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

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

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)