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

RK
Answered by Rachel K. Maths tutor

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