Question

The value of $\frac{(\sqrt{2}+1)198}{\pi }{\int }_{\pi /4}^{3\pi /4}\frac{\varphi }{1+\sin \varphi }d\varphi$ is

Answer 1

Let $I={\int }_{\pi /4}^{3\pi /4}\frac{\varphi }{1+\sin \varphi }d\varphi ={\int }_{\pi /4}^{3\pi /4}\frac{(\pi -\varphi )}{1+\sin (\pi -\varphi )}d\varphi =\pi {\int }_{\pi /4}^{3\pi /4}\frac{1}{1+\sin \varphi }d\varphi -{\int }_{\pi /4}^{3\pi /4}\frac{\varphi }{1+\sin \varphi }d\varphi$
$\Rightarrow 2I=\pi {\int }_{\pi /4}^{3\pi /4}\frac{1}{1+\sin \varphi }d\varphi \Rightarrow I=\frac{\pi }{2}{\int }_{\pi /4}^{3\pi /4}\frac{1}{1+\sin \varphi }d\varphi =\frac{\pi }{2}{\int }_{\pi /4}^{3\pi /4}\frac{1}{1+\cos (\frac{\pi }{2}-\varphi )}d\varphi =\frac{\pi }{4}{\int }_{\pi /4}^{3\pi /4}{\sec }^2(\frac{\pi }{4}-\frac{\varphi }{2})d\varphi =\frac{\pi }{4}\times -2{\left[\tan (\frac{\pi }{4}-\frac{\varphi }{2})\right]}_{\pi /4}^{3\pi /4}=-\frac{\pi }{2}[\tan (-\frac{\pi }{8})-\tan \left(\frac{\pi }{8}\right)]=\pi (\sqrt{2}-1)$
Therefore
$\frac{(\sqrt{2}+1)198}{\pi }{\int }_{\pi /4}^{3\pi /4}\frac{\varphi }{1+\sin \varphi }d\varphi =\frac{(\sqrt{2}+1)198}{\pi }\times \pi (\sqrt{2}-1)=198$