Evaluate the following integral:
The denominator is factorized, so let separate the fraction through partial fraction, hence let
We need to solve for A and B.
We will equate similar terms, we get.
2A + B = 0 ⇒ B = – 2A
And A + B = 2 cos x
Substituting the value of B, we get
A – 2A = 2 cos x ⇒ A = – 2 cos x
Hence B = – 2A = – 2( – 2 cos x)
⇒ B = 4cos x
We put the values of A and B values back into our partial fractions in equation (i) and replace this as the integrand. We get
Split up the integral,
Let substitute
so the above equation becomes,
Now substitute
v = 1 + u ⇒ dv = du
z = 2 + u ⇒ dz = du
So above equation becomes,
On integrating we get
Substituting back, we get
Applying logarithm rule, we get
Note: the absolute value signs account for the domain of the natural log function (x>0).
Hence,