I need this answer urgently!! ∫tan^(1/3)x dx = ? Read the description?

Question

ie. ∫ (cube root of tanx ) dx
I think i can not make it clearer.

Wal C. you used the reduction formula. So the answer came in the form of an infinite series.
Since you got nearest to the solution ,I tell you what, an infinite series is not regarded as the answer of a integral. It's the convention.

Answer 1

∫tan^(1/3)x dx
I can predict that this'll involve the use of partial fractions.

Let tanx=u^3

=>(sec x)^2 dx=3u^2 du

now, (sec x)^2=1+ (tan x)^2=1+u^6

=>dx=(3u^2)/1+ u^6

therefore, ∫tan^(1/3)x dx=3∫[u X u^2/(1+u^6)]du

=>3∫[u^3/(1+u^6)]du
let u^2=t
=>(2u) du=dt

=>3∫[u^3/(1+u^6)]du=(3/2)∫[t/(1+t^3)]dt

Now, partial fractions will be used
we know that, t/(1+t^3)=t/(1+t)(1+t^2-t)...........since a^3 + b^3=(a+b)(a^2 + b^2 + ab)

let, t/(1+t)(1+t^2-t)=A/(1+t) + (Bt + C)/(1+t^2-t)

=>t/(1+t)(1+t^2-t)=[A(1+t^2 -t) + (Bt + C)(1+t)]/(1+t)(1+t^2 -t)................taking LCM

equating coefficients of u, u^2, constants on both sides
=>1=-A+B+C...................equating coefficients of u
0= A+B .......................equating coefficients of u^2
0= A+C .......................equating coefficients of constant
=>A=-1/3, B=1/3, C=1/3

=> (3/2)∫[t/(1+t^3)]dt=
(3/2)∫[(1/3)/(1+t)]dt + (3/2)∫[(1/3)(1+t)/(1+t^2-t)]dt


=>(-1/2)log|1+t| + (1/2∫[(1+t)/(1+t^2-t)]dt

=>(-1/2)log|1+t| + (1/4)∫[(2+2t)/(1+t^2-t)]dt

we know, d/dt(1+ t^2 - t)=2t -1
=>(-1/2)log|1+t| + (1/4∫[(2+2t+1-1)/(1+t^2-t)]dt

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| + (3/4)∫1/(1+t^2-t)dt

we know, 1+t^2-t=(t-1/2)^2 +3/4

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| + (3/4)∫1/[(t-1/2)^2 + 3/4]dt

=>(-1/2)log|1+t| + (1/4)log|1+t^2 -t| +
(3/4)1/(2/sqrt(3) arctan[2(t-1/2)/sqrt(3)


substituting original values t=u^2

=>(-1/2)log|1+u^2| + (1/4)log|1+u^4 -u^2| +
(3/4)1/(2/sqrt(3) arctan[2(u^2-1/2)/sqrt(3)

substituting original values u=(tanx)^1/3

=>(-1/2)log|1+(tanx)^2/3| + (1/4)log|1+(tanx)^4/3 -(tanx)^2/3| +
(3/4)1/(2/sqrt(3) arctan[2((tanx)^2/3 - 1/2)/sqrt(3) + C

where arctan(x) means tan inverse x
results used=
1)∫1/(1+x)dx= log|1+x| + c
2)∫1/(a^2+x^2)dx=
(1/a)arctan(x/a) + C...........'a' is a constant


P.S. Extremely sorry for the length of the solution but it is correct and each and every explanation is given.

Answer 2

Use substitution. Been a long time since I've done this so you may want to look for the reference in your book to see if I'm right.
∫ tan ^(1/3) x dx
u = tan x
du = sec^2 x dx or [1 / cos^2 x] dx

∫ tan ^(1/3) x dx
∫ u^(1/3) du
1/4 u^(4/3) sec^2 x
1/4 tan x^(4/3) * sec^2 x

1*sin^(4/3) [over]
4 *cos^(4/3) * cos^2

sin^(4/3) / 4cos^(11/3) or
sin^(4/3) * (1/4)sec^(11/3) or
1/4 tan x^(4/3) * sec^2 x

Hope this helps.

Answer 3

Well, let me reduce this to integrating a rational function.
First, let u = tan x, x = arctan u, dx = du/(u² + 1).
This gives
int[ u^(1/3)/(u² +1)] du.
Next, let u = t^3, du = 3t^2 dt
and it becomes
int[ 3t³/(t^6 +1)]dt
To continue from here, use partial fractions.
The denominator factors as
(t² +1)(t^4 -t² +1).
I'll let you carry on from here.
Good luck!

Answer 4

∫tan^(⅓)x dx
= ∫tan²x.tan^(-5/3)x dx
= ∫(sec²x - 1).tan^(-5/3)x dx
= ∫tan^(-5/3)x sec²xdx -∫tan^(-5/3)x dx
= -3/2 tan^(-2/3)x -∫tan^(-5/3)x dx
ie I(⅓) = -3/2 tan^(-2/3)x - I(-5/3)
I{n/3} = 3/(n-3)tan^(n - 3/3)x - I{(n - 6)/3}

So ∫tan^(⅓)x dx = -3/2tan^(-2/3)x + 3/8tan^(-8/3)x - 3/14tan^(-14/3) + .... +(-1)^n 3/(6n - 4) tan^(-(6n - 4)/3)x - ...... +c

Answer 5

(-2*Sqrt[3]*ArcTan[Sqrt[3] - 2*Tan[x]^(1/3)] - 2*Sqrt[3]*ArcTan[Sqrt[3] + 2*Tan[x]^(1/3)] + Log[-1 + Sqrt[3]*Tan[x]^(1/3) - Tan[x]^(2/3)] - 2*Log[1 + Tan[x]^(2/3)] + Log[1 + Sqrt[3]*Tan[x]^(1/3) + Tan[x]^(2/3)])/4

the link below will solve it for you

Answer 6

(-2*Sqrt[3]*ArcTan[Sqrt[3] - 2*Tan[x]^(1/3)] - 2*Sqrt[3]*ArcTan[Sqrt[3] + 2*Tan[x]^(1/3)] + Log[-1 + Sqrt[3]*Tan[x]^(1/3) - Tan[x]^(2/3)] - 2*Log[1 + Tan[x]^(2/3)] + Log[1 + Sqrt[3]*Tan[x]^(1/3) + Tan[x]^(2/3)])/4

Answer 7

= (tanx)^(2/3)(d(tanx)/dx)/(2/3)
= 2[(tanx)^(2/3)][(secx)^2]/3
because ∫(f(x))^ndx = [f(x)^(n+1)]{df(x)/dx}/(n+1)

Answer 8

wish i knew sorry