Evaluate : _ - ^ x(1+ x) 1+ ^2 x d x - Vidyadip

Question

Evaluate : $\int_{-\pi}^\pi \frac{x(1+\sin x)}{1+\cos ^2 x} \cdot d x$

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Answer

Let $\begin{aligned} \mathrm{I} & =\int_{-\pi}^\pi \frac{x(1+\sin x)}{1+\cos ^2 x} \cdot d x \\ & =\left[\left(\int_{-\pi}^\pi \frac{x}{1+\cos ^2 x} \cdot d x\right)+\left(\int_{-\pi}^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x\right)\right]\end{aligned}$
The function $\frac{x}{1+\cos ^2 x}$ is odd function and the function $\frac{x \cdot \sin x}{1+\cos ^2 x}$ is even function.
$
\begin{aligned}
& \int_{-a}^a f(x) d x=2 \cdot \int_0^a f(x) d x \text {, if } f(x) \text { even function } \\
& =0 \quad, \text { if } f(x) \text { is odd function } \\
& \therefore \mathrm{I}=0+2 \cdot \int_0^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& \therefore \mathrm{I}=2 \cdot \int_0^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& =2 \cdot \int_0^\pi \frac{(\pi-x) \cdot \sin (\pi-x)}{1+[\cos (\pi-x)]^2} \cdot d x \\
& =2 \cdot \int_0^\pi \frac{(\pi-x) \cdot \sin x}{1+(-\cos x)^2} \cdot d x \\
& =2 \pi \cdot \int_0^\pi \frac{\pi \cdot \sin x-x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& =2 \pi \cdot \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-2 \cdot \int_0^\pi \frac{x \cdot \sin x}{1+\cos ^2 x} \cdot d x \\
& \mathrm{I}=2 \pi \cdot \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-\mathrm{I} \quad \ldots \text { by eq.(i) } \\
& \mathrm{I}+\mathrm{I}=2 \pi \cdot \int_0^\pi \frac{\sin x}{1+\cos ^2 x} \\
& \text { put } \cos x=t \quad \therefore \quad-\sin x \cdot d x=+d t \\
\end{aligned}
$
As varies from 0 to $\pi, t$ varies from 1 to -1
$
\begin{aligned}
& 2 \mathrm{I}=2 \pi \cdot \int_{-1}^1 \frac{-1}{1+t^2} \cdot d t \\
& \mathrm{I}=\pi \cdot 2 \int_0^1 \frac{1}{1+t^2} \cdot d t \quad \text { (where } \frac{1}{1+t^2} \text { is even function.) }
\end{aligned}
$

$\begin{aligned} & \mathrm{I}=2 \pi \cdot\left[\tan ^{-1} t\right]_0^1 . \\ & =2 \pi\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\ & =2 \pi\left(\frac{\pi}{4}-0\right)=\frac{\pi^2}{2} \\ & \therefore \quad \int_{-\pi}^\pi \frac{x(1+\sin x)}{1+\cos ^2 x} \cdot d x=\frac{\pi^2}{2} \\ & \end{aligned}$

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