Question

$\underset{0}{\overset{2a}{\int }}\frac{x^{\frac{9}{2}}}{\sqrt{2a-x}}dx=$

• A. $8\pi a^5$
• B. $\frac{63}{8}\pi a^5$
• C. $\frac{64}{8}\pi a^5$
• D. none of these.

Answer 1

Answer: (B)

Let $I={\int }_0^{2a}\frac{x^{92}}{\sqrt{2a-x}}dx$
Put $x=2asi{n}^2\theta$
$\therefore dx=4asin\theta cos\theta d\theta$
when $x=0,\theta =0$, when $x=2a.\theta =\frac{\pi }{2}$.
$\therefore I=\underset{0}{\overset{\pi /2}{\int }}\frac{{\left(2asi{n}^2\theta \right)}^{9/2}4asin\theta cos\theta d\theta }{\sqrt{2a-2asi{n}^2\theta }}$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{{\left(2a\right)}^{9/2}si{n}^9\theta .4asin\theta cos\theta d\theta }{{\left(2a\right)}^{1/2}cos\theta }$
$=\underset{0}{\overset{\pi /2}{\int }}{\left(2a\right)}^4\left(4a\right)si{n}^{10}\theta d\theta$
$=64a^5\underset{0}{\overset{\pi /2}{\int }}si{n}^{10}\theta d\theta$
$=64a^5\cdot \frac{9\cdot 7\cdot 5\cdot 3\cdot 1}{10\cdot 8\cdot 6\cdot 4\cdot 2}\cdot \frac{\pi }{2}$
$=\frac{63\pi }{8}{a}^5$