How Do You Integrate tan(x)sec^2(x)?

[scorpa]

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.

 

[OlderDan]

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.

 

[Justin Lazear]

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

--J

 

[scorpa]

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

 

[Justin Lazear]

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

 

[scorpa]

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

 

[OlderDan]

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

If you did, what would du be?

 

[scorpa]

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

 

[OlderDan]

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.

 

[Justin Lazear]

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

Stick with your original substitution.

--J

 

[dextercioby]

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

Daniel.

 

[p53ud0 dr34m5]

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*

 

[OlderDan]

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

 

[scorpa]

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