Issues computing $\int\cos^2x~\mathrm{d}x$ without using $\sin^2(x)=1-\cos^2(x)$ I was working on the integral
$$\int\cos^2x~\mathrm{d}x$$
and using the identity $\sin^2(x)=1-\cos^2(x)$ after the first partial integration to rewrite the remainder of the integration yields $\int\cos^2x~\mathrm{d}x = 1/2(\cos x\sin x+x)$. If I don't do this step of using the identity and do two partial integrations I get the obviosuly true expression
$$\int\cos^2x~\mathrm{d}x=\cos x\sin x-\cos x\sin x+\int\cos^2x~\mathrm{d}x$$
which is basically $0=0$. Why is it so that two partial integrations don't help at all?

Hint: 
$$\color{blue}{\cos^2x=\frac 1 2 \cos(2x)+\frac 1 2}$$
$$=\frac 1 2\int \cos (2x)dx+\frac 1 2 \int 1 dx$$
