# Find $$\int \frac{cos^{-1}x}{\sqrt{1-x^2}} dx$$.

Category: QuestionsFind $$\int \frac{cos^{-1}x}{\sqrt{1-x^2}} dx$$.
Editor">Editor Staff asked 11 months ago

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) –$$\frac{(cos^{-1}⁡x)^2}{2}+C$$

This interesting question is from Methods of Integration-1 in division Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

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