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∫2cos⁡2x dx(sin⁡x+cos⁡x)2\int \frac{2\cos 2x \, dx}{(\sin x+\cos x)^2} is equal to:(where C is an arbitrary constant)

Solution

✅ Correct Option: 4

Write cos⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos 2x=(\cos x-\sin x)(\cos x+\sin x) and cancel one factor:

integrand =2(cos⁡x−sin⁡x)sin⁡x+cos⁡x=\frac{2(\cos x-\sin x)}{\sin x+\cos x}.

Put u=sin⁡x+cos⁡xu=\sin x+\cos x, so du=(cos⁡x−sin⁡x)dxdu=(\cos x-\sin x)dx, giving 2∫duu=2log⁡∣sin⁡x+cos⁡x∣+C2\int\frac{du}{u}=2\log|\sin x+\cos x|+C.

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