Anybody to solve this definite integral?

Question

Can someone help me with this definite integral? $$ \int _{-\pi}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x} dx $$

Answer 1

Split the integral into two parts: $$I = \int _{-\pi}^{0}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x} dx + \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x} dx$$

Applying the substitution $x \to -x$ on the first integral: $$I = \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^{-x}} dx + \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x} dx$$ Multiply top and bottom of first integral by $2^x$: $$I = \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{2^x}{1+2^x} dx + \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)}\frac{1}{1+2^x} dx$$ Sum the two integrals: $$I= \int _{0}^{\pi}\frac{\sin(3x)}{\sin(x)} dx$$ Use the fact that $\sin(3x) = 3\sin(x)-4\sin^3(x)$ by the triple-angle formula: $$I= \int _{0}^{\pi}3-4\sin^2(x) dx$$ Using the fact that $\int_{0}^{\pi}\sin^2(x)dx = \frac{\pi}{2}:$ $$I = 3\pi-4\left(\frac{\pi}{2}\right) = \pi$$