Integration - Trigonometric Functions, 6

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 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 sin 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 - cos x, we have

But

and the above equation becomes

Therefore,

   

Derivative - Trigonometric Functions, 6

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  

Therefore,