The Intergral of $\sin^4x$ without using reduction formula

Question

So I've been trying to compute $$\int\sin^4(x)\mathrm{d}x$$ and everywhere they use the reduction formula which we haven't learned yet so I've been wondering if theres another way to do it? Thanks in advance.

Answer 1

Performing integration by parts,

$\begin{align} \int_0^x\sin^2 t\,dt&=\Big[-\cos t\sin t\Big]_0^x+\int_0^x\cos^2 t\,dt\\ &=-\cos x\sin x+\int_0^x(1-\sin^2 t)\,dt\\ &=-\cos x\sin x+\int_0^x 1\,dt-\int_0^x \sin^2 t\,dt\\ &=-\cos x\sin x+x-\int_0^x \sin^2 t\,dt\\ \end{align}$

Therefore,

$\displaystyle \int_0^x \sin^2 t\,dt=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x$

$\begin{align} \int_0^x\sin^4 t\,dt&=\int_0^x(1-\cos^2)\sin^2 t\,dt \\ &=\int_0^x\sin^2 t\,dt-\int_0^x \cos^2 t\sin^2 t\,dt\\ &=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\int_0^x \cos^2 t\sin^2 t\,dt\\ \end{align}$

Since, for $t$ real,

$\sin(2t)=2\sin t\cos t$

then,

$\begin{align} \int_0^x\sin^4 t\,dt&=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\frac{1}{4}\int_0^x \sin^2(2t)\,dt\\ \end{align}$

In the latter integral perform the change of variable $y=2t$,

$\begin{align} \int_0^x\sin^4 t\,dt&=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\frac{1}{8}\int_0^{2x} \sin^2(y)\,dy\\ &=-\frac{1}{2}\cos x\sin x+\frac{1}{2}x-\frac{1}{8}\left(-\frac{1}{2}\cos (2x)\sin(2x)+\frac{1}{2}\times 2x\right)\\ &=-\frac{1}{4}\sin(2x)+\frac{1}{2}x+\frac{1}{32}\sin(4x)-\frac{1}{8}x\\ &=-\frac{1}{4}\sin(2x)+\frac{3}{8}x+\frac{1}{32}\sin(4x)\\ \end{align}$

Therefore,

$\displaystyle \boxed{\int \sin^4 x\,dx=\frac{3}{8}x+\frac{1}{32}\sin(4x)-\frac{1}{4}\sin(2x)+C}$

($C$ a real constant)

Answer 2

Use this $$\sin^2x = \frac{1-\cos (2x)}2 \implies \sin^4x = \left( \frac{1-\cos 2x}2 \right)^2=\frac{1+\cos^2 (2x) -2 \cos 2x}{4}$$

And then,

$$\cos^2 (2x)=\frac{1+\cos (4x) }{2}$$

Answer 3

An alternative way might be to use the fact that $$\sin(x)=\frac{1}{2}i \left(e^{-ix}-e^{ix}\right)$$ $$\sin^4(x)=\frac{1}{16} \left(e^{-ix}-e^{ix}\right)^4$$ And you just need to expand and integrate it.