Official Rep
Back
to top
Integrate sec^2(x)
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
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?
I know the answer is tan(x) but I want to be able to prove it...How do I prove this?
0
reply
Report
#4
).
0
reply
Report
#5
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.
I = ∫(secx)^2 * dx
let u = tgx
∴du = (secx)^2 * dx
I = ∫du
=u + c
=tgx + c
Hopefully that helps.
0
reply
Report
#6
(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.
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.
6
reply
Report
#7
I've been trying to do this all day with no success. Any hints? It's beginning to irritate me.
0
reply
Report
#8
(Original post by Zuzuzu)
I've been trying to do this all day with no success. Any hints? It's beginning to irritate me.
I've been trying to do this all day with no success. Any hints? It's beginning to irritate me.
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.
0
reply
Report
#9
Thanks. Of all the methods I tried, I can't believe I didn't see u=secx or u=sec^2x.
0
reply
Report
#10
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
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
0
reply
Report
#11
(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 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.
0
reply
Report
#13
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)
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)
0
reply
Report
#14
(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)
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)
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)).
Last edited by simon0; 1 year ago
1
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
