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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 |
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Answer» CORRECT CHOICE is (c) \(\frac{(cos^{-1}X)^2}{2}+C\) For explanation I would say: Let cos^-1x=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 \(\int \frac{cos^{-1}x}{\sqrt{1-x^2}} dx=\frac{(cos^{-1}x)^2}{2}+C\) |
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