Integrate sin^2 x cos^5 x dx. I have the answer but I have no idea how to get to it.?

∫ sin^2(x)cos^5(x)dx , recall that cos^2(x) = 1 - sin^2(x), so we can write the integrand as follows,

= ∫ sin^2(x)((1-sin^2(x))^2)cos(x)dx

= ∫ sin^2(x)cos(x)(1 - 2sin^2(x) + sin^4(x))dx , now

= ∫ sin^2(x)cos(x)dx - 2 ∫ sin^4(x)cos(x)dx + ∫ sin^6(x)cos(x)dx ,well we now have sine functions and its derivative , if you cannot see directly, use substitution

let sin(x) = u and cos(x)dx = du , right ? so we have

= ∫ (u^2)du - 2 ∫ (u^4)du + ∫ (u^6)du , so

= (u^3)/3 - 2((u^5)/5) + (u^7)/7 + c , now use back substitution to have the result in terms of "x" ,

= sin^3(x)/3 - 2sin^5(x)/5 + sin^7(x)/7 + c

Please read :

cos^5 (x) = cos^4 (x) * cosx =
= (1-sin^2(x).)^2 * cosx
= (1-2 sin^2 (x) +sin ^4 (x)) *cosx.
sin ^2 (x) * cos ^5(x) =
sin ^2 (x) (1-2sin^2 (x) +sin^4 (x) ) cosx=
(sin ^2 (x)-2 sin ^4 (x) +sin ^6 (x) cosx.

Sostituzione
t= sin x
dt=cosx dx
integ ( (t^2-2t^4+t^6) dt)=
=t^3/3+2t^5/5 +t^7/7 +c

Integ (f (x) dx =
= (sin^3 (x))/3+2 sin ^5 (x)/5+
+sin ^7 (x)/7 +c.