Evaluate the following integrals: ^ _0 x 1+ dx

Question

Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}$

✓

Answer

Let $\text{I}=\int\limits^{\pi}_0\frac{\text{x}\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin\text{x}}{1+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\pi}_0(\text{x}+\pi-\text{x})\frac{\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi\sin\text{x}}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\frac{1+\sin\text{x}-1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{1}{1+\sin\text{x}}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}\text{ dx}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\frac{(1-\sin\text{x})}{1-\sin}$
$=\pi\int\limits^{\pi}_0\text{dx}-\pi\int\limits^{\pi}_0\big(\sec^2\text{x}-\sec\text{x}\tan\text{x}\big)\text{dx}$
$=\pi\big[\text{x}\big]^{\pi}_0-\pi\big[\tan\text{x}-\sec\text{x}\big]^{\pi}_0$
$=\pi^2-\pi(0+1-0+1)$
$=\pi^2-2\pi$
Hence, $\text{I}=\pi\Big(\frac{\pi}{2}-1\Big)$

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