$\displaystyle\int_{\pi/4}^{3\pi/4}\dfrac{x}{1+\sin x}dx$
Step-by-Step Solution
Key Concept: Use $\int_a^b xf(x)dx=\frac{a+b}{2}\int_a^b f(x)dx$ when $f(a+b-x)=f(x)$
$I=\frac{\pi/4+3\pi/4}{2}\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\sin x}=\frac{\pi}{2}\int_{\pi/4}^{3\pi/4}\frac{dx}{1+\sin x}$. $\int\frac{dx}{1+\sin x}=\int\frac{1-\sin x}{\cos^2 x}dx=\tan x-\sec x+C$. Evaluate: $[\tan x-\sec x]_{\pi/4}^{3\pi/4}=(−1+\sqrt{2})−(1-\sqrt{2})=2\sqrt{2}−2$. $I=\frac{\pi}{2}(2\sqrt{2}-2)=\pi(\sqrt{2}-1)$.
Correct Answer: 4