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

1 Answer
Noah G
Feb 19, 2017

int sinx/(cosx(cosx - 1)) dx = ln|(cosx)/(cosx - 1)| +C∫sinxcosx(cosx−1)dx=ln∣∣∣cosxcosx−1∣∣∣+C

Explanation:

Let u = cosxu=cosx. Then du = -sinxdx -> dx= (du)/(-sinx)du=−sinxdx→dx=du−sinx.

int sinx/(u(u - 1)) * (du)/(-sinx)∫sinxu(u−1)⋅du−sinx

-int1/(u(u - 1))du−∫1u(u−1)du

Now use partial fractions.

A/u + B/(u - 1) = 1/(u(u - 1))Au+Bu−1=1u(u−1)

A(u -1) + B(u) = 1A(u−1)+B(u)=1

Au - A + Bu = 1Au−A+Bu=1

(A + B)u - A = 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!

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