Is it possible to solve $\int \sin^2(\pi t) dt$ without using trig identities? Perhaps by using integration by parts? Is it possible to solve $\int \sin^2(\pi t) dt$ without using trig identities? Perhaps by using integration by parts? I've tried the latter, but It seems that you get an infinite loop?

I would greatly appreciate it if people could please take the time to demonstrate this.
Integrating by parts with $u=\sin(x),\ dv=\sin(x)dx$:
$$\int \sin^2(x)dx=-\cos(x)\sin(x)+\int\cos^2(x)dx=-\cos(x)\sin(x)+x-\int\sin^2(x)dx$$
We deduce:
$$\int\sin^2(x)dx=\frac{x}{2}-\frac{\cos(x)\sin(x)}{2}$$
This can be done by using the reduction formula derived from integration by parts...                                                                   $\int \sin^n xdx =\frac{-1}{n} \sin^{n-1}x\cos x\ + \frac{n-1}{n}\int\sin^{n-2}x\ {d}x$   now by doing a little substitution letting $u=πt$ gives $du=πdt$  solving for $dt$ gives $dt=\frac{du}{π}$ with this substitution our problem looks like this... $\int\sin^2u\frac{1}{π}\ {d}u   \ = \frac{1}{π}\int(\sin^2u)\ {d}u$      now by using the reduction formula gives $\frac{-1}{2}\sin u\cos u\ +\frac{1}{2}\int\ {d}x$ and finally the answer becomes $\frac{-1}{2}\sin(πt)\cos(πt)\ + \frac{1}{2}(πt)$ or we could rewrite as                  $[\frac{-\sin(2πt)}{4}\ +\frac{(πt)}{2}+\ c ]$ final answer hope it helps...