InterviewSolution
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InterviewSolution
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Evaluate: \(\int\limits_0^1\cfrac{tan^{-1}}{(1+x^2)}dx\)∫ tan-1x/(1+x2)dx, x ∈[0,1] |
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Answer» Let \(I=\int\limits_0^1\cfrac{tan^-1x}{1+x^2}dx\) Let tan-1x=t \(\Rightarrow\cfrac{1}{1+x^2}dx=dt.\) Also, when x=0, t=0 and when x=1,t=\(\cfrac{\pi}{4}\) Hence, \(I= \int\limits_0^{\pi/4}t\,dt=\cfrac{1}{2}t^2|_0^{\frac{\pi}{4}}=\cfrac{\pi^2}{32}\) |
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