\[\int \sin^{3}x \, dx\]
1
Use Pythagorean Identities: sin⁡2x=1−cos⁡2x\sin^{2}x=1-\cos^{2}x.
∫(1−cos⁡2x)sin⁡x dx\int (1-\cos^{2}x)\sin{x} \, dx

2
Use Integration by Substitution.
Let u=cos⁡xu=\cos{x}, du=−sin⁡x dxdu=-\sin{x} \, dx

3
Using uu and dudu above, rewrite ∫(1−cos⁡2x)sin⁡x dx\int (1-\cos^{2}x)\sin{x} \, dx.
∫−(1−u2) du\int -(1-{u}^{2}) \, du

4
Use Power Rule: ∫xn dx=xn+1n+1+C\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C.
u33−u\frac{{u}^{3}}{3}-u

5
Substitute u=cos⁡xu=\cos{x} back into the original integral.
cos⁡3x3−cos⁡x\frac{\cos^{3}x}{3}-\cos{x}

6
Add constant.
cos⁡3x3−cos⁡x+C\frac{\cos^{3}x}{3}-\cos{x}+C

Done
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