1
$\begingroup$

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.

$\endgroup$
4
  • 4
    $\begingroup$ Don't begin by trying to improve the world. Instead, improve yourself: learn the identities.:) $\endgroup$
    – avs
    Apr 8 '19 at 21:53
  • 1
    $\begingroup$ But improving yourself is selfish. :( But yeah, you are right. I should learn them, to use them once in a while. $\endgroup$
    – Cornman
    Apr 8 '19 at 21:55
  • 5
    $\begingroup$ With Euler's formula: $\sin t=\dfrac{\mathrm e^{it}-\mathrm e^{-it}}{2i}$. $\endgroup$
    – Bernard
    Apr 8 '19 at 22:04
  • 2
    $\begingroup$ That's a trigonometric identity :) $\endgroup$ Apr 8 '19 at 22:21
1
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ I'll leave the OP to decide whether $\sin^2 t+\cos^2 t=1$ is also a disallowed identity (but really, who studies these functions without learning the Pythagorean theorem?), but $\sin t$ can differ from $\sqrt{1-\cos^2t}$ by a sign, so it's better to Integrate by parts with $u=\sin t,\,v=-\cos t$, viz. $$\int\sin^2 tdt=-\sin t\cos t+\int (1-\sin^2 t)dt\implies\int\sin^2 tdt=\frac{1}{2}(t-\sin t\cos t)+C.$$ $\endgroup$
    – J.G.
    Apr 9 '19 at 6:34
  • $\begingroup$ Thanks for the insight @J.G. Also I was wondering if that expression I've written as infinite series indeed holds. It would be great if you could check the validity of what I've written as an alternative answer. $\endgroup$ Apr 9 '19 at 6:38
  • 1
    $\begingroup$ A better approach there is to show the series squares to $(1-\cos 2x)/2$ as expected (this is an exercise in combinatorial identities, which is a good learning exercise of you haven't done it), but whether the OP is OK with using "banned" results of we first prove them isn't clear. $\endgroup$
    – J.G.
    Apr 9 '19 at 6:49
  • 1
    $\begingroup$ @J.G I accept the Pythagorean theorem. This I know by heart. $\endgroup$
    – Cornman
    Apr 9 '19 at 7:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.