# How do i figure out this integral

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

## Answers and Replies

Homework Helper
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)

Homework Helper
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.

Homework Helper
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.