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Tuesday, August 6, 2013

Integration - Trigonometric Functions, 4

Category: Integral Calculus, Trigonometry

"Published in Newark, California, USA"

Prove that 


Solution:

Consider the given equation above


Rewrite the left side of the equation as a quotient of two trigonometric functions as follows


If


then


Hence, the above equation becomes


Integrate the above equation using the integration of the reciprocal function formula, we have 



Therefore,

 

Monday, August 5, 2013

Derivative - Trigonometric Functions, 4

Category: Differential Calculus, Trigonometry

"Published in Suisun City, California, USA"

Prove that


Solution:

Consider the given equation above


Rewrite the left side of the equation as a quotient of two trigonometric functions as follows


Take the derivative of the above equation using the quotient of the two functions formula, we have 






But


Hence, the above equation becomes




Therefore,

 

Sunday, August 4, 2013

Integration - Trigonometric Functions, 3

Category: Integral Calculus, Trigonometry

"Published in Suisun City, California, USA"

Prove that


Solution:

Consider the given equation above


Rewrite the left side of the equation as a quotient of two trigonometric functions as follows


If 


then


Hence, the above equation becomes


Integrate the above equation using the integration of the reciprocal function formula, we have





Note


because the left side of the equation is a reciprocal of trigonometric function which is equal to sec x while the right side of the equation is inverse trigonometric function which is equal to an angle.



Therefore,


where C is a constant of integration.