Integration - Miscellaneous Substitution, 2

Category: Integral Calculus, Trigonometry

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above

You notice that the denominator contains trigonometric functions and we cannot integrate it by simple integration. This is a difficult one because the numerator has no trigonometric functions. If you will use the integration by parts, then the above equation will be more complicated and there will be an endless repetition of the procedure.  

For this type of a function, like the given equation above, we can integrate it by Miscellaneous Substitution. Let's proceed with the integration technique as follows 

Let

From double angle formula,

Since the given problem has Cosine function, then we can get the values of Sine and Cosine functions from Tangent function as follows 

Photo by Math Principles in Everyday Life

Using the figure above that

From the given problem

Substitute the values of dx, Sin x, and Cos x to the above equation, we have

The above equation can now be integrated by Inverse Trigonometric Function Formula as follows

but

Therefore,

  

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,