Question
Evaluate the following integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$

Answer

Let $\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{1+\cos\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\frac{\text{x}+\sin\text{x}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{2}}\bigg[\frac{\text{x}}{2\cos^2\frac{\text{x}}{2}}+\frac{\sin\text{x}}{2\cos^2\frac{\text{x}}{2}}\bigg]\text{dx}$
$=\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}\text{x}\sec^2\frac{\text{x}}{2}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\frac{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\text{ dx}$
$=\frac{1}{2}\bigg[\text{x}\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\bigg]^{\frac{\pi}{2}}_0-\frac{1}{2}\int_{0}^\limits{\frac{\pi}{2}}\frac{\tan\frac{\text{x}}{2}}{\frac{1}{2}}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}$
$=\Big[\text{x}\tan\frac{\text{x}}{2}\Big]^{\frac{\pi}{2}}_0-\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\tan\frac{\text{x}}{2}\text{ dx}$
$=\Big[\frac{\pi}{2}\tan\frac{\pi}{4}\Big]$
$=\frac{\pi}{2}\times1$
$=\frac{\pi}{2}$

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