- #1

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## 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

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- Thread starter afcwestwarrior
- Start date

- #1

- 457

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∫ 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

rock.freak667

Homework Helper

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

then try a substitution.

- #3

- 457

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

- #4

- 457

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

- #5

rock.freak667

Homework Helper

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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.

- #6

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Ok thank you.

- #7

HallsofIvy

Science Advisor

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[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- sin

[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.

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