I've tried to solve this problem with no success. Maybe I am not trying hard enough? Are there any trig identities I missed, or any way to solve this kind of problem?
$$\int_{-\pi/4}^{\pi/4}{(\sin x+2\cos x)^3}dx$$
Oh, and another question. I just graduated highschool and my math skill is not amazing. I'm working to get better at math. Is there any tips on how to get better at it? Thank you before. :)
Just expand the binomial $$(\sin x+2\cos x)^3=\sin ^3x+6\sin^2 x\cos x+12\sin x\cos^2 x+8\cos ^3x$$ Now note that $$\int_{-\pi/4}^{\pi/4} \sin^3 x\ dx = 0 =\int_{-\pi/4}^{\pi/4} 12\sin x\cos^2 x\ dx$$ (because the functions inside the integrals are odd functions and since we are in a symmetric interval). The rest of the integrals are easy, since $$\int_{-\pi/4}^{\pi/4}\big(6\sin^2 x\cos x+8\cos ^3x\big)dx=\int_{-\pi/4}^{\pi/4}\big(-2\sin^2 x+8\big)\cos x\ dx$$ and if we set $u=\sin x$, the integral is much easier : $$\int_{-\pi/4}^{\pi/4}\big(-2\sin^2 x+8\big)\cos x\ dx=\int_{-1/\sqrt{2}}^{1/\sqrt{2}}(-2u^2+8)du=\frac{23\sqrt 2}{3}$$