Question

If $I={\int }_0^{1/\sqrt{3}}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$ then $I$ is equal to

• A. $\pi /2$
• B. $\pi /2\sqrt{2}$
• C. $\pi /4\sqrt{2}$
• D. $\pi /4$

Answer 1

Answer: (C)

$\pi /4\sqrt{2}$

Let $I=\int \frac{1}{(1+x^2)\sqrt{1-x^2}}dx$
Substitute $x=\sin t\Rightarrow dx=\cos tdt$
$I=\int \frac{1}{{\sin }^{2}t+1}dt$
Multiply numerator and denominator by ${\csc }^{2}t$
$I=\int \frac{{\csc }^{2}t}{{\csc }^{2}t+1}dt=\int \frac{{\csc }^{2}t}{{\cot }^{2}t+2}dt$
Substitute $u=\cot t\Rightarrow du=-{\csc }^{2}tdt$
$I=-\int \frac{1}{u^2+2}du=-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{u}{\sqrt{2}}$
$=-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\cot t}{\sqrt{2}}=-\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{\cot {\sin }^{-1}x}{\sqrt{2}}$
Therefore
${\int }_0^{1/\sqrt{3}}\frac{1}{(1+x^2)\sqrt{1-x^2}}dx=\frac{\pi }{4\sqrt{2}}$