Question
Evaluate : $\int_0^{\pi / 2} \frac{\cos x}{1+\cos x+\sin x} \cdot d x$
✓
Answer
$
\begin{aligned}
\text { Let I } & =\int_0^{\pi / 2} \frac{\cos x}{1+\cos x+\sin x} \cdot d x \\
& =\int_0^{\pi / 2} \frac{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}{2 \cos ^2\left(\frac{x}{2}\right)+2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right)} \cdot d x \\
& =\int_0^{\pi / 2} \frac{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}{2\left[\cos \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]} \cdot d x \\
& =\int_0^{\pi / 2}\left[\frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right] \cdot d x=\int_0^{\pi / 2}\left[1-\tan \left(\frac{x}{2}\right)\right] \cdot d x
\end{aligned}
$
$\begin{aligned} & =\frac{1}{2} \cdot\left[x-\log \left(\sec \frac{x}{2}\right) \cdot \frac{1}{\frac{1}{2}}\right]_0^{\pi / 2} \\ & =\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \cdot \log \left(\sec \frac{\pi}{4}\right)-(0-2 \log \sec 0)\right] \\ & =\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \log \sqrt{2}-0+2(0)\right] \quad=\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \log \sqrt{2}\right] \quad=\frac{\pi}{4}-\log \sqrt{2} \\ \therefore \quad & \int_0^{\pi / 2} \frac{\sec ^2 x}{1+\cos x+\sin x} \cdot d x=\frac{\pi}{4}-\log \sqrt{2}\end{aligned}$
\begin{aligned}
\text { Let I } & =\int_0^{\pi / 2} \frac{\cos x}{1+\cos x+\sin x} \cdot d x \\
& =\int_0^{\pi / 2} \frac{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}{2 \cos ^2\left(\frac{x}{2}\right)+2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right)} \cdot d x \\
& =\int_0^{\pi / 2} \frac{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}{2\left[\cos \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]} \cdot d x \\
& =\int_0^{\pi / 2}\left[\frac{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)}\right] \cdot d x=\int_0^{\pi / 2}\left[1-\tan \left(\frac{x}{2}\right)\right] \cdot d x
\end{aligned}
$
$\begin{aligned} & =\frac{1}{2} \cdot\left[x-\log \left(\sec \frac{x}{2}\right) \cdot \frac{1}{\frac{1}{2}}\right]_0^{\pi / 2} \\ & =\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \cdot \log \left(\sec \frac{\pi}{4}\right)-(0-2 \log \sec 0)\right] \\ & =\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \log \sqrt{2}-0+2(0)\right] \quad=\frac{1}{2} \cdot\left[\frac{\pi}{2}-2 \log \sqrt{2}\right] \quad=\frac{\pi}{4}-\log \sqrt{2} \\ \therefore \quad & \int_0^{\pi / 2} \frac{\sec ^2 x}{1+\cos x+\sin x} \cdot d x=\frac{\pi}{4}-\log \sqrt{2}\end{aligned}$
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