$\int \sqrt[3]{\tan{x}} \, dx$

Note by Monojit Kamilya
4 years, 11 months ago

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Make the substitution $\tan x = t^{3/2}$

I believe we then need to just find out $\int \frac{dt}{1+t^3}$

which should be doable by partial fraction methods.

- 4 years, 11 months ago

That's right. Note that $$1 + t^3 = (1+t)(1-t+t^2)$$

Also,

$$\frac{1}{3} \left(\frac{2-t}{t^2-t+1}+\frac{1}{t+1}\right)=\frac{1}{t^3+1}$$

The second part of the integral can now be solved by a simple substitution.

For the first part, use

$$\frac{2-t}{t^2-t+1}=-\frac{2 t-4}{2 \left(t^2-t+1\right)}=-\frac{1}{2} \left(\frac{2 t-1}{t^2-t+1}-\frac{3}{t^2-t+1}\right)$$

The first integral with $$\frac{2 t-1}{t^2-t+1}$$ can be solved by a simple substitution, $$u = 1 - t + t^2$$

For the second one, note that $$t^2-t+1=a+(t-b)^2$$ with $$a = \frac{3}{4}$$ and $$b = \frac{1}{2}$$

So,

$$\frac{1}{t^2-t+1}=\frac{1}{\left(t-\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{4}{3} \frac{1}{\left(\sqrt{\frac{4}{3}} \left(t-\frac{1}{2}\right)\right)^2+1}$$

Now make one final substitution here to arrive at the well-known $$\int \frac{1}{z^2+1} \, dz$$ . Your final solution looks like this:

$$\frac{1}{4} \left(-2 \sqrt{3} \tan ^{-1}\left(2 \sqrt[3]{\tan (x)}+\sqrt{3}\right)-2 \sqrt{3} \tan ^{-1}\left(\sqrt{3}-2 \sqrt[3]{\tan (x)}\right)-2 \log \left(\tan ^{\frac{2}{3}}(x)+1\right)+\log \left(\tan ^{\frac{2}{3}}(x)+\sqrt{3} \sqrt[3]{\tan (x)}+1\right)+\log \left(\tan ^{\frac{2}{3}}(x)-\sqrt{3} \sqrt[3]{\tan (x)}+1\right)\right)$$

Phew.

post not guaranteed to be error free

- 4 years, 11 months ago