^3uover2}+{sec^3uover2}={sec^3uover2}+{( an^2u+1)sec uover 2}cr &={sec^3uover2}+{sec u an^2 uover2}+{sec uover 2}= {sec^3u+sec u an^2uover 2}+{sec uover 2}.cr }]

We already know how to integrate (sec u), so we just need the first quotient. This is "simply'' a matter of recognizing the product rule in action: $$int sec^3u+sec u an^2u,du=sec u an u.]

So putting these together we get

$$ intsec^3u,du={sec u an uover2}+{ln|sec u + an u| over2}+C, ]

and reverting to the original variable (x):

$$eqalign{ intsqrt{1+x^2},dx&={sec u an uover2}+{ln|sec u + an u|over2}+Ccr &={sec(arctan x) an(arctan x)over2} +{ln|sec(arctan x) + an(arctan x)|over2}+Ccr &={ xsqrt{1+x^2}over2} +{ln|sqrt{1+x^2} +x|over2}+C,cr }]

using ( an(arctan x)=x) and (sec(arctan x)=sqrt{1+ an^2(arctan x)}=sqrt{1+x^2}).


Trigonometric Substitution

We use trigonometric substitution in cases where applying trigonometric identities is useful. In particular, trigonometric substitution is great for getting rid of pesky radicals. For example, if we have (sqrt) in our integrand (and (u) -sub doesn't work) we can let (x= an heta.) Then we get

In this case, we used the identity (1+ an^2 heta = sec^2 heta) .

Here are the three types of trigonometric substitutions:

SUBSTITUTION IDENTITY USE WHEN
(x=asin heta) (cos^2 heta + sin^2 heta = 1) (sqrt)
(x=a an heta) (1 + an^2 heta = sec^2 heta) (sqrt)
(x=asec heta) (1 + an^2 heta = sec^2 heta) (sqrt)


8.4: Trigonometric Substitutions - Mathematics

We have already seen that recognizing the product rule can be useful, when we noticed that $int sec^3u+sec u an^2u,du=sec u an u.$ As with substitution, we do not have to rely on insight or cleverness to discover such antiderivatives there is a technique that will often help to uncover the product rule.

Start with the product rule: $f(x)g(x)=f'(x)g(x)+f(x)g'(x).$ We can rewrite this as $f(x)g(x)=int f'(x)g(x),dx +int f(x)g'(x),dx,$ and then $int f(x)g'(x),dx=f(x)g(x)-int f'(x)g(x),dx.$ This may not seem particularly useful at first glance, but it turns out that in many cases we have an integral of the form $int f(x)g'(x),dx$ but that $int f'(x)g(x),dx$ is easier. This technique for turning one integral into another is called integration by parts, and is usually written in more compact form. If we let $u=f(x)$ and $v=g(x)$ then $du=f'(x),dx$ and $dv=g'(x),dx$ and $int u,dv = uv-int v,du.$ To use this technique we need to identify likely candidates for $u=f(x)$ and $dv=g'(x),dx$.

Example 8.4.2 Evaluate $dsint xsin x,dx$. Let $u=x$ so $du=dx$. Then we must let $dv=sin x,dx$ so $v=-cos x$ and $int xsin x,dx=-xcos x-int -cos x,dx= -xcos x+int cos x,dx=-xcos x+sin x+C.$

Example 8.4.4 Evaluate $dsint x^2sin x,dx$. Let $u=x^2$, $dv=sin x,dx$ then $du=2x,dx$ and $v=-cos x$. Now $ds int x^2sin x,dx=-x^2cos x+int 2xcos x,dx$. This is better than the original integral, but we need to do integration by parts again. Let $u=2x$, $dv=cos x,dx$ then $du=2$ and $v=sin x$, and $eqalign< int x^2sin x,dx&=-x^2cos x+int 2xcos x,dxcr &=-x^2cos x+ 2xsin x - int 2sin x,dxcr &=-x^2cos x+ 2xsin x + 2cos x + C.cr >$

Such repeated use of integration by parts is fairly common, but it can be a bit tedious to accomplish, and it is easy to make errors, especially sign errors involving the subtraction in the formula. There is a nice tabular method to accomplish the calculation that minimizes the chance for error and speeds up the whole process. We illustrate with the previous example. Here is the table:

To form the first table, we start with $u$ at the top of the second column and repeatedly compute the derivative starting with $dv$ at the top of the third column, we repeatedly compute the antiderivative. In the first column, we place a "$-

8.4: Trigonometric Substitutions - Mathematics

As we have done in the last couple of sections, let’s start off with a couple of integrals that we should already be able to do with a standard substitution.

