How do you integrate cosx sin2x?

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The discussion focuses on integrating the function cos(x)sin(2x) given the derivative dy/dx = cos(x)sin(2x) and the initial condition y = 1 when x = π/2. The user attempted integration using u-substitution and integration by parts (IBP) but encountered difficulties. Ultimately, the correct approach involves substituting u = sin(x) leading to the integral ∫u² du, resulting in y = (sin(x)³)/3. The user confirmed the solution by applying the initial condition.

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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!
 
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Substitute u=sin(x)?
 


\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:

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