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Intergrate sec x

Updated: 4/28/2022
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16y ago

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Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|

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How do you find sec x cos x?

sec x = 1/cos x so sec x * cos x = 1


How do you prove that the derivative of sec x is equal to sec x tan x?

Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.


What is sec x cos x?

sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1


What is integration of secx tanx?

Will try integration by parts. uv - int[v du] u = sec(x)----------------du = sec(x) tan(x) dv = tan(x)---------------v = ln[sec(x)] sec(x) ln[sex(x)] - int[lnsec(x) dx] = sec(x) ln[sec(x)] - xlnsec(x) - x + C ===========================


How do you put sec cubed in a calculator?

sec x = 1/cos x → sec³ x = 1/cos³ x or sec³ x = (cos x)^-3 Therefore to enter sec³ x on a calculator: Newer, "natural" calculators: mathio: sec³ x → [x-power] [cos] [<angle>] [)] [navigate →] [(-)] [3] [=] lineio: sec³ x → [(] [cos] [)] [)] [x-power] [(-)] [3] [)] [=] Older, function acts on displayed number calculators: sec³ x → [angle] [cos] [x-power] [3] [±] [=]


What is the derivative of secant x?

The derivative of sec(x) is sec(x) tan(x).


What is the second derivative of ln(tan(x))?

f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2


What is the derivative of y equals sec x?

sec(x)tan(x)


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ln|sec x + tan x| + C.


How do you solve Sin x sec x equals tan x?

Cos x = 1 / Sec x so 1 / Cos x = Sec x Then Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x


What is the derivative of secxtanx?

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What is the integral of tan squared x?

Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant