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Evaluate the following integrals int(tan^2theta+1)/(tan^2theta-1)d theta |
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Answer» Solution :`int(tan^2theta+1)/(tan^2theta-1)d THETA` PUT `TANTHETA=t` `sec^2theta d theta=DT` `int(sec^2thetad theta )/(tan^2theta-1)` `int(dt)/(t^2-1)=(1/2)Inabs(t-1)/(t+1)+C` `(1/2)Inabs((tantheta-)/(tantheta+1)) +C` |
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