Integration - Algebraic Substitution, 5

Category: Integral Calculus, Algebra

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above

Since the denominator consists of a binomial and a radical equation, then we cannot integrate it by simple integration. We need to simplify first the given equation by eliminating the radical sign by algebraic substitution.

Let

then it follows that

Hence, the above equation becomes

The denominator at the right side of the equation can be factored as follows

Since the denominator is already factored, then we can rewrite the above equation into partial fractions as follows

Consider

Multiply both sides of the equation by their Least Common Denominator (LCD), we have

 

 

Equate u:

 

Equate u⁰:

but

The above equation becomes

Substitute the values of A and B to the original equation, we have

 

 

 

 

but

Therefore,

Integration - Trigonometric Functions, 9

Category: Integral Calculus, Trigonometry

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above

The given equation above has both trigonometric functions at the numerator and the denominator. The denominator contains logarithmic function of a trigonometric function. We need to assign u by considering the complicated part in the equation.

If 

then it follows that

 

Since the numerator has the same du, then the above equation can be integrated by simple integration as follows

Integration - Exponential Functions, 2

Category: Integral Calculus, Algebra

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above

Since the numerator is greater than the denominator, then we have to do the division of polynomial as follows

The above equation becomes

If

then

Therefore,