Both of these used the substitution (u = 25 - 4) and at this point should be pretty easy for you to do. However, let’s take a look at the following integral.

In this case the substitution (u = 25 - 4) will not work (we don’t have the (x,dx) in the numerator the substitution needs) and so we’re going to have to do something different for this integral.

It would be nice if we could reduce the two terms in the root down to a single term somehow. The following substitution will do that for us.

Do not worry about where this came from at this point. As we work the problem you will see that it works and that if we have a similar type of square root in the problem we can always use a similar substitution.

Before we actually do the substitution however let’s verify the claim that this will allow us to reduce the two terms in the root to a single term.

Now reduce the two terms to a single term all we need to do is recall the relationship,

Using this fact the square root becomes,

So, not only were we able to reduce the two terms to a single term in the process we were able to easily eliminate the root as well!

Note, however, the presence of the absolute value bars. These are important. Recall that

There should always be absolute value bars at this stage. If we knew that ( an heta ) was always positive or always negative we could eliminate the absolute value bars using,

Without limits we won’t be able to determine if ( an heta ) is positive or negative, however, we will need to eliminate them in order to do the integral. Therefore, since we are doing an indefinite integral we will assume that ( an heta ) will be positive and so we can drop the absolute value bars. This gives,

So, we were able to reduce the two terms under the root to a single term with this substitution and in the process eliminate the root as well. Eliminating the root is a nice side effect of this substitution as the problem will now become somewhat easier to do.

Let’s now do the substitution and see what we get. In doing the substitution don’t forget that we’ll also need to substitute for the (dx). This is easy enough to get from the substitution.

[x = frac<2><5>sec heta hspace <0.5in>Rightarrow hspace<0.25in>,,,,,dx = frac<2><5>sec heta an heta ,d heta ]

Using this substitution the integral becomes,

With this substitution we were able to reduce the given integral to an integral involving trig functions and we saw how to do these problems in the previous section. Let’s finish the integral.

So, we’ve got an answer for the integral. Unfortunately, the answer isn’t given in (x)’s as it should be. So, we need to write our answer in terms of (x). We can do this with some right triangle trig. From our original substitution we have,

This gives the following right triangle.

From this we can see that,

We can deal with the ( heta ) in one of any variety of ways. From our substitution we can see that,

While this is a perfectly acceptable method of dealing with the ( heta ) we can use any of the possible six inverse trig functions and since sine and cosine are the two trig functions most people are familiar with we will usually use the inverse sine or inverse cosine. In this case we’ll use the inverse cosine.

So, with all of this the integral becomes,

We now have the answer back in terms of (x).

Wow! That was a lot of work. Most of these won’t take as long to work however. This first one needed lots of explanation since it was the first one. The remaining examples won’t need quite as much explanation and so won’t take as long to work.

However, before we move onto more problems let’s first address the issue of definite integrals and how the process differs in these cases.

The limits here won’t change the substitution so that will remain the same.

Using this substitution the square root still reduces down to,

However, unlike the previous example we can’t just drop the absolute value bars. In this case we’ve got limits on the integral and so we can use the limits as well as the substitution to determine the range of ( heta ) that we’re in. Once we’ve got that we can determine how to drop the absolute value bars.

Here’s the limits of ( heta ) and note that if you aren’t good at solving trig equations in terms of secant you can always convert to cosine as we do below.

