How do you use partial fractions to find the integral #int (sinx)/(cosx+cos^2x)dx#?

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Answer 1 · 2016-12-21T18:18:23

#ln|cosx + 1| - ln|cosx| + C#

We start with a substitution. Let #u = cosx#. Then #du = -sinxdx -> dx = (du)/-sinx#.

#=>intsinx/(u + u^2) xx (du)/-sinx#

#=> int -1/(u + u^2)du#

We now factor the denominator to #u(1 + u)#. Let's now write #-1/(u + u^2)# in 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#

Now write a system of equations:

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

Solving, we get: #A = -1# and #B = 1#.

Thus, the partial fraction decomposition is #-1/u + 1/(u + 1)#. The integral becomes #int(1/(u + 1) - 1/u)du#.

We can now integrate using the rule #int(1/u)du = ln|u| + C#.

#=> ln|u + 1| - ln|u| + C#

Finally, reinsert the value of #u# to make the function defined by #x# instead of #u#:

#=>ln|cosx + 1| - ln|cosx| +C#

Hopefully this helps!