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

1 Answer
Feb 19, 2017

#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!