Now, we know from solving trig equations, that there are in fact an infinite number of possible answers we could use. In fact, the more “correct” answer for the above work is,

[ heta = 0 + 2pi n = 2pi nhspace<0.25in>& hspace <0.25in> heta = frac <3>+ 2pi nhspace <0.25in>n = 0, pm 1, pm 2, pm 3, ldots ]

So, which ones should we use? The answer is simple. When using a secant trig substitution and converting the limits we always assume that ( heta ) is in the range of inverse secant. Or,

Note that we have to avoid ( heta = frac<2>) because secant will not exist at that point. Also note that the range of ( heta ) was given in terms secant even though we actually used inverse cosine to get the answers. This will not be a problem because even though inverse cosine can give ( heta = frac<2>) we’ll never get it in our work above because that would require that we started with the secant being undefined and that will not happen when converting the limits as that would in turn require one of the limits to also be undefined!

So, in finding the new limits we didn’t need all possible values of ( heta ) we just need the inverse cosine answers we got when we converted the limits. Therefore, if we are in the range (frac<2> <5>le x le frac<4><5>) then ( heta ) is in the range of (0 le heta le frac<3>) and in this range of ( heta )’s tangent is positive and so we can just drop the absolute value bars.

Let’s do the substitution. Note that the work is identical to the previous example and so most of it is left out. We’ll pick up at the final integral and then do the substitution.

Note that because of the limits we didn’t need to resort to a right triangle to complete the problem.

Let’s take a look at a different set of limits for this integral.

Again, the substitution and square root are the same as the first two examples.

Let’s next see the limits ( heta ) for this problem.

Remember that in converting the limits we use the results from the inverse secant/cosine. So, for this range of (x)’s we have (frac<<2pi >> <3>le heta le pi ) and in this range of ( heta ) tangent is negative and so in this case we can drop the absolute value bars, but will need to add in a minus sign upon doing so. In other words,

So, the only change this will make in the integration process is to put a minus sign in front of the integral. The integral is then,

In the last two examples we saw that we have to be very careful with definite integrals. We need to make sure that we determine the limits on ( heta ) and whether or not this will mean that we can just drop the absolute value bars or if we need to add in a minus sign when we drop them.

Before moving on to the next example let’s get the general form for the secant trig substitution that we used in the previous set of examples and the assumed limits on ( heta ).

Let’s work a new and different type of example.

Now, the terms under the root in this problem looks to be (almost) the same as the previous ones so let’s try the same type of substitution and see if it will work here as well.

Using this substitution, the square root becomes,

So, using this substitution we will end up with a negative quantity (the tangent squared is always positive of course) under the square root and this will be trouble. Using this substitution will give complex values and we don’t want that. So, using secant for the substitution won’t work.

However, the following substitution (and differential) will work.

[x = 3sin heta hspace<0.5in>hspace<0.25in>dx = 3cos heta ,d heta ]

With this substitution the square root is,

[sqrt <9 - > = 3sqrt <1 - <^2> heta > = 3sqrt <<^2> heta > = 3left| ight| = 3cos heta ]

We were able to drop the absolute value bars because we are doing an indefinite integral and so we’ll assume that everything is positive.

In the previous section we saw how to deal with integrals in which the exponent on the secant was even and since cosecants behave an awful lot like secants we should be able to do something similar with this.

Now we need to go back to (x)’s using a right triangle. Here is the right triangle for this problem and trig functions for this problem.

We aren’t going to be doing a definite integral example with a sine trig substitution. However, if we had we would need to convert the limits and that would mean eventually needing to evaluate an inverse sine. So, much like with the secant trig substitution, the values of ( heta ) that we’ll use will be those from the inverse sine or,

Here is a summary for the sine trig substitution.

There is one final case that we need to look at. The next integral will also contain something that we need to make sure we can deal with.

First, notice that there really is a square root in this problem even though it isn’t explicitly written out. To see the root let’s rewrite things a little.

This terms under the root are not in the form we saw in the previous examples. Here we will use the substitution for this root.

With this substitution the denominator becomes,

Now, because we have limits we’ll need to convert them to ( heta ) so we can determine how to drop the absolute value bars.

As with the previous two cases when converting limits here we will use the results of the inverse tangent or,

So, in this range of ( heta ) secant is positive and so we can drop the absolute value bars.

There are several ways to proceed from this point. Normally with an odd exponent on the tangent we would strip one of them out and convert to secants. However, that would require that we also have a secant in the numerator which we don’t have. Therefore, it seems like the best way to do this one would be to convert the integrand to sines and cosines.

We can now use the substitution (u = cos heta ) and we might as well convert the limits as well.

