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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?
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#4
).
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#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.
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#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.
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#7
I've been trying to do this all day with no success. Any hints? It's beginning to irritate me.
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#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.
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#9
Thanks. Of all the methods I tried, I can't believe I didn't see u=secx or u=sec^2x.
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#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
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#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.
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#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)
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#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 month ago
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