Trigonometric substitution
Trigonometric Substitution
Not all integrals can be solved by using usubstitution. For example let’s say we want to integrate
Are we able to use usubstitution here? Well let us set
If we are to substitute this, our integral will become:
As you can see, this is a problem because our integral is not all in terms of u. So you would probably have to try something else. However no matter how hard you try, it will never work and usubstitution fails. This seems to be the case for a lot of functions with square roots. This is why we introduce a new method called trigsubstitution. Trig substitutions help us integrate functions with square roots in them.
Trig Substitution Rules
As explained earlier, we want to use trigonometric substitution when we integrate functions with square roots. However, there are many different cases of square root functions. So how exactly do we know what type of trig we use as a substitution? Well here are 3 types you will most likely encounter, and these are the substitutions you would use:
where a is a constant.
How to do trig substitution
The reason why we use these substitutions for each case is because they make use of the following trig identities:
For example in case 1, using the substitution gives:
Using the identity
We get
We can see that the substitution cleans up the function really well here, and can be easily integrated.
Similarly in case 2:
Now using the identity:
We have:
Again, the function is cleaned up and this can be easily integrated. Lastly in case 3,
Using the trig identity,
We now have
If this was an integral, then we would find the integral of tangent with coefficient a to find the answer.
Now in all cases, make sure to also find the derivative of the substitution because you are also going to substitute dx. So
In summary, here is a trigsubstitution table that shows you the strategy in using trigsubstitution.
Trig Substitution Table
Trigonometric substitution integrals
Let’s start by finding the integral of $\sqrt{1  x^{2}}$. In other words,
Question 1: Integrate
 Where do we start here? Let’s rewrite the integral to
 Notice that this looks really similar to $\sqrt{a^{2}  x^{2}}$, except $a = 1$. So that means we need to use the substitution $x = (1) \sin \theta$. So we set:
 So substituting gives:
 Now this is just an integral of a trig function. Notice that we need to use the identity:
 We are going to integrate that. This is one of the famous trig identities integrals you have probably done before. So our trig integrals become:
 We may have finished integrating this trig function, but we are not done yet. Notice that our original integral was in terms of $x$, so we have to make sure our answer is in terms of $x$ as well. We need to know how to convert $\theta, \sin \theta$, and $\cos \theta$ in terms of $x$. See that earlier our substitution was
 Isolating for $\theta$ gives:
 However, what is $\cos \theta$ exactly? Well notice that:
 We can use this to draw a triangle where the opposite length is $x$ and hypotenuse is $1$.
 Notice that the length of the adjacent is:
 In addition, we know that
 Hence substituting our answer back in terms of $x$ gives:
which is answer for the integral of $\sqrt{1  x^{2}}$. Now let’s look at another question similar to this.
Question 2: Integrate $\frac{1}{\sqrt{1  x^{2}}}$. In other words, integrate
 This is very similar to the integral in question 1, but now the square root actually appears in the denominator. Now this question is actually pretty easy if you know about the integrals of inverse trig functions. Recall that:
 So right away, you would know what the answer is. However, what if you don’t spot this right away? Well, you can still use trigonometric substitution to obtain this answer! Notice from the denominator that:
 Again, the denominator is in the form $\sqrt{a^{2}  x^{2}}$, except $a = 1$. So that means we need to use the substitution $x = (1) \sin \theta$. So we set:
 Remember we also need to use the trigonometric identity:
 So substituting gives:
 Notice that
 So converting the answer back in terms of $x$ gives:
which matches the answer above. Now don’t we try to do another similar question like this?
Question 3: Integrate
 Now again, these are one of those inverse trig integrals where:
 However, what if I did not know that in the first place? Again, use trigonometric substitution! First we are going to get rid of the absolute value and set a restriction for ourselves.
 Note that the square root in the denominator is in the form $\sqrt{x^{2}  a^{2}}$, except $a = 1$. So we use the substitution
 In addition we will use the trigonometric identity:
 So substituting gives us:
 Now remember we have to convert $\theta$ back in terms of $x$. Notice from substitution earlier
 So our final answer becomes:
which again is the exact answer as the above. Let’s now do some harder ones that require to use the triangle to convert back in terms of $x$ again.
Question 4: Integrate
 Notice in the denominator that it is in the form $\sqrt{x^{2} + a^{2}}$, except $a = 3$. So that means we need to use the substitution $x = (3) \tan \theta$. We set:
 Remember we also need to use the trigonometric identity:
 So substituting gives:
 Again, this is just an integral of trig function. We know the integral of $\sec$ is $\ln \sec \theta + \tan \theta + C$. So
 Remember that we need to convert back in terms of $x$. So we need to find what how to express $\sec \theta$ in terms of $x$. Know that:
 Drawing the triangle we see that:
 This means
 So our final answer is
Let’s look at a slightly harder question that requires us to use case 3 of trigonometric substitution rule.
Question 5: Integrate
 First we need to play around the inside of the square root. Notice that:
 Note the square root in the denominator is in the form $\sqrt{x^{2}  a^{2}}$, except $a = 2$ and $x$→$3x$. So we use the substitution
 In addition we will use the trigonometric identity:
 So substituting gives us:
 Again, this is just an integral of a trig function. So,
 Now we need to figure out how to convert $\sin \theta$ back in terms of $x$. Notice that:
 So drawing the triangle to find the opposite gives us:
 Notice that:
 Hence our final answer gives us that:
If you would like to do more trig substitution integrals, I suggest you look at this link.
http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx
There are a lot more integral substitution questions here, and Paul Lamar gives a stepby step solution each question! Have fun with trig substitution integration!
Trigonometric substitution
Basic concepts:
 Combination of SohCahToa questions
Lessons
Notes:
Prerequisite: Trigonometry Ratio: “SOHCAHTOA”

a)
$\int \frac{1}{\sqrt{9x^2}}xdx$

b)
$\int \frac{1}{\sqrt{9x^2}}dx$


2.
Evaluate the integral (Type A: $\sqrt{a^2x^2}$).

b)
$\int \frac{\sqrt{49x^2}}{x^2}dx$

3.
Evaluate the integral (Type B: $\sqrt{a^2+x^2}$).

b)
$\int \frac{x^3}{(16x^2+25)^{\frac{3}{2}}}dx$