Trigonometric Substitution

Rather than trying to find an appropriate u this time, let’s substitute for x instead. Let’s use a function that can be written many different ways, like sinθ.

Note: We can use cosθ as well, but sinθ will make calculating the integral a little easier, as we will see.

$x=\sin \theta$

$dx=\cos \theta d\theta$

What’s another way we can write cosθ? Recall from our trig identities that

$\cos \theta =\sqrt{1-{\sin }^2\theta }$

Let’s also account for the intervals of integration:

As $x:0\to \frac{1}{2},$

$\sin \theta :0\to \frac{\pi }{6}$​
We are assuming that θ lies in quadrants I and IV, which is the standard domain for the function sinθ.

Hopefully you are starting to see how this will work out. If not, perhaps rewriting the integral this way will help

$={\int }_0^{0.5}\frac{1}{\sqrt{(1-x^2}{)}^3}dx$

Now let’s begin substituting:

$={\int }_0^{\frac{\pi }{6}}\frac{\cos \theta }{(\sqrt{1-{\sin }^2\theta }{)}^3}d\theta$​
​
Remember that $\cos \theta =\sqrt{1-{\sin }^2\theta }$​

$={\int }_0^{\frac{\pi }{6}}\frac{\cancel{\cos \theta }}{(\cos \theta {)}^{\cancel{3}2}}d\theta$

(Did you see what just happened?!?)

$={\int }_0^{\frac{\pi }{6}}{\sec }^2\theta d\theta$

Which function has a derivative of ${\sec }^2(x)$? Now, before you go ahead and start deriving trigonometric derivatives using implicit differentiation, it’s tan(x).

$=[\tan \theta {]}_0^{\frac{\pi }{6}}$

$=\tan \frac{\pi }{6}-\tan 0$

$=\frac{1}{\sqrt{3}}-0$$=\frac{\sqrt{\mathbf{3}}}{\mathbf{3}}$

Proof: $\int \text{sec x dx}$

This is one of the trickiest proofs we will ever come across. We have to solve it using u-substitution by substituting $u=\sec x+\tan x$. Then we get:

$du=\sec x\tan x+{\sec }^{2}xdx$

$du=\sec x(\sec x+\tan x)dx$

Now the reason for choosing this substitution becomes more apparent, because it allows for the substitution of $\sec xdx$ (highlighted in green) back into the original integral.

How do we substitute $\sec x+\tan x$, though...? We do something clever.

$\int \sec x\cdot \frac{\sec x+\tan x}{\sec x+\tan x}dx$

We simply multiplied the integral by a special “1,” leaving us with a denominator that is equal to u.​
Making these substitutions gives us

$\int \frac{du}{u}=\int \frac{1}{u}du$

​
$=\mathbf{l}\mathbf{n}\left|\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{x}+\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\right|+\mathbf{C}$

This may seem like a relatively simple integral at first, but any time we see $-x^2$, especially inside a radical, it should alert our brains to the possibility that solving the integral will require trig substitution. Get ready for a tough problem.

In the previous problem, Example 9.2.1, we substituted x = sin(x) and converted $1-x^2$ into cos(x). This is a common strategy in trig substitution. Our job here is to try to change the function within the radical into the form

${\int }_{-1}^2\sqrt{\text{constant - }{\text{(function)}}^2}dx$

The way we do that is by completing the square.

${\int }_{-1}^2\sqrt{8+A-(x^2+2x+A)}dx$​
We want to select A so that $x^2-2x+A$ will become a perfect square. Notice that the value of A does not change the function inside the radical because the A is added, then later subtracted (you can check for yourself by distributing out the negative sign).

The rules of completing the square tell us that since the coefficient of $x^2$ is 1,

$A={\left(\frac{2}{2}\right)}^2=1$

The green 2 in the numerator is because it is the coefficient of x. We now know that A = 1, and the radical can be rewritten as

${\int }_{-1}^2\sqrt{9-(x+1{)}^2}dx$​
The problem now is the 9. How can we change that into a 1 for easier substitution? Let’s just factor it out!

