Given that y=π/6 at x=0 solve the differential equation,dy/dx=(e^x)cosec2ycosecy

This question challenges a students skills with integration and trigonometry a very common question on Core 4 papers that can leave students struggling. First we use our knowledge that cosecy=1/siny and multiply both sides by sin2ysiny to give us sin2ysinydy/dx=e^x. Next we integrate both sides with respect to x so, int[sin2ysinydy/dx]dx=int[e^x]dx. Here due to the nature of integration the 1/dx and dx cancel out on the left side leave us with an integration with respect to y. Next we use the double angle formula to expand out the left side meanwhile we know from the definition of e^x that e^x is equal to the differential and integral of itself, int[2cosy*(siny)²]dy=e^x. Here 2cosy(siny)² is of the form f'(x)f(x)ⁿ where f(x)=siny and f'(x)=cosy. So to integrate the left side we increase the power then divide my the new power and divide by f'(x), in this case cos x. This gives us (2/3)sin(y)³=e^x+c (since this is an indefinite integral we have to add plus c once finished integrating).Now finally we can sub in our values of y=π/6 and x=0 to calculate our c. (2/3)sin(π/6)³=e⁰+c (2/3)(1/2)³=1+c therefore c=1/12-1=-11/12. So our answer becomes sin³y=e^x-11/12