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Consider $I=\int_{0}^{\pi} \frac{x d x}{1+\sin x}$
What is $\int_{0}^{\pi} \frac{(\pi-x) d x}{1+\sin x}$ equal to?
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What is $\int_{0}^{\pi} \frac{(\pi-x) d x}{1+\sin x}$ equal to?
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Let $\mathrm{I}_{1}=\int_{0}^{\pi} \frac{(\pi-\mathrm{x}) \mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{[\pi-(\pi-\mathrm{x})] \mathrm{d} \mathrm{x}}{1+\sin (\pi-\mathrm{x})}$
$\left[\because \int_{0}^{a} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$\mathrm{I}_{1}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}=\pi \quad[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Given, $\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\left[\because \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin \mathrm{x}} \mathrm{dx}$
$[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Adding eqs. (i) and (ii), we get
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$\begin{aligned} \Rightarrow 2 \mathrm{I}=2 \pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\sin \mathrm{x}} \\ &\left[\because \int_{0}^{2 \mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=2 \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}, \text { if } \mathrm{f}(2 \mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})\right] \end{aligned}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \mathrm{x}}{1+\left(\frac{2 \tan \frac{\mathrm{x}}{2}}{1+\tan ^{2} \frac{\mathrm{x}}{2}}\right)}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}}{\tan ^{2} \frac{\mathrm{x}}{2}+1+2 \tan \frac{\mathrm{x}}{2}}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\left(\sec ^{2} \frac{\mathrm{x}}{2}\right) \mathrm{dx}}{\left(\tan \frac{\mathrm{x}}{2}+1\right)^{2}}$
Let $\tan \frac{x}{2}+1=t$
$\Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}$
$\Rightarrow \sec ^{2} \frac{x}{2} \mathrm{dx}=2 \mathrm{dt}$
When $\mathrm{x}=0$, then $\mathrm{t}=1$ and when $\mathrm{x}=\frac{\pi}{2}$, then $\mathrm{t}=2$
$\therefore \mathrm{I}=2 \pi \int_{1}^{2} \frac{\mathrm{dt}}{\mathrm{t}^{2}}=2 \pi\left[\frac{\mathrm{t}^{-2+1}}{-2+1}\right]_{1}^{2}=-2 \pi\left[\frac{1}{\mathrm{t}}\right]_{1}^{2}$
$=-2 \pi\left[\frac{1}{2}-1\right]$
$=-2 \pi\left(-\frac{1}{2}\right)=\pi$
$=\int_{0}^{\pi} \frac{[\pi-(\pi-\mathrm{x})] \mathrm{d} \mathrm{x}}{1+\sin (\pi-\mathrm{x})}$
$\left[\because \int_{0}^{a} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$\mathrm{I}_{1}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}=\pi \quad[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Given, $\mathrm{I}=\int_{0}^{\pi} \frac{\mathrm{x} \mathrm{dx}}{1+\sin \mathrm{x}}$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin (\pi-\mathrm{x})} \mathrm{dx}$
$\left[\because \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{a}-\mathrm{x}) \mathrm{d} \mathrm{x}\right]$
$=\int_{0}^{\pi} \frac{(\pi-\mathrm{x})}{1+\sin \mathrm{x}} \mathrm{dx}$
$[\because \sin (\pi-\mathrm{x})=\sin \mathrm{x}]$
Adding eqs. (i) and (ii), we get
$2 \mathrm{I}=\pi \int_{0}^{\pi} \frac{\mathrm{d} \mathrm{x}}{1+\sin \mathrm{x}}$
$\begin{aligned} \Rightarrow 2 \mathrm{I}=2 \pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\sin \mathrm{x}} \\ &\left[\because \int_{0}^{2 \mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=2 \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}, \text { if } \mathrm{f}(2 \mathrm{a}-\mathrm{x})=\mathrm{f}(\mathrm{x})\right] \end{aligned}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\mathrm{d} \mathrm{x}}{1+\left(\frac{2 \tan \frac{\mathrm{x}}{2}}{1+\tan ^{2} \frac{\mathrm{x}}{2}}\right)}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}}{\tan ^{2} \frac{\mathrm{x}}{2}+1+2 \tan \frac{\mathrm{x}}{2}}$
$\Rightarrow \mathrm{I}=\pi \int_{0}^{\frac{\pi}{2}} \frac{\left(\sec ^{2} \frac{\mathrm{x}}{2}\right) \mathrm{dx}}{\left(\tan \frac{\mathrm{x}}{2}+1\right)^{2}}$
Let $\tan \frac{x}{2}+1=t$
$\Rightarrow \sec ^{2} \frac{x}{2} \cdot \frac{1}{2} \mathrm{~d} x=\mathrm{dt}$
$\Rightarrow \sec ^{2} \frac{x}{2} \mathrm{dx}=2 \mathrm{dt}$
When $\mathrm{x}=0$, then $\mathrm{t}=1$ and when $\mathrm{x}=\frac{\pi}{2}$, then $\mathrm{t}=2$
$\therefore \mathrm{I}=2 \pi \int_{1}^{2} \frac{\mathrm{dt}}{\mathrm{t}^{2}}=2 \pi\left[\frac{\mathrm{t}^{-2+1}}{-2+1}\right]_{1}^{2}=-2 \pi\left[\frac{1}{\mathrm{t}}\right]_{1}^{2}$
$=-2 \pi\left[\frac{1}{2}-1\right]$
$=-2 \pi\left(-\frac{1}{2}\right)=\pi$
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