need help. I know i got the wrong answer, because i checked in the back of the book, but i have no idea where i messed up. help please
sinx/((1+cosx)^2)dx from the intervals 0 to pie/3
It's hard to know where you messed up without seeing your work, but I would try the substitution:
$\displaystyle u=1+\cos(x)\,\therefore\,du=-\sin(x)\,dx$ and the integral becomes:
$\displaystyle -\int_{u(0)}^{u\left(\frac{\pi}{3} \right)}u^{-2}\,du=\int_{\frac{3}{2}}^2 u^{-2}\,du$
Can you proceed from here?
ok so to get the 2 you need to plug the 0 to the equation 1+cosx and that gives you 2. to get the 3/2 you plug in pi/3 to the same formula and that gives you 3/2. the reason that the 2 and the 3/2 switch is because there is a negative in the front and that is a rule.