# Integrating indefinite and improper integrals

I am given the integral $$\int_0^\infty\frac{\sin^4(x)}{x^2}dx$$ And I must compute it. I know that the answer is $\frac{\pi}{4}$, but I don't really know how to begin solving this.

I am thinking of taking $f(x)=\frac{\sin^4(x)}{x^2}$ and then trying to find its antiderivative and evaluating it at the required limits, but I'm unsure of how to find the antiderivative of such a function.

Integration by parts leads to: $$\int_{0}^{+\infty}\frac{\sin^4 x}{x^2}\,dx = \int_{0}^{+\infty}\frac{4\cos x\sin^3 x}{x}\,dx=\int_{0}^{+\infty}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{x}\,dx$$ so we just need to exploit: $$\forall \alpha > 0,\quad \int_{0}^{+\infty}\frac{\sin(\alpha x)}{x}\,dx=\frac{\pi}{2}$$ to prove the claim.
• Sorry...can you explain how this integration by parts is happening? I'm not seeing this in the form of the traditional $\int udv=uv-\int vdu$ – Taylor May 4 '15 at 17:34
• $v=-1/x$ and $u=\sin^4 x$ is the answer to your question. – mickep May 4 '15 at 17:40