Integration - Hyperbolic Functions

Category: Integral Calculus, Trigonometry

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above

There are two functions in the given equation which are trigonometric and hyperbolic functions. Since there are two functions in the integration, then we have to integrate the given equation by using Integration by Parts. 

If

then 

If

then

Using Integration by Parts,

Since the second term of the above equation have two functions, then we have to use the Integration by Parts again. 

Consider

If

then

If 

then

Again, using by Integration by Parts,

From the first integration by Integration by Parts,

but

Therefore, the final answer is