Mathematics - Definite Integration Question with Solution | TestHub

Let $I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{1+\sin x}dx$ also let $K=\frac{x}{1+\sin x}$ Multiplying numerator and denominator by $(1-\sin x)$, we get; $$\begin{array}{rl}K& =\frac{x(1-\sin x)}{1-(\sin x{)}^2}=\frac{x(1-\sin x)}{(\cos x{)}^2}\\ & =x(1-\sin x){\sec }^{2}x\\ & =x{\sec }^{2}x-x\sin x{\sec }^{2}x=x{\sec }^{2}x-x\tan \\ x& \sec x\end{array}$$ Now, $I={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}x{\sec }^{2}xdx-{\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}x\sec x\tan xdx$ $$={\left[x\tan x-\int \frac{dx}{dx}\tan xdx\right]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}-{\left[x\sec x-\int \frac{dx}{dx}\sec xdx\right]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$$ $$\begin{array}{rl}& =[x\tan x-\ln |\sec x|{]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\\ & -[x\sec x-\ln |\sec x+\tan x|{]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}+c\\ & \Rightarrow I=\{[\frac{3\pi }{4}\tan \frac{3\pi }{4}-\ln \left|\frac{3\pi }{4}\right|\\ & -\left[\frac{3\pi }{4}\sec \frac{3\pi }{4}-\ln \left|\sec \frac{3\pi }{4}+\tan \frac{3\pi }{4}\right|\right]\}\\ & -\{[\frac{\pi }{4}\tan \frac{\pi }{4}-\ln \left|\frac{\pi }{4}\right|\\ & -\left[\frac{\pi }{4}\sec \frac{\pi }{4}-\ln \left|\sec \frac{\pi }{4}+\tan \frac{\pi }{4}\right|\right]\}\\ & =\frac{\pi }{2}(\sqrt{2}+1)\end{array}$$