Question

${\int }_0^{\frac{1}{\sqrt{2}}}\frac{{\sin }^{-1}x}{{(1-x^2)}^{\frac{3}{2}}}=A-\frac{1}{2}\log 2$ then A = ..........

Answer 1

L.H.S is, ${\int }_0^{\frac{1}{\sqrt{2}}}\frac{{\sin }^{-1}x}{{(1-x^2)}^{\frac{3}{2}}}dx$

put, ${\sin }^{-1}x=t⟺x=\sin t$

differentiating on both sides

$⟹\frac{1}{\sqrt{1-x^2}}dx=dt$

Now, the L.H.S becomes ${\int }_0^{\frac{\pi }{4}}\frac{t}{(1-{\sin }^{2}t)}dt$

$={\int }_0^{\frac{\pi }{4}}t{\sec }^{2}tdt$

Applying the UV formula, we get (here U=$t$ and V=${\sec }^{2}t$)

$={[t\int {\sec }^{2}tdt]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\int {\sec }^{2}tdtdt$

$={\left[t\tan t\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\tan tdt$

$={\left[t\tan t\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\frac{\sin t}{\cos t}dt$

put, $\cos t=m⟹-sintdt=dm$

$={\left[t\tan t\right]}_0^{\frac{\pi }{4}}+{\int }_1^{\frac{1}{\sqrt{2}}}\frac{1}{m}dm$

$={\left[t\tan t\right]}_0^{\frac{\pi }{4}}+{\left[\ln \left|m\right|\right]}_1^{\frac{1}{\sqrt{2}}}$

$=\frac{\pi }{4}-\frac{1}{2}\log 2$

Comparing, this result with the R.H.S,

we obtain, A=$\frac{\pi }{4}$.