Question

If $\int sin^{-1} x cos^{-1} x \,dx = f' \left(x\right)$
$ \left[ \frac{\pi}{2} x- xf' \left(x\right) - 2 \sqrt{1-x^{2}}\right] \frac{\pi}{2} \sqrt{1-x^{2}} +2x + C $, then

• A. $ f \left(x\right) = sin\, x$
• B. $ f \left(x\right) = cos\, x$
• C. $ f \left(x\right) = tan\, x$
• D. None of these

Answer 1

Answer: (A)

Let $I=\int \sin ^{-1} x \cos ^{-1} x \,d x$
$=\int \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right) d x$
$=\frac{\pi}{2} \int \sin ^{-1} x d x-\int\left(\sin ^{-1} x\right)^{2} d x$
$=\frac{\pi}{2}\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]-\left[x \cdot\left(\sin ^{-1} x\right)^{2}\right.$
$\left.+2 \sin ^{-1} x \sqrt{1-x^{2}}-2 x\right]+C$
[using integration by parts]
$=\sin ^{-1} x\left(\frac{\pi}{2} x-x \sin ^{-1} x-2 \sqrt{1-x^{2}}\right)$
$+\frac{\pi}{2} \sqrt{1-x^{2}}+2 x+C $
$ \therefore \, f^{-1}(x)=\sin ^{-1} x$
$ \Rightarrow f(x)= \sin x $