Question

Calc: Integrals with Trig

integral (-pi)/(4) to (3pi)/(4) 8sec(theta)tan(theta) ?

I know I first need to find the anti-derv
but can I rewrite sec(theta) and tan(theta) using identities?

Answer 1

\[\large \int\limits_{ -\frac{\pi}{4}}^{\frac{3\pi}{4}} 8\sec(\theta)\tan(\theta)d(\theta)\] let \(\ u = (\theta)\) and factor out the 8.

Answer 2

\[\large 8\int\limits_{ -\frac{\pi}{4}}^{\frac{3\pi}{4}} \sec(u)\tan(u)d(u)\]

Answer 3

Or...you can remember that \(\dfrac{d}{d\theta}\sec\theta=\sec\theta\tan\theta \implies \displaystyle \int\sec\theta\tan\theta \,d\theta= \sec\theta + C\).

There really isn't a need to change the variable of integration.

Answer 4

Then you've got your identity \[\ \int sec(u)tan(u)du = sec(u)\] So now just integrate \[ \sec(\theta) ]_{-\frac{\pi}{4}}^{\frac{3\pi}{4}}\]

Answer 5

Well, I wasn't changing the variable, but just using it to fit the identity. :P

Answer 6

@mathstudent55

Answer 7

how do I Factor out the 8 ?

Answer 8

You can pull any constants out of an integral.

Answer 9

As long as they're multiplied to one function throughout. in this case they act as a scaling factor to sin(theta)cos(theta) therefore do not need to be integrated along with the function.

Therefore they can be dealt with later once you integrate your function.

Answer 10

ok so what exactly would be the anti-derivative in this case

Answer 11

\[8\sec (\Theta)+C\]

Answer 12

@ArkGoLucky, and then I can plug in the values on the integral? in this case, -pi/4 and 3pi/4?
But shouldn't i change the sec(theta) to 1/cos(theta) ?

Answer 13

@Jhannybean, thank you :)

Answer 14

Yw :) and no you don't need to change it to 1/cos(\(\theta\)) since you've already integrated your function :)

Answer 15

ok thank you !