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

Note by Monojit Kamilya 4 years 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. – Peiyush Jain · 4 years ago

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@Peiyush Jain – 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 – Ivan Stošić · 4 years ago

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## Comments

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TopNewestMake 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. – Peiyush Jain · 4 years ago

Log in to reply

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– Ivan Stošić · 4 years agoLog in to reply