Question

Evaluate the integrals? (question will be displayed ASAP)

Answer 1

\[\int\limits_{0}^{\pi} \frac{ \sin t }{ 2-\cos t }dt\]

Answer 2

u = cos(t)
du = -sin(t)dt

Answer 3

\[
\int\limits_{0}^{\pi} \frac{ \sin t }{ 2-\cos t }dt= -\int_1^{-1} \frac {du}{2-u}
\]

Answer 4

why did you turn the pi and 0 into -1 and 1?

Answer 5

cos(pi) =-1
cos(0) =1

Answer 6

Did you understand it?

Answer 7

wait I'm still ingesting ^_^

Answer 8

\[
\int\limits_{0}^{\pi} \frac{ \sin t }{ 2-\cos t }dt= -\int_1^{-1} \frac {du}{2-u}=\\
\int_{-1}^{1} \frac {du}{2-u}=-\left [ \ln(2-u)\right ]_{-1}^1=\ln(3)
\]

Answer 9

When you make a change of variables, you have to change the bound accordingly.

Answer 10

haha I think I get it now. Really? what do you mean changing the bounds accordingly?

Answer 11

does it mean we have to convert the bounds, for example, from pi to cos(pi) when we make the the u become cos(t) ?

Answer 12

you could also skip changing the limits, evaluate the integral in terms of u then put back the original variable; that is replace every u by Cos(x). Then you can evaluate the expression you get with the original limits. (Remember that the integration is already done so its just the last step you have to do with x)

Answer 13

Your original bounds were not just pi and 0, they were t=pi and t=0. When you switch to u, you are using the same bounds, but it would be meaningless to integrate with bounds that are values for t, so we use those original values to find bounds for u.

Answer 14

alright guys, I think I understand this loud and clear :) thanks for explaining!

Answer 15

\[\int\limits_{x=0}^{x=\pi}(-cosx/sinx)dx\]
u=cosx
dx=-du/sinx
\[\int\limits_{x=0}^{x=\pi}udu=\int\limits_{u(x=0)}^{u(x=\pi)}udu\]