# Evaluate the integral using trigonometric substitution integral^3_{fraction {3}{2} } x^2 square...

## Question:

Evaluate the integral using trigonometric substitution

{eq}\displaystyle \int^3_{\frac {3}{2} } x^2 \sqrt {x^2 - 9} \ dx {/eq}

## Definite Integration:

Consider a function F(x) such that

{eq}F(x) = \int f(x) dx {/eq}

Then the definite integral can be written as

{eq}\int_a^b f(x) dx = F(b)-F(a) {/eq}

Trigonometric substitution helps us to solve complex integration problems.

{eq}\int x^a dx = \frac{x^{a+1}}{a+1} + c \quad \text{[c is the constant of integration]} {/eq}

{eq}\displaystyle \int^3_{\frac {3}{2}} x^2 \sqrt {x^2 - 9} \ dx {/eq}

We have to integrate the function using trigonometric substitution.

So, let

{eq}x = 3\sin z\\ dx= -3\cos z\\ {/eq}

working on the limits,

when

{eq}x = 3\\ 3\sin z = 3\\ \sin z =1\\ z = \sin^{-1}1\\ z = \dfrac{\pi}{2}\ (upper\ limit)\\ {/eq}

when

{eq}x = \dfrac{3}{2} \\ 3\sin z = \dfrac{3}{2} \\ \sin z =\dfrac{1}{2}\\ z = \sin^{-1}\dfrac{1}{2}\\ z = \dfrac{\pi}{6}\ (lower\ limit)\\ {/eq}

Plugging these in the given function:

{eq}=\displaystyle \int^3_{\frac {3}{2}} x^2 \sqrt {x^2 - 9} dx \\ =\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} \left ( (3\sin z)^2 \sqrt { (3\sin z)^2 - 9} \right ) (-3\cos z dz) \\ =-\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} 9\sin^2z\times 3\cos z \times 3\cos z dz\\ =- 81\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} \sin^2z\times \cos^2 z dz\\ =\dfrac{-81}{4}\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} 4\sin^2z\times \cos^2 z dz {/eq}

since we know that the

{eq}\sin 2x = 2 \sin x \cos x\\ \therefore \sin^2 2x = 4 \sin^2 x \cos^2 x\\ {/eq}

Hence we get simplified function as:

{eq}=\dfrac{-81}{4}\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} 4\sin^2z\times \cos^2 z dz\\ = \dfrac{-81}{4}\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} \sin^2 2z dz\\ = \dfrac{-81}{4}\displaystyle \int^{\frac{\pi}{2}}_{\frac {\pi}{6}} \dfrac{1-\cos 4z}{2}dz\\ = \dfrac{-81}{4}\left ( \left [ \dfrac{z}{2}-\dfrac{\sin 4z}{2\times 4} \right ]^{\frac{\pi}{2}}_{\frac {\pi}{6}} \right ) \\ = \dfrac{-81}{4} \left (\left [ \dfrac{z}{2}-\dfrac{\sin 4z}{8} \right ]^{\frac{\pi}{2}}_{\frac {\pi}{6}} \right ) \\ {/eq}

Now plugging the values of the limits, we get:

{eq}\dfrac{-81}{4} \left (\left [ \dfrac{z}{2}-\dfrac{\sin 4z}{8} \right ]^{\frac{\pi}{2}}_{\frac {\pi}{6}} \right ) \\ = \dfrac{-81}{4} \left [ \left ( \dfrac{\pi}{2\times 2}-\dfrac{\sin 4\frac{\pi}{2}}{8} \right ) - \left ( \dfrac{\pi}{2\times 6}-\dfrac{\sin 4\frac{\pi}{6}}{8} \right ) \right ] \\ = \dfrac{-81}{4} \left [ \left ( \dfrac{\pi}{4}-\dfrac{\sin 2\pi}{8} \right ) - \left ( \dfrac{\pi}{12}-\dfrac{\sin \frac{2\pi}{3}}{8} \right ) \right ] \\ = \dfrac{-81}{4} \left [ \left ( \dfrac{\pi}{4}-0 \right ) - \left ( \dfrac{\pi}{12}-\dfrac{0.866}{8} \right ) \right ] \\ = \dfrac{-81}{4} \left [ 0.785- \left ( 0.261-0.1085 \right ) \right ] \\ = \dfrac{-81}{4} \left [ 0.785-0.1525 \right ] \\ = \dfrac{-81}{4} \left ( 0.6325 \right ) \\ = -12.81 {/eq}