Question

${\int }_{\frac{5}{2}}^5\frac{\sqrt{(25-x^2{)}^3}}{x^4}dx$ is equal to

• A. $\frac{\pi }{6}$
• B. $\frac{2\pi }{3}$
• C. $\frac{5\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (D)

$\frac{\pi }{3}$

$I={\int }_{\frac{5}{2}}^5\frac{\sqrt{(25-x^2{)}^3}}{x^4}dx$
Let $x=5\sin \theta$
$\therefore dx=5\cos \theta d\theta$
$\therefore I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{5^3{\cos }^3\theta .5\cos \theta }{5^4{\sin }^4\theta }d\theta$
$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\cot }^2\theta (cose{c}^2\theta -1)d\theta$
$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\cot }^2\theta cose{c}^2\theta d\theta -{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\cot }^2\theta d\theta$
$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\cot }^2\theta cose{c}^2\theta d\theta -{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}(cose{c}^2\theta -1)$
$={[-\frac{{\cot }^3\theta }{3}+\cot \theta +\theta ]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}$
$=-0+0+\frac{\pi }{2}-(-\frac{3\sqrt{3}}{3}+\sqrt{3}+\frac{\pi }{6})$ $=\frac{\pi }{3}$