Integral of $\sin^2 \pi x$

Question

Evaluate

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

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

Answer 1

Hint

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

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

Added

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

Answer 2

Hint :

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

put $\theta=\pi x$

Answer 3

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

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

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

$$\frac1\pi\frac{\frac{\pi}4-\frac12}2=\frac18-\frac1{4\pi}$$

Answer 4

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

A simple system.

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