Evaluate the following integrals: ^ _0 x 1+ dx,0< <

Question

Evaluate the following integrals:
$\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx},0<\alpha<\pi$

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Answer

We have,
$\text{I}=\int\limits^{\pi}_0\frac{\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin(\pi-\text{x})}\text{ dx}$
$=\int\limits^{\pi}_0\frac{\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get,
$2\text{I}=\int\limits^{\pi}_0\frac{\text{x}+\pi-\text{x}}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$\Rightarrow\text{I}=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\sin\text{x}}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1}{1+\cos\alpha\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{1+\tan^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\pi}_0\frac{\sec^{2}\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}+2\cos\alpha\tan\frac{\text{x}}{2}}\text{ dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\text{x dx}=\text{dt}$
When $\text{x}\rightarrow0;\text{t}\rightarrow0$
and $\text{x}\rightarrow\pi;\text{t}\rightarrow\infty$
$\therefore\ \text{I}=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{1+\text{t}^2+2\cos\alpha\text{t}}\text{ dt}$
$=\frac{\pi}{2}\int\limits^{\infty}_0\frac{2}{(\text{t}+\cos\alpha)^2-\cos^2\alpha+1}\text{ dt}$
$={\pi}\int\limits^{\infty}_0\frac{1}{(\text{t}+\cos\alpha)^2+\sin^2\alpha}\text{ dt}$
$=\pi\bigg[\frac{1}{\sin\alpha}\tan^{-1}\Big(\frac{1+\cos\alpha}{\sin\alpha}\Big)\bigg]^1_0$
$=\frac{\pi}{\sin\alpha}\Big[\tan^{-1}(\infty)-\tan^{-1}(\cot\alpha)\Big]$
$=\frac{\pi}{\sin\alpha}\bigg[\frac{\pi}{2}-\tan^{-1}\Big(\tan\Big(\frac{\pi}{2}-\alpha\Big)\Big)\bigg]$
$=\frac{\pi\alpha}{\sin\alpha}$