Question

Integrate:
sin^3(x) dx

Answer 1

= 3sin²x * cosx

Answer 2

First, we should break-up the sin^3(x) because that isn't useful at all:

\[\int\limits_{}^{} \sin^3(x) dx = \int\limits_{}^{} \sin^2(x) \sin(x) dx\]

By the Pythagorean Identity, we know Sin^2(x) = 1 - Cos^2(x), so:

\[\int\limits_{}^{} Sin^2(x) Sin(x) dx = \int\limits_{}^{} (1 - Cos^2x) Sin(x) dx\]

now, do you have any ideas of what you should do? Hmm... let's go ahead and try a u-substitution:

u = Cos(x)

du = - Sin(x) dx

- du = Sin(x) dx

\[\int\limits_{}^{} (1 - Cos^2x) Sin(x) dx = \int\limits_{}^{} (1 - u^2)(-du)\]

\[= -\int\limits\limits_{}^{} (1 - u^2)(du)\]

now lets integrate that to get:

\[- (u - (1/3)u^3)\]

Now back substituting:

\[= - (Cos(x) - (1/3)Cos^3 x)\]

\[= - Cos(x) + (1/3)Cos^3 (x)\]

Answer 3

thank you!