How do you integrate (sinx)^3 dx?

Firstly, we must use the special property of trig functions which means that (sinx)ⁿ is the same as sinⁿx. Likewise, (cosx)ⁿ=cosⁿx, (tanx)ⁿ=tanⁿx and so on. In this case, we need to start by replacing (sinx)³ with sin³x. Our question has now changed from ∫(sinx)³ dx to ∫sin³x dx which we can then split up to become ∫(sinx)(sin²x)dx. We now need to use integration by substitution and we will start by using u=cosx. By differentiating this, we get to du/dx= -sinx. Remember the aim of integration by substitution is to replace all x variables with a new variable (in this case u) so we need to start by rearranging du/dx= - sinx to dx=(-1/sinx)du. We now need to substitute this into our question and we get ∫(sinx)(sin²x)(-1/sinx)du and the sinx at the start will now cancel with the sinx on the denominator of the third term to leave us with ∫(sinx)(sin²x)(-1/sinx)du = ∫(sin²x)(-1) du. Before we can integrate we have to remember that we still need to get everything in terms of u (which w defined as cosx) so to do this, we can use the trig identity sine²x+cos²x=1 and rearrange to sin²x=1-cos²x. If we now replace this into our question, we get ∫(1-cos²x)(-1) du which equals ∫(1-u²)(-1) du. We are now ready to integrate as everything is in terms of u but to make it easier for ourselves, we should take the -1 outside the front of the integral sign and split the question up to -1∫(1 du -1∫-u²du . We can now finally integrate! After integration we get -1u+(1/3)u³ + c. Remember to use +c as the question was an indefinite integral (an integration question with no limits). The final step is to get everything back in terms of x so all we need to do is replace any 'u's with cosx which gets us -cosx+(1/3)cos³x + c