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Question

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

−1e2x+1- \frac{1}{e^{2x} + 1}

A

∫cos⁡2x(cos⁡x+sin⁡x)2 6mudx=\int_{}^{}{\frac{\cos 2x}{(\cos x + \sin x)^{2}}\mspace{6mu} dx =}

B

log⁡cos⁡x+sin⁡x+c\log\sqrt{\cos x + \sin x} + c

C

log⁡(cos⁡x−sin⁡x)+c\log(\cos x - \sin x) + c

D

None of these

Answer

∫cos⁡2x(cos⁡x+sin⁡x)2 6mudx=\int_{}^{}{\frac{\cos 2x}{(\cos x + \sin x)^{2}}\mspace{6mu} dx =}

Explanation

Solution

cosecx−cot⁡x+c\text{cosec}x - \cot x + c

sec⁡x−cosecx+c\sec x - \text{cosec}x + c

∫(cos⁡x2−sin⁡x2)2dx=\int_{}^{}{\left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)^{2}dx =}, (Putting x+cos⁡x+cx + \cos x + c in first part)