Home > Calculus > Integrals of Trigonometric Functions

How do you find the integral of #int 1/(1 + sec(x))#?

Answer 1

Ezekiel Alexander

#int dx/(1+secx ) = x -tan(x/2)+C #

Take note of this:

#1/(1+secx ) = 1/(1+1/cosx) #

Apply the parametric formula now:

#cosx = (1-tan^2(x/2))/(1+tan^2(x/2))#

#1/(1+secx ) = 1/(1+(1+tan^2(x/2))/(1-tan^2(x/2))) #

#1/(1+secx ) = (1-tan^2(x/2))/((1-tan^2(x/2))+(1+tan^2(x/2))) #

#1/(1+secx ) = (1-tan^2(x/2))/2 #

#1/(1+secx ) = 1/2 - 1/2(sec^2(x/2) -1) #

#1/(1+secx ) = 1 - 1/2sec^2(x/2) #

Then:

#int dx/(1+secx ) = int (1 - 1/2sec^2(x/2))dx #

#int dx/(1+secx ) = int dx - int sec^2(x/2)d(x/2) #

#int dx/(1+secx ) = x -tan(x/2)+C #

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezra Adams

To find the integral of ( \int \frac{1}{1 + \sec(x)} ), you can use the substitution method. Let ( u = \tan(\frac{x}{2}) ). Then ( \sec(x) = \frac{1}{\cos(x)} = \frac{1}{\frac{1 - u^2}{1 + u^2}} = \frac{1 + u^2}{1 - u^2} ). Rewrite the integral in terms of ( u ). After simplifying, you will get a form that can be integrated using standard techniques.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.