How do you Integrate?

Question

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

997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-04T17:13:38

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

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#.