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$$\int_0^{1/4} \sin^2 \pi x \; dx$$

Can someone please explain what to do if theres a power and how to do it in general thanks

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You can use integration by parts as well. – Ali Caglayan Aug 7 '13 at 12:52
up vote 4 down vote accepted


$$\sin^2 x=\frac{1}{2}(1-\cos(2x))$$

And in general $$\sin^p x=\left(\frac{1}{2i}(e^{ix}-e^{-ix})\right)^p$$ and use the binomial formula.


$$\int_0^{1/4} \sin^2 \pi x \; dx=\int_0^{1/4}\frac{1}{2}(1-\cos(2\pi x))dx=\frac{1}{8}-\frac{1}{4\pi}\sin(2\pi x)\Big|_0^{1/4}=\frac{1}{8}-\frac{1}{4\pi}$$

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Where do I input on the pi ? – Red Queen10101 Aug 7 '13 at 11:49
You can of course write $\sin^2(\pi x)=\cdots$ so just replace $x$ by $\pi x$. – user63181 Aug 7 '13 at 11:50
Alright, thanks. Ill try and see if I can do it. – Red Queen10101 Aug 7 '13 at 11:52
Do I expand the 1/2? or leave it alone? – Red Queen10101 Aug 7 '13 at 11:53
Yes you can expand $1/2$. – user63181 Aug 7 '13 at 11:56

Hint :

$$\sin^2 \theta=\frac{1-\cos(2 \theta )}{2}$$

put $\theta=\pi x$

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$$\int \sin^2x\,dx=\frac{x-\sin x\cos x}2\;,\;\;\text {so}\;\;\int f'(x)\sin^2f(x)\,dx=\frac{f(x)-\sin f(x)\cos f(x)}2\implies$$

$$\implies \int\sin^2\pi x\,dx=\frac1\pi\int (\pi dx)\sin \pi x=\frac1\pi\frac{\pi x-\sin \pi x\cos \pi x}{2}$$

Check now that on $\,[0,\,1/4]\;$ , the value of the above integral is


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Hey this is a different question, but how would you further integrate (1-cos2(x/2))/2 with top bound pi/6 and lower bound 2? if it is possible could we start a chat? – Red Queen10101 Aug 7 '13 at 12:37
@RedQueen10101, why won't you open a new thread? Besides this, I shall be back in some 40 minutes more. – DonAntonio Aug 7 '13 at 12:46
Will do ill be under the tag integration – Red Queen10101 Aug 7 '13 at 12:48

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

A simple system.

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

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Sorry for the bad editing – BinaryBurst Aug 7 '13 at 12:50

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