[egin heta & = 0 & hspace <0.75in>& u = cos 0 = 1 heta & = frac<4>& hspace <0.75in>& u = cos frac <4>= frac<><2>end]

Here is a summary for this final type of trig substitution.

Before proceeding with some more examples let’s discuss just how we knew to use the substitutions that we did in the previous examples.

The main idea was to determine a substitution that would allow us to reduce the two terms under the root that was always in the problem (more on this in a bit) into a single term and in doing so we were also able to easily eliminate the root. To do this we made use of the following formulas.

[egin25 - 4 & hspace <0.25in>Rightarrow hspace<0.25in> heta - 1 = < an ^2> heta 9 - & hspace <0.25in>Rightarrow hspace<0.25in>1 - heta = heta 36 + 1 & hspace <0.25in>Rightarrow hspace<0.25in>< an ^2> heta + 1 = heta end]

If we step back a bit we can notice that the terms we reduced look like the trig identities we used to reduce them in a vague way.

For instance, (25 - 4) is something squared (i.e. the (25)) minus a number (i.e. the 4) and the left side of formula we used, ( heta - 1), also follows this basic form. So, because the two look alike in a very vague way that suggests using a secant substitution for that problem. We can notice similar vague similarities in the other two cases as well.

If we keep this idea in mind we don’t need the “formulas” listed after each example to tell us which trig substitution to use and since we have to know the trig identities anyway to do the problems keeping this idea in mind doesn’t really add anything to what we need to know for the problems.

So, in the first example we needed to “turn” the 25 into a 4 through our substitution. Remembering that we are eventually going to square the substitution that means we need to divide out by a 5 so the 25 will cancel out, upon squaring. Likewise, we’ll need to add a 2 to the substitution so the coefficient will “turn” into a 4 upon squaring. In other words, we would need to use the substitution that we did in the problem.

The same idea holds for the other two trig substitutions.

Notice as well that we could have used cosecant in the first case, cosine in the second case and cotangent in the third case. So, why didn’t we? Simply because of the differential work. Had we used these trig functions instead we would have picked up a minus sign in the differential that we’d need to keep track of. So, while these could be used they generally aren’t to avoid extra minus signs that we need to keep track of.

Next, let’s quickly address the fact that a root was in all of these problems. Note that the root is not required in order to use a trig substitution. Instead, the trig substitution gave us a really nice of eliminating the root from the problem. In this section we will always be having roots in the problems, and in fact our summaries above all assumed roots, roots are not actually required in order use a trig substitution. We will be seeing an example or two of trig substitutions in integrals that do not have roots in the Integrals Involving Quadratics section.

Finally, let’s summarize up all the ideas with the trig substitutions we’ve discussed and again we will be using roots in the summary simply because all the integrals in this section will have roots and those tend to be the most likely places for using trig substitutions but again, are not required in order to use a trig substitution.

Now, we have a couple of final examples to work in this section. Not all trig substitutions will just jump right out at us. Sometimes we need to do a little work on the integrand first to get it into the correct form and that is the point of the remaining examples.

In this case the quantity under the root doesn’t obviously fit into any of the cases we looked at above and in fact isn’t in the any of the forms we saw in the previous examples. Note however that if we complete the square on the quadratic we can make it look somewhat like the above integrals.

Remember that completing the square requires a coefficient of one in front of the (). Once we have that we take half the coefficient of the (x), square it, and then add and subtract it to the quantity. Here is the completing the square for this problem.

[2left( <- 2x - frac<7><2>> ight) = 2left( <- 2x + 1 - 1 - frac<7><2>> ight) = 2left( << ight)>^2> - frac<9><2>> ight) = 2 ight)^2> - 9]

Now, this looks (very) vaguely like ( heta - 1) (i.e. something squared minus a number) except we’ve got something more complicated in the squared term. That is okay we’ll still be able to do a secant substitution and it will work in pretty much the same way.

[x - 1 = frac<3><>sec heta hspace<0.25in>x = 1 + frac<3><>sec heta hspace<0.25in>dx = frac<3><>sec heta an heta ,d heta ]

Using this substitution the root reduces to,

[sqrt <2- 4x - 7> = sqrt <2< ight)>^2> - 9> = sqrt <9<^2> heta - 9> = 3sqrt <<< an >^2> heta > = 3left| < an heta > ight| = 3 an heta ]

Note we could drop the absolute value bars since we are doing an indefinite integral. Here is the integral.