${\int }_{-1}^2\sqrt{9\left(1-\frac{(x+1{)}^2}{9}\right)}dx$​
The rules of radicals and exponents tell us that the above expression is the same as

${\int }_{-1}^2\sqrt{9}\cdot \sqrt{1-\frac{(x+1{)}^2}{3^2}}dx$​
${\int }_{-1}^{2}3\sqrt{1-{\left(\frac{x+1}{3}\right)}^2}dx$

Aaaaah, now we can begin substituting. We want the integral to have $1-{\sin }^2\theta$ in it.

$\sin \theta =\frac{x+1}{3}$

$\cos \theta d\theta =\frac{1}{3}dx$

$3\cos \theta d\theta =dx$

Plugging these substitutions into the integral, we get

$={\int }_1^{2}3\sqrt{1-{\sin }^2\theta }\cdot 3\cos \theta d\theta$​
​
Since $1-{\sin }^2\theta ={\cos }^2\theta$, the square root of which equals cosθ, we now have in our integrand

${\int }_{-1}^{2}9{\cos }^2\theta d\theta$

Before we integrate this, notice we are missing something. Can you guess what it is? The intervals of integration. They are still in the form of x ranging from -1 to 2. So we ask ourselves, when x ranges from -1 to 2, what is the range of θ? From our original substitution,

$\theta ={\sin }^{-1}\left(\frac{x+1}{3}\right)$

${\sin }^{-1}(0)=0$

${\sin }^{-1}(1)=\frac{\pi }{2}$

Plugging this into our intervals, we get

$={\int }_0^{\frac{\pi }{2}}9{\cos }^2\theta d\theta$

What to do about ${\cos }^2\theta$, though? We want to reduce its order to 1 somehow.

Let’s seek guidance from our trigonometric identities, more specifically, the double angle formulas. Can you spot a ${\cos }^2\theta$? It states that $\cos (2\theta )=2{\cos }^2\theta -1$. Going from that, we can see that

$={\cos }^2\theta =\frac{\cos (2\theta )+1}{2}$​
​
We can substitute this into our integral to reduce the power of the integrand by 1, giving us

$={\int }_0^{\frac{\pi }{2}}\frac{9}{2}\left(\cos (2\theta )+1\right)d\theta$

We can actually go ahead and evaluate this integral because we know that the integral of $\cos (2\theta )$ is equal to $0.5\sin (2\theta )$ .

$=\frac{9}{2}\left({\left[\frac{\sin (2\theta )}{2}\right]}_0^{\frac{\pi }{2}}+[\theta {]}_0^{\frac{\pi }{2}}\right)$​
$=\frac{9}{2}\left[\left(\frac{\sin (\pi )}{2}-\frac{\sin (0)}{2}\right)+\left(\frac{\pi }{2}-0\right)\right]$​
Both $\sin (\pi )$ and sin(0) are 0, so the first term is 0, leaving us with the integral being equal to

$=\frac{9}{2}\cdot \frac{\pi }{2}$​

$=\frac{\mathbf{9}\pi }{\mathbf{4}}$​

Recall that the purpose of an integral is to calculate the area under the curve of the integrand. So, if we know what kind of function the integrand is, then perhaps that will help us calculate its area. Let’s call the integrand y so that

​
$y=\sqrt{-x^2-2x+8}$

$y^2=-x^2-2x+8$​
$x^3+2x+y^2=8$​
Completing the square for the x-terms on the left side gives

$x^2+2x+1+y^2=9$

Notice that the equation still stays balanced if you subtract 1 from both sides.

$(x+1{)}^2+y^2=3^2$

This equation should look familiar. It’s the equation of a circle of radius 3, with a center at (-1, 0). However, due to the square root sign in the original function, we are only looking at the upper half of the circle (if the area in question was the lower half, there would be a negative sign in front of the square root).​

Now, what is the area of a semi-circle with radius 3 and a center at (-1, 0) from x = -1 to x = 2. It’s equal to half the area of the semi-circle!

$A=\frac{1}{4}\pi (3{)}^2$

$=\frac{\mathbf{9}\pi }{\mathbf{4}}$