How do you Integrate?

int((1-sinx)/(1-cosx))e^xdx

1 Answer
May 4, 2018

-e^xcot(x/2)+C.

Explanation:

Let, I=int((1-sinx)/(1-cosx))e^xdx.

:. I=int((1-sinx)/(2sin^2(x/2)))e^xdx,

=int{(1/(2sin^2(x/2))-sinx/(2sin^2(x/2))}e^xdx,

=int(1/2csc^2(x/2)-(2sin(x/2)cos(x/2))/(2sin^2(x/2))}e^xdx.

:. I=int{1/2csc^2(x/2)-cot(x/2)}e^xdx.

Now, have a look at the following useful Result (R) :

R : int{f(x)+f'(x)}e^xdx=e^xf(x)+c.

We have, int{f(x)+f'(x)}e^xdx,

=intf(x)e^xdx+intf'(x)e^xdx.

=f(x)e^x-intf'(x)e^xdx+intf'(x)e^xdx...[IBP, u=f(x), v'=e^x],

=f(x)e^x+c.

Hence, the Result R.

Applying R to I,

with f(x)=-cot(x/2), f'(x)=-csc^2(x/2)*1/2,

:. I=-e^xcot(x/2)+C.