How would I integrate following trigonometric function $\displaystyle\int \sin^3(x) \,\operatorname{d}x$?

Question

How to evaluate $$\displaystyle\int \sin^3(x) \,\operatorname{d}x$$ without using integration by parts?

Answer 1

HINT:

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

Set $\cos x=u$

Answer 2

Hint. You may just write $$ \sin^3 x=\sin x \times \sin^2 x=\sin x \times (1- \cos^2 x). $$

Answer 3

Hint:

$$\int \sin^3(x) \,\operatorname{d}x=\int \left[\sin(x)\right]\left[1-\cos^2(x)\right] \,\operatorname{d}x$$

Then, use the substitution, $t=\cos(x)$.

Answer 4

Let $t:=\cos x$ $$\implies \int\sin^3x dx=\int (t^2-1)dt=t^3/3-t+c=\frac13\cos^3x-\cos x+c.$$