I have this integral:
$$\int\sin^3t \, dt$$
I have tried partial integration with $\sin t \cdot \sin^2t$, but then I get another integral to evaluate which needs partial integration:
$$\dots \int \cos^2t \cdot \sin t \, dt \dots$$
Which gives me another integral that needs partial integration and so on. I get stuck in a partial integration loop.
How should I evaluate this?
Answer
$$\int \sin^{3}(t)dt=\int \sin(t)(1-\cos^{2}(t))dt$$
$$=\int \sin(t)dt-\int \sin(t)\cos^{2}(t)dt$$
$$=-\cos(t)-\int \sin(t)\cos^{2}(t)dt$$
Try $u=\cos(t)$ so that $du=-\sin(t)dt$ and the second integral is equivalent to
$-\int u^{2}du=-\frac{1}{3}u^{3}=-\frac{1}{3}\cos^{3}(t)$
Which gives us a final answer of
$$\int \sin^{3}(t)dt=-\cos(t)+\frac{1}{3}\cos^{3}(t)+C$$
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