Evaluate the following integrals:
Question:

Evaluate the following integrals:

$\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{3 / 2}} d x$

Solution:

$\int \frac{x^{2}}{\left(a^{2}-x^{2}\right)^{3 / 2}} d x$

PUT $x=a \sin \theta$, so $d x=a \cos \theta d \theta$ and $\theta=\sin ^{-}(x / a)$

Above equation becomes,

$=\int \frac{a^{2} \sin ^{2} \theta}{\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{3 / 2}}(a \cos \theta d \theta)=\int \frac{a^{2} \sin ^{2} \theta}{\left(a^{2}\right)\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{3 / 2}}(a \cos \theta d \theta)\left\{\right.$ take a $^{2}$ outside $)$

$=\int \frac{\mathrm{a}^{2} \sin ^{2} \theta}{\left(\mathrm{a}^{2}\right) 3 / 2\left(\mathrm{a}^{2}-\mathrm{a}^{2} \sin ^{2} \theta\right)^{3 / 2}}(\operatorname{acos} \theta \mathrm{d} \theta)=\int \sin ^{2} \theta * \frac{\cos \theta}{\cos ^{2} \theta} \mathrm{d} \theta$

$=\int \frac{\sin ^{2} \theta}{\cos ^{2} \theta} \mathrm{d} \theta=\int \tan ^{2} \theta \mathrm{d} \theta=\int\left(\sec ^{2} \theta-1\right) \mathrm{d} \theta\left(\sec ^{2} \theta-1=\tan ^{2} \theta\right)$

$=\int \sec ^{2} \theta \mathrm{d} \theta-\int \theta \mathrm{d} \theta=\tan \theta+\mathrm{c}-\theta$

$=\tan \theta-\theta+\mathrm{c}$

Put $\theta=\sin ^{-}(x / a)$

$=\tan \theta * \sin ^{-}\left(\frac{x}{a}\right)-\sin ^{-}\left(\frac{x}{a}\right)+c$