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?

score 2 · Accepted Answer · answered Dec 31 '15 at 10:00

2

HINT:

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

Set $\cos x=u$

answered Dec 31 '15 at 10:00

lab bhattacharjee

score 2 · Answer 2 · edited Dec 31 '15 at 10:31

2

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, $t=\cos(x)$.

edited Dec 31 '15 at 10:31

Vincent

answered Dec 31 '15 at 10:01

Emilio Novati

Alternatively you could multiply out the terms in the integral and note that $-\sin x \cos^2 x = 1/3 d/dx ( \cos^3 x )$ and then use the FTC for that part of the integral. – okrzysik Dec 31 '15 at 10:07

score 1 · Answer 3 · answered Dec 31 '15 at 10:00

1

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

answered Dec 31 '15 at 10:00

Olivier Oloa

score 0 · Answer 4 · edited Oct 26 '19 at 16:24

0

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

edited Oct 26 '19 at 16:24

callculus42

answered Dec 31 '15 at 10:02

RE60K

4 Answers4