Question

If $I={\int }_{1/2}^1\frac{dx}{x^4\sqrt{1-x^2}}$ then $I$ is equal to

• A. $2$
• B. $2/3$
• C. $1/2$
• D. $\sqrt{3}$

Answer 1

Answer: (C)

$1/2$

Let $I={\int }_{\frac{1}{2}}^1\frac{dx}{x^4\sqrt{1-x^2}}$

Put $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$

$\therefore I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{1}{{\sin }^4\theta }d\theta ={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\text{cosec}}^4\theta d\theta$

Using $\int {\text{cosec}}^{m}xdx=-\frac{\cos x{\cos }^{m-1}x}{m-1}+\frac{m-2}{m-1}\int {\text{cosec}}^{m-2}dx$

$\therefore I={\left[\frac{-1}{3}\cot \theta {\text{cosec}}^2\theta \right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}+\frac{2}{3}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}{\text{cosec}}^2\theta d\theta$

$={\left[\frac{-1}{3}\cot \theta ({\text{cosec}}^2\theta +2)\right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}=\frac{1}{2}$