Trig Integrals: In this Youtube video (http://www.youtube.com/watch?feature=player_embedded&v=y_CA5btuoQk#!) from 4:20 to 5:20 the Instructor makes some substitutions which I can't quite follow. At 4:34 he claims that BOTH sinx's will be replaced by du and thus disappear. Is this really the case? Next: u=cosx, but he substitute cosx^-2 with -u^-2. Shouldn't this be u^-2? Finally, he ends up with: -u^-2+1du which becomes u^-1+u+c when integrating. Where does this 1 come from?
the integral is (cosx)^-2 * sin(x) - sin(x) dx -sin(x)dx [ -(cosx)^-2 + 1] u = cosx du/dx = -sin(x) du =-sin(x)dx so : du [ -u^2 + 1]
yeah
you're right, the instructor was just using notation that confused me here
du [ -u^-2 + 1]
im sorry my last line in the first response isnt correct it should be du [ -u^-2 + 1]
so integrating this gives us : -u^(-1) / -1 + u = u^(-1) + u
yeah, i got it though @coolsector, thx ^^
:)
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