How Do You Solve the Integral of 1/cos(theta)?

[afcwestwarrior]

Homework Statement

∫ 1/cos theta
before it was ∫ sec theta but i know that sec theta is equal to 1/cos theta

would i write it as 1/2 ∫ cos theta

and then 1/2 sin theta

 

[rock.freak667]

try multiplying sec(theta) by

[tex]\frac{sec\theta +tan\theta}{sec\theta +tan\theta}[/tex]


then try a substitution.

 

[afcwestwarrior]

where did you get sec theta + tan theta/ sec theta+ tan theta from

 

[afcwestwarrior]

is this what i do (sec theta) (sec theta+ tan theta)/ Sec theta( sec theta + tan theta)

 

[rock.freak667]

is this what i do (sec theta) (sec theta+ tan theta)/ Sec theta( sec theta + tan theta)

You should get

[tex]\frac{sec\theta(sec\theta +tan\theta)}{sec\theta + tan\theta}[/tex]


The expand out the numerator.


It's an easier way to find the integral faster rather than another method.

 

[afcwestwarrior]

Ok thank you.

 

[HallsofIvy]

By the way, [itex]\int dx/cos(x)[/itex] involves an odd power of cosine so the "standard" method for that situation will work. Multiply both numerator and denominator by cos(x) to get
[tex]\int \frac{dx}{cos x}= \int \frac{cos x dx}{cos^2 x}= int \frac{cos x dx}{1- sin^2 x}[/tex]

Now let u= sin x so du= cos x dx and 1- sin² x= 1- u²

[tex]\int \frac{dx}{cos x}= \int \frac{du}{1- u^2}= \int \frac{du}{(1- u)(1+ u)}[/itex]

and you can use "partial fractions" to integrate that.