# 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.

• What do you mean with 'solve' $\sin^2(\pi t)$? Jun 13, 2017 at 22:46
• @Fimpellizieri I forgot to include the integral symbol. My apologies. Jun 13, 2017 at 22:47
• Am I at least allowed to use $\sin^2+\cos^2=1$? Jun 13, 2017 at 22:53
• @Fimpellizieri Yes, that one is ok. Jun 13, 2017 at 22:53
• Oct 5, 2019 at 15:02

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: $$\int\sin^2(x)dx=\frac{x}{2}-\frac{\cos(x)\sin(x)}{2}$$

• Where is the constant?
Jun 13, 2017 at 22:55

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.

• Err. $\int 1 dt= t + c$. Jun 13, 2017 at 22:57
• Yea, I brainfarted right there. Jun 13, 2017 at 22:57
• Wow, it was a mess. Should be fixed now. Jun 13, 2017 at 23:07

$\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\ \sin^2 x = \frac {e^{2ix} + e^{-2ix}-2}{-4} = -\frac {e^{2ix}}{4} - \frac {e^{-2ix}}{4} +\frac 12$

$\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$

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...