Integrate: (sinx)^2 * (cosx)^3 dx.?

Answer 1

∫sin² x cos³ x dx

First, extract a factor of cos² x:

∫sin² x (cos² x) cos x dx

Using the Pythagorean theorem:

∫sin² x (1 - sin² x) cos x dx

Distributing:

∫(sin² x - sin⁴ x) cos x dx

Now we just make the substitution u=sin x, du=cos x dx to obtain:

∫u² - u⁴ du
u³/3 - u⁵/5 + C
1/3 sin³ x - 1/5 sin⁵ x + C

And we are done.

Answer 2

let int be the integral sign...

the given may be written as
int(sin^2x*cos^3x)dx
=int((sin^2x*cos^2x*cosx)dx
{{{from the identity that sin^2x + cos^2x=1, we may write cos^2x=1-sin^2x}}}, then,
=int((sin^2x(1-sin^2x)*cosx)dx
=int[(sin^2x-sin^4x)cosx]dx
=int(sin^2x cosx)dx - int(sin^4x cosx)dx
we let u=sinx, then du=cosxdx, then
=int(u^2 du) - int(u^4 du)
=(u^3) / 3 - (u^5) / 5 + C
since u=sinx
= (sin^3x)/3 - (sin^5x)/5 + C