Question

How to integrate integral cube root tanx.

Answer 1

\begin{array}{l} \oint \tan^{1/3} x \, dx

Take $\tan x = z$; $\sec^2 x \, dx = dz$ i.e. $dx = dz / (1 + z^2)$

$$\begin{array}{l} = \int z^{1/3} / (1 + z^2) \, dz \\ = \int 3t^3 / (1 + t^6) \, dt \\ = 3L \end{array}$$

Take $t^3 = z$; $dz = 3t^2 \, dt$

where, $L = \int t^3 / (1 + t^6) \, dt$

Now, by partial fractions,

$$\begin{array}{l} t^3 / (1 + t^6) = t^3 / \left[ (1 + t^2)(t^4 - t^2 + 1) \right] \\ = [Ax + B] / [1 + t^2] + [Cx + D] / [t^4 - t^2 + 1] \end{array}$$

{since, $(t^2 + 1) \& (t^4 - t^2 + 1)$ have no real solution}

Now, equivalently,

$$t^3 \equiv At^5 + Bt^4 + (C - A)t^3 + (D - B)t^2 + (C + A)t + (B + D)$$

Putting $t = 0$, in the above,

$$B + D = 0 \quad \text{(1)}$$

Putting $t = 1$,

$$A + 2C + D = 1 \quad \text{(2)}$$

Putting $t = -1$,

$$A + 2C - D = 1 \quad \text{(3)}$$

Using (2) & (3), $D = 0$

Therefore, $B = 0$ using (1)

$$A + 2C = 1 \quad \text{(4)}$$

So, Now,

$$t^3 \equiv At^5 + (C - A)t^3 + (C + A)t$$

Putting, $t = 2$,

$$13A + 5C = 4 \quad \text{(5)}$$

Solving, (4) & (5), $C = 3/7$ & $A = 1/7$

Now,

$$\begin{array}{l} L = \int t^3 / (1 + t^6) \, dt = 1/7 \int t / (1 + t^2) \, dt + 3/7 \int t / (t^4 - t^2 + 1) \, dt \\ = 1/14 \cdot \log(1 + t^2) + C + 3/7K \end{array}$$

where $C$ is an arbitrary integration constant and $K = \int t / (t^4 - t^2 + 1) \, dt$

$$\begin{array}{l} K = \int t / (t^4 - t^2 + 1) \, dt = \int t / \left[ (t^2 - 1/2)^2 + (\sqrt{3} / 2)^2 \right] \, dt \\ \text{Take, } t^2 - 1/2 = m; \, 2t \, dt = dm \\ K = 1/2 \int dm / \left[ m^2 + (\sqrt{3} / 2)^2 \right] = 1/\sqrt{3} \tan^{-1}(2m / \sqrt{3}) + C_1 \\ = 1/\sqrt{3} \tan^{-1}(2(t^2 - 1/2) / \sqrt{3}) + C_1 \end{array}$$

Therefore,

$$L = 1/14 \cdot \log(1 + t^2) + C + \sqrt{3/7} \tan^{-1}(2(t^2 - 1/2) / \sqrt{3}) + C_1$$

Therefore,

$$\int \tan^{1/3} x \, dx = 3L$$ $$= 3 \left[ 1 / 14 \cdot \log \left(1 + t^{2}\right) + C + \sqrt{3/7} \tan^{-1} \left(2 \left(t^{2} - 1/2\right) / \sqrt{3}\right) + C_{1} \right],$$

where $t = \tan^{1/3} x$.