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Question: If I = \(\int_{- \pi}^{\pi}\frac{2x(1 + \sin x)}{1 + \cos^{2}x}\)dx, then I equals-...

If I = ∫−ππ2x(1+sin⁡x)1+cos⁡2x\int_{- \pi}^{\pi}\frac{2x(1 + \sin x)}{1 + \cos^{2}x}dx, then I equals-

A

−2π- \sqrt{2}\pi

B

2π\sqrt{2}\pi

C

π2\frac{\pi}{\sqrt{2}}

D

–π2\frac{\pi}{\sqrt{2}}

Answer

2π\sqrt{2}\pi

Explanation

Solution

We can write

I = I1 + I2

where I1 = ∫−ππ2x1+cos⁡2x\int_{- \pi}^{\pi}\frac{2x}{1 + \cos^{2}x}dx = 0

and I2 = 4∫0πxsin⁡x1+cos⁡2x4\int_{0}^{\pi}\frac{x\sin x}{1 + \cos^{2}x}

= 4∫0π(π−x)sin⁡(π−x)1+(cos⁡(π−x))24\int_{0}^{\pi}\frac{(\pi - x)\sin(\pi - x)}{1 + (\cos(\pi - x))^{2}}dx

= 4π∫0πdx1+cos⁡2x−I24\pi\int_{0}^{\pi}\frac{dx}{1 + \cos^{2}x} - I_{2}

Ž 2π∫0πcos⁡ec2xcos⁡ec2x+cot⁡2x2\pi\int_{0}^{\pi}\frac{\cos ec^{2}x}{\cos ec^{2}x + \cot^{2}x}dx

= 2π∫0πcos⁡ec2xdx1+2cot⁡2x2\pi\int_{0}^{\pi}\frac{\cos ec^{2}xdx}{1 + 2\cot^{2}x}

=  −2π2tan⁡−1(2cot⁡x)]0π\left. \ \frac{- 2\pi}{\sqrt{2}}\tan^{- 1}\left( \sqrt{2}\cot x \right) \right\rbrack_{0}^{\pi}

= – 2(−π2−π2)\sqrt{2}\left( - \frac{\pi}{2} - \frac{\pi}{2} \right)= 2π\sqrt{2}\pi