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) –\(\frac{(cos^{-1}⁡x)^2}{2}+C\)
 
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This interesting question is from Methods of Integration-1 in division Integrals of Mathematics – Class 12
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1 Answers
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