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,



11