Let Q=∫log⁡(1+tan⁡x)dxsec⁡2x then Q is equal to

Let Q=∫log⁡(1+tan⁡x)dxsec⁡2xQ=∫log1+tanxcos2xdx I II Using integration by partsQ=log⁡(1+tan⁡x)∫cos⁡2xdx−∫1(1+tan⁡x)sec2⁡xsin⁡2x2dxQ=log⁡(1+tan⁡x)sin⁡2x2−∫tan1+tan⁡xdx=log⁡(1+tan⁡x)sin⁡2x2−∫(1+tan⁡x−1)1+tan⁡xdx=log⁡(1+tan⁡x)sin⁡2x2−∫1−11+tan⁡xdxQ=12sin⁡2xlog⁡(1+tan⁡x)−x+∫cos⁡xdxsin⁡x+cos⁡x=12sin⁡2xlog⁡(1+tan⁡x)−x+12∫(sin⁡x+cos⁡x)+(cos⁡x−sin⁡x)(sin⁡x+cos⁡x)dxQ=12sin⁡2xlog⁡(1+tan⁡x)−x+12x+12log⁡|sin⁡x+cos⁡x|+C

Let Q=∫log⁡(1+tan⁡x)dxsec⁡2x then Q is equal to

Let Q=∫log⁡(1+tan⁡x)dxsec⁡2xQ=∫log1+tanxcos2xdx I II Using integration by partsQ=log⁡(1+tan⁡x)∫cos⁡2xdx−∫1(1+tan⁡x)sec2⁡xsin⁡2x2dxQ=log⁡(1+tan⁡x)sin⁡2x2−∫tan1+tan⁡xdx=log⁡(1+tan⁡x)sin⁡2x2−∫(1+tan⁡x−1)1+tan⁡xdx=log⁡(1+tan⁡x)sin⁡2x2−∫1−11+tan⁡xdxQ=12sin⁡2xlog⁡(1+tan⁡x)−x+∫cos⁡xdxsin⁡x+cos⁡x=12sin⁡2xlog⁡(1+tan⁡x)−x+12∫(sin⁡x+cos⁡x)+(cos⁡x−sin⁡x)(sin⁡x+cos⁡x)dxQ=12sin⁡2xlog⁡(1+tan⁡x)−x+12x+12log⁡|sin⁡x+cos⁡x|+C

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$Let Q = \int \frac{\log (1 + \tan x) d x}{\sec 2 x} then Q is equal to$

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$\sin 2 x \log (1 + \tan x) - \frac{x}{2} \log |\sin x + \cos x| + C$

$\frac{1}{2} \sin 2 x \log (1 + \tan x) - x + \frac{1}{2} x + \frac{1}{2} \log |\sin x + \cos x| + C$

$\sin 2 x \log (1 + \tan x) + x + \log |\sin x + \cos x| + C$

$\sin 2 x \log (1 + \tan x) - x - \log |\sin x + \cos x| + C$

answer is A.

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Detailed Solution

$Let Q = \int \frac{\log (1 + \tan x) d x}{\sec 2 x}$

$\begin{array}{l} Q = \int \log (1 + \tan x) \cos 2 x d x \\ I I I \end{array}$

Using integration by parts

$\begin{array}{l} Q = \log (1 + \tan x) \int \cos 2 x d x - \int \frac{1}{(1 + \tan x)} \sec^{2} x \frac{\sin 2 x}{2} d x \\ Q = \log (1 + \tan x) \frac{\sin 2 x}{2} - \int \frac{\tan}{1 + \tan x} d x \end{array}$

$\begin{array}{l} = \log (1 + \tan x) \frac{\sin 2 x}{2} - \int \frac{(1 + \tan x - 1)}{1 + \tan x} d x \\ = \log (1 + \tan x) \frac{\sin 2 x}{2} - \int [1 - \frac{1}{1 + \tan x}] d x \\ Q = \frac{1}{2} \sin 2 x \log (1 + \tan x) - x + \int \frac{\cos x d x}{\sin x + \cos x} \\ = \frac{1}{2} \sin 2 x \log (1 + \tan x) - x + \frac{1}{2} \int \frac{(\sin x + \cos x) + (\cos x - \sin x)}{(\sin x + \cos x)} d x \\ Q = \frac{1}{2} \sin 2 x \log (1 + \tan x) - x + \frac{1}{2} x + \frac{1}{2} \log | \sin x + \cos x | + C \end{array}$