The Student Room Group
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.
Reply 7
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 5 years ago)