And here is the right triangle for this problem.

This doesn’t look to be anything like the other problems in this section. However it is. To see this we first need to notice that,

Upon noticing this we can use the following standard Calculus I substitution.

We do need to be a little careful with the differential work however. We don’t have just an (<<f>^x>) out in front of the root. Instead we have an (<<f>^<4x>>). So, we’ll need to strip one of those out for the differential and then use the substitution on the rest. Here is the substitution work.

This is now a fairly obvious trig substitution (hopefully). The quantity under the root looks almost exactly like (1 + < an ^2> heta ) and so we can use a tangent substitution. Here is that work.

[u = an heta hspace<0.25in>du = heta ,d heta hspace<0.5in>sqrt <1 + > = sqrt <1 + << an >^2> heta > = sqrt <<^2> heta > = left| ight|]

Because we are doing an indefinite integral we can assume secant is positive and drop the absolute value bars. Applying this substitution to the integral gives,

We’ll finish this integral off in a bit. Before we get to that there is a “quicker” (although not super obvious) way of doing the substitutions above. Let’s cover that first then we’ll come back and finish working the integral.

We can notice that the (u) in the Calculus I substitution and the trig substitution are the same (u) and so we can combine them into the following substitution.

We can then compute the differential. Just remember that all we do is differentiate both sides and then tack on (dx) or (d heta ) onto the appropriate side. Doing this gives,

With this substitution the square root becomes,

Again, we can drop the absolute value bars because we are doing an indefinite integral. The integral then becomes,

So, the same integral with less work. However, it does require that you be able to combine the two substitutions in to a single substitution. How you do this type of problem is up to you but if you don’t feel comfortable with the single substitution (and there’s nothing wrong if you don’t!) then just do the two individual substitutions. The single substitution method was given only to show you that it can be done so that those that are really comfortable with both kinds of substitutions can do the work a little quicker.

Now, let’s finish the integral work.

Here is the right triangle for this integral.

This was a messy problem, but we will be seeing some of this type of integral in later sections on occasion so we needed to make sure you’d seen at least one like it.

So, as we’ve seen in the final two examples in this section some integrals that look nothing like the first few examples can in fact be turned into a trig substitution problem with a little work.


Exercises

Integrate each of the given functions:

This question is in the form of the first substitution suggestion in this section, that is,

So we have `a=4`, `x= 4 sin &theta`, and `dx = 4 cos &theta d&theta`.

Substituting and simplifying the square root part first:

`sqrt(16-x^2) =sqrt(16-16 sin^2 theta)`

Substituting into the integral gives:

`intsqrt(16-x^2) dx =int4 cos theta(4 cos theta d theta) `

`=16int1/2(cos 2 theta+1)d theta`

`=8(sin theta cos theta + theta)+K`

The second-last step comes from drawing a triangle, using `sin theta = x/4` in this case, as follows:

Triangle to find `theta`, `sin theta` and `cos theta` in terms of `x`.

Quite often we can get different forms of the same final answer! That is, math software (or another human) can produce an answer which is actually correct, but in a different formto the one given here since. If this happens, don't panic! Just check your solution perhaps by substituting various values for `x`, or (better), drawing the graph using software.

This contains a `sqrt(a^2-x^2)` term, so we will use a substitution of `x =a sin theta`.

So `a=2`, and we let `x = 2 sin &theta`, so `dx = 2 cos &theta d&theta`.

Substituting and simplifying the square root gives:

This time our triangle will use `sin theta = x/2`, as follows:

Triangle to find `csc theta` and `cot theta` in terms of `x`.

