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detailed solution
Correct option is A
Let I=∫tanθdθ Let tanθ=x2 . Then, d(tanθ)=dx2 or sec2θdθ=2xdx or dθ=2x⋅dxsec2θ=2xdx1+tan2θ=2xdx1+x4I=∫x2×2xdx1+x4=∫2x2x4+1dx=∫2x2+1/x2dx=∫1+1/x2+1−1/x2x2+1/x2dx=∫1+1/x2x2+1/x2dx+∫1−1/x2x2+1/x2dx=∫1+1/x2(x−1/x)2+2dx+∫1−1/x2(x+1/x)2−2dxputting x−1x=u in the first integral and x+1x=v in the second integral we getI=∫duu2+(2)2+∫dvv2−(2)2=12tan−1u2+122logv−2v+2+C=12tan−1x−1/x2+122logx+1/x−2x+1/x+2+C=12tan−1x2−1x2+122logx2−x2+1x2+x2+1+C=12tan−1tanθ−12tanθ=12tan−1tanθ−12tanθ+122logtanθ−2tanθ+1tanθ+2tanθ+1+CSimilar Questions
is equal to
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