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?

Question

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?

anonymous · Answer

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]

anonymous · Answer

yeah

anonymous · Answer

you're right, the instructor was just using notation that confused me here

anonymous · Answer

du [ -u^-2 + 1]

anonymous · Answer

im sorry my last line in the first response isnt correct it should be 
du [ -u^-2 + 1]

anonymous · Answer

so integrating this gives us :
-u^(-1) / -1 + u  =
u^(-1) + u

anonymous · Answer

yeah, i got it though @coolsector, thx ^^

anonymous · Answer

:)