How do you integrate cosx sin2x?

[IntegrateMe]

If dy/dx = cosx sin²x and if y = 1 when x = pi/2, what is the value of y when x = 0?

So, i separate the derivative into dy = cosx sin²x dx

Then integrate both sides...this is where I'm stuck.

How do you integrate cosx sin²x? I've tried u-sub and IBP but failed both times!

 

[Dick]

Substitute u=sin(x)?

 

[IntegrateMe]

$$\int cosxsin^2x dx$$

u = sinx
du = cosx dx

$$\int u^2 du$$

= $$\frac{u^3}{3}$$

= $$\frac{sinx}{3}$$

plugging in 0 for x gets me $$\frac{1}{3}$$

Correct?

CORRECTION: Nevermind, i got it. They gave me an initial condition.