Question

Evaluate: $\underset{0}{\overset{1\sqrt{2}}{\int }}\frac{si{n}^{-1}x}{\left(1-x^2\right)3/2}dx$

• A. $\frac{\pi }{4}-\frac{1}{2}log2$
• B. $\frac{\pi }{4}-log2$
• C. $\frac{\pi }{2}-log2$
• D. $\frac{\pi }{2}-\frac{1}{2}log2$

Answer 1

Answer: (A)

Let $si{n}^{-1}x=\theta \Rightarrow x=sin\theta .Then,$

$dx=cos\theta d\theta$

Now, $x=0\Rightarrow \theta =0$

and $x=\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}$

$\therefore I=\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}\frac{si{n}^{-1}x}{{\left(1-x^2\right)}^{\frac{3}{2}}}dx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\theta cos\theta d\theta }{{\left(co{s}^2\theta \right)}^{\frac{3}{2}}}$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\theta se{c}^2\theta d\theta ={\left[\theta tan\theta \right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}1\cdot tan\theta d\theta$

$=\left(\frac{\pi }{4}-0\right)+{\left[logcos\theta \right]}_0^{\frac{\pi }{4}}=\frac{\pi }{4}-\frac{1}{2}log2$