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Question: The value of the integral \(\int_{}^{}{\frac{x}{1 + x\tan x}dx}\) is equal to...

The value of the integral ∫x1+xtan⁡xdx\int_{}^{}{\frac{x}{1 + x\tan x}dx} is equal to

A

log⁡∣xcos⁡x+sin⁡x∣+c\log|x\cos x + \sin x| + c

B

log⁡∣cos⁡x_+x∣\log|\cos x\_ + x|+ c

C

log⁡∣cos⁡x+xsin⁡x∣+C\log|\cos x + x\sin x| + C

D

None of these

Answer

log⁡∣cos⁡x+xsin⁡x∣+C\log|\cos x + x\sin x| + C

Explanation

Solution

I=∫x1+xtan⁡xdx=∫xcos⁡xcos⁡x+xsin⁡xdxI = \int_{}^{}{\frac{x}{1 + x\tan x}dx = \int_{}^{}{\frac{x\cos x}{\cos x + x\sin x}dx}}Put (cos⁡x+xsin⁡x)=t(\cos x + x\sin x) = t ⇒ (−sin⁡x+xcos⁡x+sin⁡x)dx=dt( - \sin x + x\cos x + \sin x)dx = dt ⇒ xcos⁡xdx=dtx\cos xdx = dt

I=∫dtt=log⁡∣t∣+c=log⁡∣cos⁡x+xsin⁡x∣+cI = \int_{}^{}{\frac{dt}{t} = \log|t| + c = \log|\cos x + x\sin x| + c}