∫0π xtan⁡xsec⁡x+tan⁡xdx=

I=∫0π xtan⁡xsec⁡x+tan⁡xdx=∫0π (π−x)sin⁡(π−x)1+sin⁡(π−x)dx =∫0π (π−x)sin⁡x1+sin⁡xdx =∫0π sin⁡x1+sin⁡xdx−I⇒2I=π∫0π/2 sin⁡x1+sin⁡xdx =2π∫0π/2 sin⁡x1+sin⁡xdx =2π∫0π/2 1−11+sin⁡xdx =2π∫0π/2 1−1−sin⁡xcos2⁡xdx=2π∫0π/2 1−sec2⁡x+sec⁡xtan⁡xdx =2π[x−tan⁡x+sec⁡x] =2π[x+sec⁡x−tan⁡x]=2πx+1−sin⁡xcos⁡x0π/2 =2πx+cos⁡x1+sin⁡x0π/2 =2ππ2+0−1 ⇒I=ππ2−1.

∫0π xtan⁡xsec⁡x+tan⁡xdx=

I=∫0π xtan⁡xsec⁡x+tan⁡xdx=∫0π (π−x)sin⁡(π−x)1+sin⁡(π−x)dx =∫0π (π−x)sin⁡x1+sin⁡xdx =∫0π sin⁡x1+sin⁡xdx−I⇒2I=π∫0π/2 sin⁡x1+sin⁡xdx =2π∫0π/2 sin⁡x1+sin⁡xdx =2π∫0π/2 1−11+sin⁡xdx =2π∫0π/2 1−1−sin⁡xcos2⁡xdx=2π∫0π/2 1−sec2⁡x+sec⁡xtan⁡xdx =2π[x−tan⁡x+sec⁡x] =2π[x+sec⁡x−tan⁡x]=2πx+1−sin⁡xcos⁡x0π/2 =2πx+cos⁡x1+sin⁡x0π/2 =2ππ2+0−1 ⇒I=ππ2−1.

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$\int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x =$

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$\frac{π (π + 2)}{2}$

$\frac{π (π - 2)}{2}$

$\frac{π - 2}{2}$

$\frac{π + 2}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} I = \int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \sin (π - x)} d x = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \sin x} d x = \int_{0}^{π} \frac{\sin x}{1 + \sin x} d x - I \\ \Rightarrow 2 I = π \int_{0}^{π / 2} \frac{\sin x}{1 + \sin x} d x = 2 π \int_{0}^{π / 2} \frac{\sin x}{1 + \sin x} d x = 2 π \int_{0}^{π / 2} (1 - \frac{1}{1 + \sin x}) d x = 2 π \int_{0}^{π / 2} (1 - \frac{1 - \sin x}{\cos^{2} x}) d x \\ = 2 π \int_{0}^{π / 2} (1 - \sec^{2} x + \sec x \tan x) d x = 2 π [x - \tan x + \sec x] = 2 π [x + \sec x - \tan x] \\ = 2 π {[x + \frac{1 - \sin x}{\cos x}]}_{0}^{π / 2} = 2 π {[x + \frac{\cos x}{1 + \sin x}]}_{0}^{π / 2} = 2 π [\frac{π}{2} + 0 - 1] \Rightarrow I = π (\frac{π}{2} - 1) . \end{array}$