Substituting everything into the integral gives:

`int(3 dx)/(xsqrt(4-x^2)) = int(3(2 cos theta d theta))/((2 sin theta)(2 cos theta))`

`=3/2int(d theta)/(sin theta)`

`=3/2intcsc theta d theta`

`=3/2ln |csc theta-cot theta|+K`

`=3/2ln |2/x-(sqrt(4-x^2))/x|+K`

`=3/2ln|(2-sqrt(4-x^2))/x|+K`

If we put `u = x + 1`, then `du = dx` and our integral becomes:

Now, we use `u = sec &theta` and so `du = sec &theta tan &theta d&theta`

The triangle in this case starts with `x+1=sec theta` (that is, `cos theta = 1/(x+1)`), and is as follows:

Triangle to find `sec theta` and `tan theta` in terms of `x`.

Returning to our integral, we have:

`int(dx)/(sqrt(x^2+2x)) =int(du)/(sqrt(u^2-1))`

`=int(sec theta tan theta d theta)/(tan theta)`

`=int sec theta d theta`

`=ln |sec theta+tan theta|+K`

`=ln |x+1+sqrt(x^2+2x)|+K`


Exercises on Trigonometric Substitution

Exercise. Evaluate the following integrals.

$(1) quad displaystyle int sqrt<1-9t^2 >, dt$

$(4) quad displaystyle int x^3 sqrt<4-x^2>, dx$

$(5) quad displaystyle int sqrt <25-t^2>, dt$

$(6) quad displaystyle int (4-x^2)^<3/2>, dx$

$(8) quad displaystyle int e^x sqrt<4-e^<2x>>, dx$

$(9) quad displaystyle int frac<1><(1+x^2)^<3/2>>, dx$

$(10) quad displaystyle int frac<1>>, dx$

Exercise. Evaluate the following integrals.

$(1) quad displaystyle int_1^2 frac>, dx$

$(2) quad displaystyle int_4^6 frac>, dx$

$(3) quad displaystyle int_0^1 x sqrt, dx$

$(5) quad displaystyle int_0^a x^2 sqrt, dx$

$(6) quad displaystyle int_0^ <3/5>sqrt<9-25x^2>, dx$

Exercise. Find the area of the region bounded by the hyperbola $<9x^2-4y^2=36>$ and the line $x=3$.

Exercise. Find the arc length of the curve over $y=frac<1> <2>x^2$ on the interval $[0,4]$.

Exercise. Evaluate the following integrals.

$(1) quad displaystyle int frac<1><2u^2-12u+36>, dx$

$(5) quad displaystyle int_1^2 frac<1>>, dx$

$(6) quad displaystyle int_0^4 sqrt, dx$

Exercise. Verify the special integration formulas ($a > 0$).

$(2) quad displaystyle int sqrt, du =frac<1> <2>left( u sqrt -a^2 ln | u+sqrt | ight) +C$, $u > a$

$(3) quad displaystyle int sqrt, du =frac<1> <2>left( u sqrt +a^2 ln |u+sqrt| ight) +C $.


When I was an undergraduate, I attempted to derive the entire table of integrals in the back of my Thomas Calculus book. I found that trigonometric substitution helped me, or was useful in solving something like 30 of the 141 integrals, and so I would call the technique pretty useful, although I was biased as I really loved the method and had very strong trig skills, hence no fear. I ended up writing a class paper on the technique for my junior exit seminar.

In general, a substitution works when the change of variables is complete, and the resulting integral lends itself to a reversal of the process into the original variables. The best way to get an understanding of when trig subs will work is to do alot of them and see how it works so to speak.

Start by integrating all of the derivatives of the inverse trig functions. Proceed to several integrals over rational expressions involving polynomials of degree 2 or less (with and without nested radicals in various places). You may fail, or sometimes find it is not the best technique. Here is an entertaining one for you: Consider $int x , dx$. Let $x= an heta$. This is clearly a ridiculous method to use for this particular integral, but try it and see that you do not get the solution! Silly things like this will help you to spot when the technique is viable.

You need not necessarily have terms in the form $a^2-x^2$, $x^2-a^2$, and $x^2+a^2$. For example, $int frac<1> , dx$ lends itself to trig sub. You need to complete the square on the quadratic, to get $x^2+3x+5=left( x+frac<3> <2> ight)^2+frac<11><4>$, then let $left(x+frac<3><2> ight)=frac><2> an heta$. So here we are forcing the integrand to have the desired form.

You need not necessarily have quadratic polynomials to apply trig subs, for example, consider $int frac<1><1-e^<2x>>, dx$. Let $e^x = sin heta$. Do your implicit differentiation and see that the technique will work beautifully here.

