# Trigonometric substitution

## Trigonometric Substitution

Not all integrals can be solved by using u-substitution. For example let's say we want to integrate

Are we able to use u-substitution 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 u-substitution fails. This seems to be the case for a lot of functions with square roots. This is why we introduce a new method called trig-substitution. 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 trig-substitution table that shows you the strategy in using trig-substitution.

## Trig Substitution Table

Now that we know the idea behind these trigonometric substitutions, why don't we integrate some functions?

## 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 step-by step solution each question! Have fun with trig substitution integration!

### Trigonometric substitution

In this section, we will look at evaluating trigonometric functions with trigonometric substitution. A lot of people normally substitute using trig identities, which you will have to memorize. However, Dennis will use a different and easier approach. He will use the idea of Pythagoras Theorem and make a relation with sine and cosine in terms of x to substitute. In more advance questions, we will use Pythagoras Theorem to make a relation with tangent and secant to substitute. No matter what type of questions you do, you will only need to know Pythagoras Theorem and nothing else!

#### Lessons

Pre-requisite: Trigonometry Ratio: "SOH-CAH-TOA"
• 1.
u-Substitution VS. Trigonometric Substitution
Evaluate:
a)
$\int \frac{1}{\sqrt{9-x^2}}xdx$

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

• 2.
Evaluate the integral (Type A: $\sqrt{a^2-x^2}$).
a)
$\int \frac{1}{x^2\sqrt{16-x^2}}dx$

• 3.
Evaluate the integral (Type B: $\sqrt{a^2+x^2}$).
a)
$\int \frac{1}{x^2\sqrt{x^2+9}}dx$