Can you solve $\int \sin(t)^2\, dt$ without trigonometric identities?
I wanted to solve this integral and I actually had a rough time with it... First I tried a normal product integration, with $u'=\sin(t)$ and $v=\sin(t)$, which lead nowhere.
Then I tried $u'=1$ and $v=\sin(t)^2$ which was actually do able, with the identity $\sin(t)\cos(t)=2\sin(2t)$, which I had to look up...
But the easiest way seems to be, that one uses $\sin(t)^2=\frac12-\frac12\cos(2t)$
The problem is, that you have to know these identities. Which I barely do.
Is there an elementary way to solve this integral, which uses as less knowledge about these identities as possible.
The easiest one requires you to know that $\sin^2x+\cos^2x=1$. You can also derive other identities from De Moivre's formula namely $\exp i\theta=\cos\theta+i\sin\theta$.
$$\int\sin^2t\mathrm dt=\int \sqrt{1-\cos^2t}\sin t\mathrm dt=-\int\sqrt{1-u^2}\mathrm du$$
This form can be handled using Integration by Parts. Can you proceed?
Aliter:
Using Taylor series for $\sin^2 x$ and integrate term-by-term. This gives you an answer in the form of an infinite series. $$\int\sin^2t\mathrm dt=\int\left(\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n+1}}{(2n+1)!}\right)^2\mathrm dt$$
Now expand this expression namely the integrand using the Multinomial Theorem and integrate term-by-term. However, that does not look very good. Also you have to be sure if it holds for infinitely many terms.