The question of learning "when" to apply a technique boils down to experience. You really have to just apply the techniques you know as much as possible and see what happens. When you are learning the answer to your very question by application, and you find that the application was a bad idea, I wager that you have learned as much as you would if you had applied a technique and found it appropriate.

If you are really interested in the technique, after learning the three standard trig subs you find in any calculus book (the only three you should ever need), try going crazy and applying say a cosine substitution in place of a sine substitution, a cosecant instead of a secant and other such silliness just to see what happens. You will sometimes find you hit a brick wall when doing such things, and sometimes you will get to the end. In general doing these sorts of rebellious things just to see what happens will only make your skills stronger and help you to truly understand what is going on with your methods.


Absolute value in trigonometric substitutions

In general, when we are trying to remove radicals from integrals, we perform a trigonometric substitution (either a circular or hyperbolic trig function), but often this results in a radical of the form $sqrt<(f(x))^2>$, with $f$ being an arbitrary trigonometric function.

What most texts tend to do is simply take $sqrt <(f(x))^2>= f(x)$, without the absolute value of |f(x)|, and the texts do not offer any motivation as to why $sqrt <(f(x))^2>= f(x) eq |f(x)|$. I would have assumed the correct way to proceed would be $sqrt <(f(x))^2>= |f(x)|$. Why is this the case?

I'll give an example to show further explain what I'm trying to ask :

What most texts do is omit the absolute value in the last starred step. Thus the denomitor of the integral becomes $ 4sec heta $ instead of $4cdot|sec heta |$ and there is no need to break the integral up into cases. Why is that so? We have not assumed $sec heta > 0$, so how can $|sec heta | = sec heta$?


Calculus Early Transcendentals: Integral & Multi-Variable Calculus for Social Sciences

So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a “reverse” substitution, but it is really no different in principle than ordinary substitution.

Example 2.26 . Sine Substitution.

Evaluate (dsint sqrt<1-x^2>,dx ext<.>)

Let (x=sin u) so (dx=cos u,du ext<.>) Then

We would like to replace (ds sqrt) by (cos u ext<,>) but this is valid only if (cos u) is positive, since (ds sqrt) is positive. Consider again the substitution (x=sin u ext<.>) We could just as well think of this as (u=arcsin x ext<.>) If we do, then by the definition of the arcsine, (-pi/2le ulepi/2 ext<,>) so (cos uge0) and so we are allowed to continue and perform the simplification:

This is a perfectly good answer, though the term (sin(2arcsin x)) is a bit unpleasant. It is possible to simplify this. Using the identity (sin 2x=2sin xcos x ext<,>) we can write

Then the full antiderivative is

This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity (ds sin^2x+cos^2x=1) in one of three forms:

If your function contains (ds 1-x^2 ext<,>) as in the example above, try (x=sin u ext<>) if it contains (ds 1+x^2) try (x= an u ext<>) and if it contains (ds x^2-1 ext<,>) try (x=sec u ext<.>) Sometimes you will need to try something a bit different to handle constants other than inverse substitution, which is described next.

In a we let (u=u(x) ext<,>) i.e., our new variable is defined in terms of (x ext<.>) In an we let (x=g(u) ext<,>) i.e., we assume (x) can be written in terms of (u ext<.>) We cannot do this arbitrarily since we do NOT get to “choose” (x ext<.>) For example, an inverse substitution of (x=1) will give an obviously wrong answer. However, when (x=g(u)) is an invertible function, then we are really doing a (u)-substitution with (u=g^<-1>(x) ext<.>) Now the Substitution Rule applies.

Sometimes with inverse substitutions involving trig functions we use ( heta) instead of (u ext<.>) Thus, we would take (x=sin heta) instead of (x=sin u ext<.>)

We would like our inverse substitution (x=g(u)) to be a one-to-one function, and (x=sin u) is not one-to-one. In the next few paragraphs, we discuss how we can overcome this issue by using the restricted trigonometric functions.

The three common are the restricted sine, restricted tangent and restricted secant. Thus, for sine we use the domain ([-pi/2,

pi/2]) and for tangent we use ((-pi/2,

pi/2) ext<.>) Depending on the convention chosen, the restricted secant function is usually defined in one of two ways.

