what is the integral of tan^4x dx?

Answer 1

tan^4x=tan^2x*tan^2x

1+tan^2x=sec^2x
tan^2x=sec^2x-1

tan^4x=(sec^2x-1)(tan^2x)

integral(tan^4x)=integral(sec^2xtan^2x)-integral(tan^2x)
for first of those two integrals
let u = tan x, du = sec^2x
becomes integral(u^2du)= 1/3u^3=1/3tan^3x

second of those two integrals:
integral(tan^2x)=integral(sec^2x-1)=tanx-x

total becomes 1/3tan^3x-tanx+x+C

Answer 2

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Given: y = ∫ tan²(4x) /[(x-5)tan 4x + tan²(4x)]dx and h = ∫ (4x² - 20x + 25)/[4x² + (2 tan4x - 20)x + 5 ( -tan 4x + 5)] dx First, y can be reduced to y = ∫ tan(4x) /[(x-5)+ tan(4x)] dx. Now let's rearrange h. Numerator can be written as complete square 4x² - 20x + 25 = (2x - 5)². Let's factor the denominator: 4x² + (2 tan4x - 20)x + 5 ( -tan 4x + 5) = 0 D = (2 tan4x - 20)² - 4·4·5 ( -tan 4x + 5) = 4tan²4x - 80tan4x + 400 - 400 + 80tan4x = 4tan²4x x1 = (-2tan4x + 20 + 2tan4x)/8 = 20/8 = 5/2 , x2 = (-2tan4x + 20 - 2tan4x)/8 = (20 - 4tan4x)/8 = (5 - tan4x)/2 , Then 4x² + (2 tan4x - 20)x + 5 ( -tan 4x + 5) = 4(x - 5/2)(x - 5/2 + (tan4x)/2) or 4x² + (2 tan4x - 20)x + 5 ( -tan 4x + 5) = (2x - 5)(2x - 5 + tan4x) . Coming back to the integral h: h = ∫ (2x - 5)²/[(2x - 5)(2x - 5 + tan4x)] dx = ∫ (2x - 5)/(2x - 5 + tan4x) dx . So y + h = ∫ tan(4x) /[(x-5)+ tan(4x)] dx + ∫ (2x - 5)/(2x - 5 + tan4x) dx. I assume there is a typo in the denominator of the first integral, it should be [(2x-5)+ tan(4x)]. In such a case y + h = ∫ (tan4x + 2x - 5)/(2x - 5 + tan4x) dx = ∫ dx = x + C. Initial integral becomes ∫ (y + h) tan 4x dx = ∫ (x + C) tan 4x dx.

Answer 3

Integral Of Tan 4x

Answer 4

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RE:
what is the integral of tan^4x dx?

Answer 5

∫sqrt tan(4x) sec^2(4x) dx let tan(4x) = u 4sec^2(4x)dx = du sec^2(4x)dx = du/4 now the integral becomes 1/4∫(u)^1/2 du (1/4)u^(3/2) /3/2 + c =>(1/6) u^(3/2) + c substitute back u = tan(4x) (1/6)(tan(4x))^(3/2) + c

Answer 6

Integral sec^4(x)/√tg(x). Por favor alguem pode responder

Answer 7

I don't understand your question....

is: int. (tan1)^(4X)dx or int. (tan 4)^X dx or ???????