how do i figure out this integral

[afcwestwarrior]

1. The problem statement, all variables and given/known data
∫ 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

\frac{sec\theta +tan\theta}{sec\theta +tan\theta}


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

\frac{sec\theta(sec\theta +tan\theta)}{sec\theta + tan\theta}


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, \int dx/cos(x) 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
\int \frac{dx}{cos x}= \int \frac{cos x dx}{cos^2 x}= int \frac{cos x dx}{1- sin^2 x}

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.