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Question

Question: \(\log(e^{2x} + 1) - x + c\)...

log(e2x+1)x+c\log(e^{2x} + 1) - x + c

A

log(e2x+1)+c\log(e^{2x} + 1) + c

B

cosecxlogtanx26mudx=\int_{}^{}{\frac{\text{cosec}x}{{logtan}\frac{x}{2}}\mspace{6mu} dx =}

C

log(logtanx2)+c\log\left( {logtan}\frac{x}{2} \right) + c

D

None of these

Answer

cosecxlogtanx26mudx=\int_{}^{}{\frac{\text{cosec}x}{{logtan}\frac{x}{2}}\mspace{6mu} dx =}

Explanation

Solution

Putting a=π4,6mub=a = - \frac{\pi}{4},\mspace{6mu} b = we get

dxsinx+cosx=\int_{}^{}\frac{dx}{\sin x + \cos x} =

logtan(π8+x2)+c{logtan}\left( \frac{\pi}{8} + \frac{x}{2} \right) + c

logtan(π8x2)+c{logtan}\left( \frac{\pi}{8} - \frac{x}{2} \right) + c