Question

integration of cube root of tanx

Answer 1

\int (\tan x)^{1/3}dx=\frac{(\tan x)^{4/3}}{2};_2F_1\Bigl(\frac{2}{3},1;\frac{5}{3};-\tan^2x\Bigr)+C.

Answer 2

We will show one acceptable solution. One may begin by writing

$$I=\int (\tan x)^{1/3}\,dx.$$

A useful substitution is to let

$$t=\sqrt[3]{\tan x}\quad\Longrightarrow\quad \tan x=t^3.$$

Differentiate with respect to $x$:

$$\frac{d}{dx}(t^3)=3t^2\frac{dt}{dx}=\sec^2x\quad\Longrightarrow\quad dx=\frac{3t^2}{\sec^2x}\,dt.$$

Also note that

$$(\tan x)^{1/3}=t\quad \text{and}\quad \sec^2x=1+\tan^2x=1+t^6.$$

Thus the integral becomes

$$I=\int t\,dx=\int t\cdot\frac{3t^2}{1+t^6}\,dt=3\int\frac{t^3}{1+t^6}\,dt.$$

Now, we recognize the standard integral

$$\int \frac{x^{m-1}}{1+x^n}\,dx=\frac{x^m}{n}\;_2F_1\Bigl(\frac{m}{n},1;1+\frac{m}{n};-x^n\Bigr)+C,\quad \text{provided }m>0,$$

with $m-1=3$ (so $m=4$) and $n=6$.

Thus

$$\int\frac{t^3}{1+t^6}\,dt=\frac{t^4}{6}\;_2F_1\Bigl(\frac{4}{6},1;1+\frac{4}{6};-t^6\Bigr)+C =\frac{t^4}{6}\;_2F_1\Bigl(\frac{2}{3},1;\frac{5}{3};-t^6\Bigr)+C.$$

Returning to our expression for $I$, we have

$$I=3\cdot \frac{t^4}{6}\;_2F_1\Bigl(\frac{2}{3},1;\frac{5}{3};-t^6\Bigr)+C =\frac{t^4}{2}\;_2F_1\Bigl(\frac{2}{3},1;\frac{5}{3};-t^6\Bigr)+C.$$

Finally, substituting back $t=\sqrt[3]{\tan x}$ with $t^4=(\tan x)^{4/3}$ and $t^6=\tan^2x$, we obtain

$$\int (\tan x)^{1/3}dx=\frac{(\tan x)^{4/3}}{2}\;_2F_1\Bigl(\frac{2}{3},1;\frac{5}{3};-\tan^2x\Bigr)+C.$$

Minimal (core) explanation:

1. Substitute $t=\sqrt[3]{\tan x}$ so that $\tan x=t^3$ and $dx=\frac{3t^2}{1+t^6}\,dt$.
2. Then the integral becomes $3\int \frac{t^3}{1+t^6}\,dt$.
3. Use the formula $$\int \frac{x^{m-1}}{1+x^n}\,dx=\frac{x^m}{n}\;_2F_1\Bigl(\frac{m}{n},1;1+\frac{m}{n};-x^n\Bigr)+C,$$ with $m=4,\; n=6$.
4. Back-substitute $t=\sqrt[3]{\tan x}$ to get the final answer.