One convention is to restrict secant to the region ([0,

pi/2)cup(pi/2,pi]) as shown in the middle graph. The other convention is to use ([0,

3pi/2)) as shown in the right graph. Both choices give a one-to-one restricted secant function and no universal convention has been adopted. To make the analysis in this section less cumbersome, we will use the domain ([0,

3pi/2)) for the restricted secant function. Then (sec^<-1>x) is defined to be the inverse of this restricted secant function.

Typically trigonometric substitutions are used for problems that involve radical expressions. The table below outlines when each substitution is typically used along with their restricted intervals.


Contents

Examples of Case I Edit

Example 1 Edit

Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions. In that case, the antiderivative gives

Example 2 Edit

may be evaluated by letting x = a sin ⁡ θ , d x = a cos ⁡ θ d θ , θ = arcsin ⁡ x a , >,>

For example, the definite integral

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields


Integration by trigonometric substitution Calculator

Example

Solved Problems

Difficult Problems

Solved example of integration by trigonometric substitution

We can solve the integral $intsqrtdx$ by applying integration method of trigonometric substitution using the substitution

Differentiate both sides of the equation $x=2 anleft( heta ight)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if $$, then $$

The derivative of the linear function is equal to $1$

Now, in order to rewrite $d heta$ in terms of $dx$, we need to find the derivative of $x$. We need to calculate $dx$, we can do that by deriving the equation above

Substituting in the original integral, we get

Factor by the greatest common divisor $4$

The power of a product is equal to the product of it's factors raised to the same power

Applying the power of a power property

Applying the trigonometric identity: $ an(x)^2+1=sec(x)^2$

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

When multiplying exponents with same base you can add the exponents: $secleft( heta ight)secleft( heta ight)^2$

Subtract the values $3$ and $-2$

Any expression to the power of $1$ is equal to that same expression

Rewrite $secleft( heta ight)^<3>$ as the product of two secants

We can solve the integral $intsecleft( heta ight)^2secleft( heta ight)d heta$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

Taking the derivative of secant function: $fracleft(sec(x) ight)=sec(x)cdot an(x)cdot D_x(x)$

First, identify $u$ and calculate $du$

Now, identify $dv$ and calculate $v$

The integral of $sec(x)^2$ is $ an(x)$

When multiplying two powers that have the same base ($ anleft( heta ight)$), you can add the exponents

Now replace the values of $u$, $du$ and $v$ in the last formula

Solve the product $4left( anleft( heta ight)secleft( heta ight)-int anleft( heta ight)^2secleft( heta ight)d heta ight)$

Apply the formula: $intsecleft(x ight) anleft(x ight)^2dx$=intsecleft(x ight)^3dx-intsecleft(x ight)dx$, where $x= heta $

The integral of the secant function is given by the following formula, $displaystyleintsec(x)dx=lnleft|sec(x)+ an(x) ight|$

Rewrite $secleft( heta ight)^3$ as the product of two secants

We can solve the integral $intsecleft( heta ight)^2secleft( heta ight)d heta$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

First, identify $u$ and calculate $du$

Now, identify $dv$ and calculate $v$

The integral of $sec(x)^2$ is $ an(x)$

Now replace the values of $u$, $du$ and $v$ in the last formula

Solve the product $-4left( anleft( heta ight)secleft( heta ight)-int anleft( heta ight)^2secleft( heta ight)d heta ight)$

Apply the formula: $intsecleft(x ight) anleft(x ight)^2dx$=intsecleft(x ight)^3dx-intsecleft(x ight)dx$, where $x= heta $

The integral of the secant function is given by the following formula, $displaystyleintsec(x)dx=lnleft|sec(x)+ an(x) ight|$

Cancel like terms $-4lnleft(secleft( heta ight)+ anleft( heta ight) ight)$ and $4lnleft(secleft( heta ight)+ anleft( heta ight) ight)$

Simplify the integral $intsecleft( heta ight)^3d heta$ applying the reduction formula, $displaystyleintsec(x)^dx=frac>+fracintsec(x)^dx$


Watch the video: Trigonometric Substitution (December 2021).