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.
 A: 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:
$$\int\sin^2(x)dx=\frac{x}{2}-\frac{\cos(x)\sin(x)}{2}$$
A: Yes. We have
\begin{align}\int {\sin(\pi t)}^2 \,dt
&=\int \underbrace{\sin(\pi t)}_f\cdot \underbrace{\sin(\pi t)}_{g'} \,dt\\
&= \underbrace{\sin(\pi t)}_f\cdot\underbrace{\left(\frac{-\cos(\pi t)}{\pi}\right)}_g-
\int \underbrace{\pi\cos(\pi t)}_{f'}\cdot \underbrace{\left(\frac{-\cos(\pi t)}{\pi}\right)}_g \,dt\\
&= -\frac1\pi\sin(\pi t)\cdot\cos(\pi t)
+\int {\cos(\pi t)}^2 \,dt\\
&= -\frac1\pi\sin(\pi t)\cdot\cos(\pi t)
+\int 1-{\sin(\pi t)}^2 \,dt\\
&= -\frac1\pi\sin(\pi t)\cdot\cos(\pi t)
+t-\int {\sin(\pi t)}^2 \,dt
\end{align}
From which we find that
$$\int {\sin(\pi t)}^2 \,dt=\frac{t}2-\frac{\sin(\pi t)\cdot\cos(\pi t)}{2\pi},$$
up to a constant.
A: $\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\sin^2 x = \frac {e^{2ix} + e^{-2ix}-2}{-4} = -\frac {e^{2ix}}{4} - \frac {e^{-2ix}}{4} +\frac 12$
$\int \sin^2 x \ dx= $$\int-\frac {e^{2ix}}{4} - \frac {e^{-2ix}}{4} +\frac 12\ dx\\
\frac {-e^{2ix} + e^{-2ix}}{8} + \frac 12 x +C\\
-\frac {\sin 2x}{4} + \frac 12 x+C$
A: 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...
