# Integrals of Trig Functions

Hello everyone,

I am having some trouble finding the integral of tanxsec^2x. I honestly have no idea where to go with this one. I've finished all the others but this one is really screwing me up. I tried doing something with the chain rule but it didn't work out at all. I know that I am supposed to show my work, but at this point I don't have anything to show you guys. Thanks a lot for any help you can give me.

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OlderDan
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scorpa said:
Hello everyone,

I am having some trouble finding the integral of tanxsec^2x. I honestly have no idea where to go with this one. I've finished all the others but this one is really screwing me up. I tried doing something with the chain rule but it didn't work out at all. I know that I am supposed to show my work, but at this point I don't have anything to show you guys. Thanks a lot for any help you can give me.
Write everything in terms of sines and cosines. Then you can do a simple substitution.

Or just note that the derivative of tangent is the secant squared... u-substitution, anybody?

--J

Ok guys, thanks for the help, but I'm still completely lost. I still don't have a clue of what I should do.

Are you familiar with u-substitution? You're going to have to make one to evaluate the integral. You have two choices for what to let u equal, both work. What do you think u might equal?

--J

I am sort of familiar with it. Could you let u equal tanx?

OlderDan
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scorpa said:
I am sort of familiar with it. Could you let u equal tanx?
If you did, what would du be?

Ok, actually I might let u = sec^2x, then du should equal tanx+c?

OlderDan
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scorpa said:
Ok, actually I might let u = sec^2x, then du should equal tanx+c?
That is not correct

You should review the derivatives of all the trig functions.

You're integrating u to get du for some reason. You must differentiate u to get du.

--J

dextercioby
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$$\int \sec^{2} x \tan x \ dx=-\int \frac{d {}\cos x}{\cos^{3} x}=\frac{1}{2}\sec^{2}x +\mathcal{C}$$

Daniel.

shouldnt it be:
$$\int sec^2x~tanx~dx=\int\frac{sinx}{cos^3x}dx$$
then, you could do u-sub and set $u=cosx$ and go from there?

*Edit*
dextercioby, you put sinx instead of cosx. i think tanx = sinx/cosx
*Edit*

Last edited:
OlderDan
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p53ud0 dr34m5 said:
shouldnt it be:
$$\int sec^2x~tanx~dx=\int\frac{sinx}{cos^3x}dx$$
then, you could do u-sub and set $u=cosx$ and go from there?
and du would be

$$du = d cosx = -sinx dx$$

giving

$$sinx dx = -d cosx$$

which is what Dexter did to get his result, and what earlier hints were pointing to. It could also be done using

$$u= tanx$$

$$du= d tanx = sec^2x dx$$

$$sec^2x dx = d tanx$$

as suggested by Justin early on

Oh ok....I think I am starting to get it