Question

Find \(\int \frac{cos^{-1}x}{\sqrt{1-x^2}} dx\).

(a) \(\frac{(sin^{-1}⁡x)^2}{2}+C\)

(b) \(\frac{(cos^{-1}⁡x)^2}{7}+C\)

(c) \(\frac{(cos^{-1}⁡x)^2}{2}+C\)

(d) –\(\frac{(cos^{-1}⁡x)^2}{2}+C\)

I had been asked this question at a job interview.

This interesting question is from Methods of Integration-1 in division Integrals of Mathematics – Class 12

Answer 1

Correct choice is (c) \(\frac{(cos^{-1}⁡x)^2}{2}+C\)

For explanation I would say: Let cos^-1⁡x=t

Differentiating w.r.t x, we get

\(\frac{1}{\sqrt{1-x^2}} dx=dt\)

∴\(\int \frac{cos^{-1}⁡x}{\sqrt{1-x^2}} dx=\int t dt\)

=\(\frac{t^2}{2}\)

Replacing t with cos^-1x,we get

\(\int \frac{cos^{-1}⁡x}{\sqrt{1-x^2}} dx=\frac{(cos^{-1}⁡x)^2}{2}+C\)