How do i figure out this integral

1. Aug 24, 2008

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

2. Aug 24, 2008

rock.freak667

try multiplying sec(theta) by

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

then try a substitution.

3. Aug 24, 2008

afcwestwarrior

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

4. Aug 24, 2008

afcwestwarrior

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

5. Aug 24, 2008

rock.freak667

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.

6. Aug 24, 2008

afcwestwarrior

Ok thank you.

7. Aug 25, 2008

HallsofIvy

Staff Emeritus
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.