integrate tan^(3)4x dx?

Question

Don't even know where to start with this one. Any help is greatly appreciated! Thanks

Answer 1

This looks tricky, but is actually easy peasy if you make the correct substitution. Remember (a) tan x = sin x/cos x, (b) (sinx)^2 + (cosx)^2 = 1, and (c) d/dx(cos x) = -sin x.

So, let cos 4x = Y, so sin 4x dx = -1/4 dY

Thus, your integral becomes the integral of -1/4(1/Y^3 - 1/Y)dY
= 1/4(1/Y - 1Y^-3)dY
= 1/4(ln Y) + (Y^-2)/8 + c
= 1/4(ln cox 4x) + 1/(8 cos ^2 4x) + c.

Check: d/dx[1/4(ln cox 4x) + 1/(8 cos ^2 4x) + c]
= {1/4*(1/cos 4x)*4 + 1/8[cos^-3 4x]*(-2)*4}-sin 4x dx
= {1/ cos 4x - 1/cos^-3 4x}-sin 4x dx
which, with a bit more trig, is tan^(3)4x dx

Answer 2

using fact the by-fabricated from tan(x) is sec²(x), ?tan(x) sec²(x) dx = (a million/2)tan²(x) + C ---- replace: the reason which you're transforming into a good number of diverse looking solutions under is via the fact a million+tan²(x) = sec²(x). especially, think that C = a million/2 + D Then the respond may well be (a million/2)tan²(x) + a million/2 + D = (a million/2)(tan²(x) + a million) + D = (a million/2)sec²(x) + D = (a million/2)(a million/cos²(x)) + D those solutions are all equivalent. it fairly is between the few situations the place that consistent you frequently tack on after integration impacts what the result sounds like. So often, to envision if 2 integrations of the comparable function are equivalent, you're able to subtract the two effects and if the adaptation is a relentless, you recognize that the two solutions are purely as valid (or purely as incorrect, for that count).

Answer 3

There is a trick to solve integrals, which contain products of trigonometric functions. It uses the property, that the derivative of trigonometric function can be expressed in terms of the function. The integral can be simplified to a integral of a product of the function and its derivative, which can be evaluated by partial integration:
Integral (f' * f)dx = f*f -Integral(f * f')dx
=> Integral (f' * f)dx = 1/2 * f²

In that particular case use:
d(tan(4x)/dx = 4 *(1+tan²(4x))

Integral[tan³(4x)]dx
= Integral[1/4* {4(1+tan²(4x))tan(4x)} - tan(4x)]dx
=1/4*Integral[4(1+tan²(4x))tan(4x)]dx - Integral[tan(4x)]dx
First Integral can be integrated as shown above
Second integral is the Integral of the tangens, which i guess you know. Integral(tan(ax))dx= -1/a*ln(cos(ax))
Hence:
Integral[tan³(4x)]dx
= 1/4*1/2*tan²(4x) + 1/4*ln(cos(4a))+c
= 1/8*tan²(4x) + 1/4*ln(cos(4a))+c

Answer 4

hey,i am going to give you some pointers.
>tan^3(4x)=tan^2(4x)*tan4x
>put tan^2(4x)=sec^2(4x)-1
>integral [sec^2(4x)-1]tan4x
>integral sec^2(4x)tan4x-integral tan4x
then finally put tan4x=t differentiate 4sec^2(4x)d(x)=d(t)
>1/4integral td(t)-integral td(t) and then proceed.its easy.lastly substitute the value of t.
remember there can be different answers for finding value of indefinite integral but for definite the answers have to match

Answer 5

Try the substitution tan^2(x) = sec^2(x) - 1.

Steve

Answer 6

differentiate tan^2 4x and tell me what you get.

Answer 7

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