Question

$ \int \limits_{0} ^{2a} \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 $I=\int_{0}^{2a} \frac{x^{9 2}}{\sqrt{2a-x}}dx$
Put $x = 2 a sin^{2} \, \theta $
$\therefore dx=4a\, sin\,\theta\, cos\,\theta\, d\theta$
when $x = 0, \theta=0$, when $x = 2 a. \theta=\frac{\pi}{2}$.
$\therefore I=\int\limits_{0}^{\pi /2} \frac{\left(2a\,sin^{2}\, \theta\right)^{9/ 2}4a sin\,\theta \,cos\,\theta\,d\theta}{\sqrt{2a-2a sin^{2}\,\theta}}$
$=\int\limits_{0}^{\pi /2}\frac{\left(2a\right)^{9/ 2}sin^{9}\theta. 4a sin\,\theta\,cos\,\theta\,d\theta}{\left(2a\right)^{1/ 2} cos\,\theta}$
$=\int\limits_{0}^{\pi /2}\left(2a\right)^{4}\left(4a\right)sin^{10} \theta\,d\theta$
$=64a^{5} \int\limits_{0}^{\pi /2}sin^{10} \theta\,d\theta$
$=64a^{5}\cdot\frac{9\cdot7\cdot5\cdot3\cdot1}{10\cdot8\cdot6\cdot4\cdot2}\cdot\frac{\pi}{2}$
$=\frac{63 \pi}{8} a^{5}$