How do you integrate:

òsinx(1+cos² x) dx?? I have been trying to do this for quite a while now, but I must say I am really stuck! I guess you have to make some substitutions using the product formulae or the sum formulae or the trigonometric identities or things like that, but I cannot do it. I appreciate if someone can help.


Tony Ho

So du/dx=-sinx, so -du=sinx dx, and we have that:

òsinx(1+cos² x)dx

=-ò1+u² du

=-u-u³/3+ constant

=-cosx-(cos³ x)/3)+ constant

This works because you'll notice that we have (-1 times) the derivative of cosx times a function of cosx, so when we do the substitution, the sinx disappears nicely.

Hope that helps,

Pras

Thanks very much, I am ashamed I haven't noticed something so obvious.