Question

How do you use partial fractions to find the integral `int (sinx)/(cosx(cosx-1))dx`?

Answer 1

`int sinx/(cosx(cosx - 1)) dx = ln|(cosx)/(cosx - 1)| +C`

Explanation:

Let `u = cosx`. Then `du = -sinxdx -> dx= (du)/(-sinx)`.

`int sinx/(u(u - 1)) * (du)/(-sinx)`

`-int1/(u(u - 1))du`

Now use partial fractions.

`A/u + B/(u - 1) = 1/(u(u - 1))`

`A(u -1) + B(u) = 1`

`Au - A + Bu = 1`

`(A + B)u - A = 1`

Write a system of equations.

`{(A + B = 0), (-A = 1):}`

Solving, we get `A = -1` and `B = 1`.

`-int 1/(u - 1) - 1/u du`

`int1/udu - int 1/(u - 1)du`

`ln|u| - ln|u - 1| + C`

`ln|cosx| - ln|cosx - 1| + C`

`ln|(cosx)/(cosx - 1)| +C`

Hopefully this helps!