How to calculate following integration? ∫(tan⁡x+cot⁡x)dx

Question

How to calculate following integration?  ∫(tan⁡x+cot⁡x)dx

kaluitagf · Accepted Answer

I=∫(tan⁡x+cot⁡x)dx  =∫sin⁡x+cos⁡xsin⁡xcos⁡xdx  Putting sin⁡x−cos⁡x=u, du=(cos⁡x+sin⁡x)dx, u2=1−2sin⁡xcos⁡x, sin⁡xcos⁡x=u2−12  I=∫2du1−u2=2arcsin⁡u+C=2arcsin⁡(sin⁡x−cos⁡x)+C  where C is an arbitrary constant for indefinite integral.

alexandrebaud43 · Accepted Answer

Substiute tan⁡(x)=u=e2t:  ∫(tan⁡(x)+cot⁡(x))dx=∫(u12+u−12)du1+u2  =∫u12+u−12u+u−1duu  =∫cosh(t)cosh(2t)2dt  =∫2cosh(2t)dsinh(t)  =∫21+2sinh2(t)d2sinh(t)  =2arctan⁡(2sinh(t))+C  =2arctan⁡(tanh(x)−cot⁡(x)2)+C

karton · Accepted Answer

Substitute tan⁡x=t2 ∫tan⁡x+cot⁡xdx=2∫t2+1t4+1dt=2∫1+1t2(t−1t)2+2dt Now make a t−1t substitution and we get the answer.

How to calculate following integration? ∫(tan⁡x+cot⁡x)dx

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