tanx-1-3-dx-

Question Number 20686 by NECx last updated on 31/Aug/17

\int {(t a n x)}^{1 / 3} d x

Commented by NECx last updated on 01/Sep/17

p l e a s e h e l p

Answered by ajfour last updated on 01/Sep/17

l e t \tan x = \frac{1}{t^{3}} \Rightarrow \sec^{2} x d x = \frac{- 3 d t}{t^{4}}

I = \int {(\tan x)}^{1 / 3} d x = \int \frac{1}{t} \times \frac{- 3 d t}{t^{4} (1 + \frac{1}{t^{6}})}

= - 3 \int \frac{t d t}{1 + t^{6}} = - \frac{3}{2} \int \frac{2 t d t}{1 + {(t^{2})}^{3}}

l e t t^{2} = z

I = - \frac{3}{2} \int \frac{d z}{1 + z^{3}} = - \frac{3}{2} \int \frac{d z}{(1 + z) (z^{2} - z + 1)}

= - \frac{3}{2} \int [\frac{1 / 3}{1 + z} + \frac{- 1 / 3 x + 2 / 3}{z^{2} - z + 1}] d z

= - \frac{1}{2} \ln ∣ 1 + z ∣ + \frac{1}{4} \int \frac{2 z - 4}{z^{2} - z + 1} d z

= - \frac{1}{2} \ln ∣ 1 + z ∣ + \frac{1}{4} \int \frac{2 z - 1}{z^{2} - z + 1} d z

- \frac{3}{4} \int \frac{d z}{{(z - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}

= - \frac{1}{2} \ln ∣ 1 + z ∣ + \frac{1}{4} \ln ∣ z^{2} - z + 1 ∣

- \frac{3}{4} (\frac{2}{\sqrt{3}}) \tan^{- 1} (\frac{2 z - 1}{\sqrt{3}}) + C

w h e r e z = t^{2} = {(\tan x)}^{- 2 / 3} .

Answered by $@ty@m last updated on 01/Sep/17

A l t e r n a t i v e m e t h o d :

L e t t a n x = w^{\frac{3}{2}} \Rightarrow {s e c}^{2} x d x = \frac{3}{2} \sqrt{w} d w

\Rightarrow d x = \frac{3}{2} . \frac{\sqrt{w} d w}{1 + w^{3}}

∴ \int {(t a n x)}^{1 / 3} d x = \frac{3}{2} \int \frac{w}{1 + w^{3}} d w

= \frac{3}{2} \int \frac{w}{(1 + w) (1 - w + w^{2})} d w

L e t \frac{w}{(1 + w) (1 - w + w^{2})} = \frac{A}{1 + w} + \frac{B w + C}{1 - w + w^{2}}

\Rightarrow A (1 - w + w^{2}) + (B w + C) (1 + w) \equiv w

\Rightarrow A + C = 0, - A + B + C = 1, A + B = 0

\Rightarrow B = C = \frac{1}{3}, A = \frac{- 1}{3}

n o w p r o c e e d s i m i l a r t o a b o v e s o l u t i o n .

Commented by NECx last updated on 01/Sep/17

w o w \dots . i r e a l l y a p p r e c i a t e t h i s .

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