# How do you integrate cosx sin2x?

1. Apr 13, 2010

### 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!

2. Apr 13, 2010

### Dick

Re: Differentiation...

Substitute u=sin(x)?

3. Apr 13, 2010

### IntegrateMe

Re: Differentiation...

$$\int cosxsin^2x dx$$

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.

Last edited: Apr 13, 2010