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Thursday, August 8, 2013

Integration - Trigonometric Functions, 5

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 reciprocal of trigonometric function as follows


Since the above equation cannot be integrated by a simple integration, then we have to use the principles of trigonometric identities as follows


Multiply both the numerator and the denominator by cos x, we have



But



Hence, the above equation becomes



If 


then


Integrate the above equation using algebraic substitution as follows



Express the right side of the equation into partial fractions, we have


Solve for the value of A and B by equating their coefficients:





for u:



for constant:


Substitute the first equation to the second equation





and


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







But


Hence, the above equation becomes


Multiply both the numerator and the denominator by 1 + sin x, we have



But


and the above equation becomes








Therefore,

 

Wednesday, August 7, 2013

Derivative - Trigonometric Functions, 5

Category: Differential 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


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








Therefore,