# Evaluate using trig substitution. {eq}\int_{0}^{a} x^2 \sqrt{a^2 - x^2} {/eq} dx

## Question:

Evaluate using trig substitution.

{eq}\int_{0}^{a} x^2 \sqrt{a^2 - x^2} {/eq} dx

## Definite Integrals:

In the given problem, we will evaluate the given definite integral using suitable trigonometric substitutions. Using substitution method makes the integral much easier for evaluation. As we make substitutions, we will also change the limits of the integral accordingly.

{eq}\begin{align*} \ &=\int_{0}^{a} x^2 \sqrt{a^2 - x^2} \ dx \end{align*} {/eq}

Put,

{eq}\begin{align*} \ & x = a \ \sin t \end{align*} {/eq}

Differentiating both sides, we get

{eq}\begin{align*} \ & dx = a \ \cos t \ dt \end{align*} {/eq}

Also when

{eq}\begin{align*} \ & x = 0 \ , t=0 \end{align*} {/eq} and {eq}\begin{align*} \ & x = a \ , t= \frac{\pi}{2} \end{align*} {/eq}

Substituting the above values, we get

{eq}\begin{align*} \ &= \int_{0}^{\frac{\pi}{2}} a^2 \sin^2 t \sqrt{a^2 - a^2 \ sin^2 t} \ \ a \cos t \ dt \\ \\ \ &= \int_{0}^{\frac{\pi}{2}} a^3 \sin^2 t \ \cos t \sqrt{a^2(1 - \ sin^2 t)} \ dt \\ \\ \ &= \int_{0}^{\frac{\pi}{2}} a^3 \sin^2 t \ \cos t \sqrt{a^2 \ \cos^2 t} \ dt \ \ \ \ \ \ \ \ \ \ \left[ \ Because \ 1- \sin^2 x= \cos^2 x \right] \\ \\ \ &= \int_{0}^{\frac{\pi}{2}} a^4 \sin^2 t \ \cos^2 t \ dt \end{align*} {/eq}

Multilpy numerator and denominator by 4, we get

{eq}\begin{align*} \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} (4 \sin^2 t \cos^2 t) \ dt \\ \\ \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} (2 \sin t \cos t)^2 \ dt \\ \\ \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} \sin^2 2t \ dt \\ \\ \ &= \frac{a^4}{4}\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 4t}{2} \ dt\ \ \ \ \ \ \ \ \ \ \left[ \ Because \sin^2 2x= \frac{1-\cos 4x}{2} \right] \\ \\ \ &= \frac{a^4}{8}\int_{0}^{\frac{\pi}{2}} (1-\cos 4t) \ dt \\ \\ \ &= \frac{a^4}{8} \left[t-\frac{\sin 4t}{4}\right]_{0}^{\frac{\pi}{2}} \\ \\ \ &= \frac{a^4}{8} \left[\frac{\pi}{2}-\frac{\sin 2\pi}{4} - 0+ 0\right] \\ \\ \ &= \frac{a^4}{8} \left[\frac{\pi}{2}-0- 0+ 0\right] \\ \\ \ &= \frac{a^4 \pi}{16} \end{align*} {/eq}