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Question

Question: \(a = - \frac{\pi}{4},\mspace{6mu} b =\)...

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

A

∫dxsin⁡x+cos⁡x=\int_{}^{}\frac{dx}{\sin x + \cos x} =

B

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

C

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

D

12logtan(π8+x2)+c\frac{1}{\sqrt{2}}{logtan}\left( \frac{\pi}{8} + \frac{x}{2} \right) + c

Answer

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

Explanation

Solution

∫11−e2x 6mudx=\int_{}^{}{\frac{1}{\sqrt{1 - e^{2x}}}\mspace{6mu} dx =}

x−log⁡[1+1−e2x]+cx - \log\lbrack 1 + \sqrt{1 - e^{2x}}\rbrack + c