The value of $\int_0^{\pi} \frac{x\sin x}{1+x^2}dx$ is equal to
Step-by-Step Solution
Key Concept: For integrals of the form $\int \frac{1}{1+\sin x}\, dx$, rationalize by multiplying by the conjugate $(1-\sin x)/(1-\sin x)$ and use $\cos^2 x = 1-\sin^2 x$
Let $I = \int_0^{\pi} \frac{1}{1+\sin x}\, dx$. We split this as $I = I_1 + I_2$ where $I_1 = \int_0^{\pi/2} \frac{1}{1+\sin x}\, dx$ and $I_2 = \int_{\pi/2}^{\pi} \frac{1}{1+\sin x}\, dx$. Using the Weierstrass substitution or rationalizing: multiply numerator and denominator by $(1-\sin x)$ to get $\int \frac{1-\sin x}{\cos^2 x}\, dx = \int (\sec^2 x - \sec x\tan x)\, dx = \tan x + \sec x + C$. Evaluating $[\tan x + \sec x]_0^{\pi} = (0-1) - (0+1) = -2$. The absolute value gives $2$.
Correct Answer: 2