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,

 

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,

 

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.