Question

solve the initial value problem
dy/dx= 15 sin^2 xcosx y(0)=6

Answer 1

\[\huge \frac{ dy }{ dx }=15\sin^2x \cdot \cos x\]We can rewrite this as,\[\huge dy=15\sin^2x \cdot \cos x \; dx\]And from here take the integral of both sides.\[\huge \int\limits dy=15\int\limits \sin^2x \cdot \cos x dx\]
It looks like it's going to be a simple U substitution, having trouble changing it up? :D

Answer 2

\[\huge Let \; u=\sin x\]Taking the derivative of this substitution gives us..\[\huge du=\cos x dx\]Understand what to do from here? or still lost? :D

Answer 3

I'm still lost, what is the integral of sin2x? and coax?

Answer 4

Our goal is to replace all of the X terms and DX with U's and DU.
It will hopefully make it a lot easier to integrate.