Question

$\int \frac{{\sin }^{-1}x}{{\left(1-x^2\right)}^{\frac{3}{2}}}dx$

• A. $\frac{x\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}+\frac{1}{2}\log \left|\left(1-x^2\right)\right|+C.$
• B. $\frac{1}{2}\log \left|\left(1-x^2\right)\right|+C$
• C. $\frac{x\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}+C$
• D. $4+\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{x\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}+\frac{1}{2}\log \left|\left(1-x^2\right)\right|+C.$

$\begin{array}{c}Putx=\sin tso,thatdx=costdtandt=si{n}^{-1}x\\ \therefore \int \frac{{\sin }^{-1}x}{{\left(1-x^2\right)}^{\frac{3}{2}}}dx=\int \frac{t\cos t}{{\left(1-{\sin }^{2}t\right)}^{\frac{3}{2}}}dt=\int \frac{tcost}{{\cos }^{3}t}dt\\ =\int t{\sec }^{2}tdt\\ =t.\left(\tan t\right)-\int 1.\tan tdt\\ =t\left(\tan t\right)+\log \left|\cos t\right|+C\\ ={\sin }^{-1}x\frac{x}{\sqrt{1-x^2}}+\log \left|\sqrt{1-x^2}\right|+C\\ =\frac{x\left({\sin }^{-1}x\right)}{\sqrt{1-x^2}}+\frac{1}{2}\log \left|\left(1-x^2\right)\right|+C.\\ hence,theoptionAiscorrectanswer.\end{array}$