Question

Value of ${\int }_0^{2a}{x}^{3/2}{(2a-x)}^{-1/2}dx$ is

• A. $\frac{3}{4}\pi a^2$
• B. $\frac{3}{2}\pi a^2$
• C. $\pi a^2$
• D. $2\pi a^2$

Answer 1

The correct option is B $\frac{3}{2}\pi a^2$

Let $I={\int }_0^{2a}\frac{x^{\frac{3}{2}}}{\sqrt{2a-x}}dx$
Substitute $u=\sqrt{x}\Rightarrow du=\frac{1}{2\sqrt{x}}dx$
$I=2{\int }_0^{\sqrt{2a}}\frac{u^4}{\sqrt{2a-u^2}}du$
Substitute $u=\sqrt{2a}\sin t\Rightarrow du=\sqrt{2a}\cos tdt$
$I=2\sqrt{2a}{\int }_0^{\frac{\pi }{2}}\left(2\sqrt{2}{a}^{\frac{3}{2}}{\sin }^{4}t\right)dt=8a^2{\int }_0^{\frac{\pi }{4}}{\sin }^{4}tdt$
Using reduction formulae
$\int {\sin }^{m}xdx=\frac{-\cos x{\sin }^{m-1}x}{m}+\frac{m-1}{m}\int {\sin }^{m-2}xdx$
$I={[-2a^2{\sin }^{3}t\cos t]}_0^{\frac{\pi }{2}}+6a^2{\int }_0^{\frac{\pi }{2}}{\sin }^{2}tdt$
$=0+6a^2{\int }_0^{\frac{\pi }{2}}(\frac{1}{2}-\frac{1}{2}\cos 2t)dt$
$={[3a^{2}t-3a^2\sin t\cos t]}_0^{\frac{\pi }{2}}=\frac{3\pi a^2}{2}$