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

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Answer 1 · 2017-02-19T02:01:35

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

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!

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