Integrate ∫sin²xcosxdx

This indefinite integral (has no defined limits) should be evaluated by integration by substitution This is because you cannot integrate sin²xWhen doing integration by substitution, the first thing we need to look for is if the integral is of the form ∫ [f(x)]ⁿf(x) dx or ∫ f'(x) / f(x) dxLooking at this question, we can see that cosx is the derivative of sinx, this allows us to simplify this integralFirst, let u = sinx (this is our substitution)Since we now have a 'u' in the integral, we need to change dx to du, as the integral is now with respect to 'u', rather than 'x'To do this, we say that u = sinx so du/dx = cosxTo get an expression for dx, in terms of du, we rearrange this expression to give: dx = du/cosxWe can now substitute u = sinx and dx = du/cosx into the integral to give ∫ u²cosx du/cosxIt is clear that cosx cancels leaving ∫ u² duThis can easily be integrated using standard methods to give: [ u³/3] + c [remember the constant of integration + c since it is an indefinite integral!]Finally, substitute u = sinx back into the solution to get answer in terms of x: (1/3)sin³x + c