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Integrate sec^2(x)

Integrate sec^2(x)

I know the answer is tan(x) but I want to be able to prove it...How do I prove this?
Reply 1
differentiate tan x.
A t-substitution will work.
Reply 3
Glutamic Acid
A t-substitution will work.


Um, but then you're using the result that tanx differentiates to sec^2x so it's cheating. That's assuming that we can't use that result (if we can, yusufu's method is better :p:).
I guess there is, however it still uses the fact that (tgx)' = (secx)^2.

I = ∫(secx)^2 * dx
let u = tgx
∴du = (secx)^2 * dx
I = ∫du
=u + c
=tgx + c

Hopefully that helps.
Original post by anomalocaris
I guess there is, however it still uses the fact that (tgx)' = (secx)^2.

I = ∫(secx)^2 * dx
let u = tgx
∴du = (secx)^2 * dx
I = ∫du
=u + c
=tgx + c

Hopefully that helps.


Pretty sure the OP figured it out in the space of three years.
Reply 6
I've been trying to do this all day with no success. Any hints? It's beginning to irritate me.
Original post by Zuzuzu
I've been trying to do this all day with no success. Any hints? It's beginning to irritate me.


There is a stupid way that avoids knowing how to differentiate tan x.

You can make a substitution u = sec x.

You will then need to make a second substitution.

You still need to know the Pythagorean identities.
Reply 8
Thanks. Of all the methods I tried, I can't believe I didn't see u=secx or u=sec^2x.
If anyone searching for a solution to this problem comes across this page, here's how its done.

INT sec(^2)x dx

You have to use a substitution. Let u = tan x. We know tan x = sin x / cos x and using the quotient rule, du/dx = 1/sec^2 x. So rearranging, du = sec(^2)x * dx

Substitute and you have INT du

which is just u + c.

Substitute u = tan x and you have tan x + c
Original post by BrasenoseAdm
If anyone searching for a solution to this problem comes across this page, here's how its done.

INT sec(^2)x dx

You have to use a substitution. Let u = tan x. We know tan x = sin x / cos x and using the quotient rule, du/dx = 1/sec^2 x. So rearranging, du = sec(^2)x * dx

Substitute and you have INT du

which is just u + c.

Substitute u = tan x and you have tan x + c



If I am reading that correctly, you use that the derivative of tan x = sec^2 x which is itself sufficient proof.
Use integration by parts
sec^2(x) = 1/cos^2(x)
so use quotient rule with u=1 and v=cos^2(x)
the derivative of cos^2(x) can be worked out using chain rule
Put that in the quotient rule formula
you should get the answer sec(x)tan(x)
Reply 13
Original post by dj_ad_1
sec^2(x) = 1/cos^2(x)
so use quotient rule with u=1 and v=cos^2(x)
the derivative of cos^2(x) can be worked out using chain rule
Put that in the quotient rule formula
you should get the answer sec(x)tan(x)

First, this is a very old thread.

Second, this discussion is of the INTEGRAL of sec^2(x), not the derivative (and the derivative of sec^2(x) is not sec(x)tan(x)).
(edited 4 years ago)

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