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I am trying to evaluate

$$ \int \frac{\tan^{3/2}\left(x\right)} {1 - \sin\left(x\right)}\,dx \label{1}\tag{1} $$


I tried using Weierstrass substitution. > **The Weierstrass substitution**, ( named after K.Weierstrass $\left(~1815~\right)$ ), is a substitution used in order to convert trigonometric functions rational expressions to polynomial rational expressions. Integrals of this type are usually easier to evaluate.

This substitution is constructed by letting: $$t = \tan\left(\frac{x}{2}\right) \iff x = 2\arctan(t) \iff dx = \frac{2}{t^2+1}$$

Using basic trigonometric identities it is easy to prove that: $$\cos x = \dfrac{1 - t^2}{1 + t^2}$$

$$\sin x = \dfrac{2t}{1 + t^2}$$

Using this substitution we end up to this integral:

$$ 2 \int \frac{(2t)^{\frac{3}{2}}(1+t^2)}{(1-t^2)^{\frac{3}{2}}(t^2-2t+1)}\,dt$$

Which is clearly not easier to evaluate than $(1)$.

I also tried other standard trigonometric substitutions such as $u = \cos(x)$, $u = \sin(x)$, $u=\tan(x)$ with no better luck.

At last I can't see any trigonometric identities that could simplify the fraction.

Any ideas on how to evaluate this integral?

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    $\begingroup$ I am afraid that the result could be an hypergeometric function. $\endgroup$ Jul 20 '20 at 10:55
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    $\begingroup$ Yes, we'll have to calculate $\int (v^2-1)^{3/4} \,dv$ somewhere down the line. Which, according to WA can be expressed in terms of hypergeometric functions. $\endgroup$
    – Nikunj
    Jul 20 '20 at 11:33
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We can show fairly straightforwardly that this reduces to an elliptic integral, which cannot be an elementary function: put $ x = \arctan(u^2) $. Then $ dx = 2u/(1+u^4) \, du $, $\tan x = u^2$ and $\sin x = u^2/\sqrt{1+u^4}$, and rationalising implies that the integral becomes $$ \int \bigg( 2u^4 + \frac{2u^6}{\sqrt{1+u^4}} \bigg) \, du , $$ and we just need to worry about the second term. It so happens that this was one of the earliest integrals Liouville considered when he became interested in when an integral is algebraic (See Lützen's Joseph Liouville 1809–1882 pp. 374ff. for the details). An integration by parts reduces us to $ \int \frac{u^2}{\sqrt{1+u^4}} \, du $, which is known to not be elementary (see either Liouville's work, or Ritt's book Integration in finite terms). Thus the "elementary part" is $$ \frac{2}{5} ( u^5 + u^3 \sqrt{1+u^4}) , $$ while the non-elementary part is the elliptic integral $$ - \frac{6}{5} \int \frac{u^2}{\sqrt{1+u^4}} \, du = \frac{6}{5}\sqrt{i} ( F(\arcsin(\sqrt{i}u) \mid -1) - F(\arcsin(\sqrt{i}u) \mid -1) . $$ One could write in terms of $x$ again, but there doesn't seem much point.

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  • $\begingroup$ It is sure that you provided the best substitution. Howver, I have the feeling that you may have a few typo's. Cheers and thanks for your answes and $\to +1$. $\endgroup$ Jul 21 '20 at 2:43
  • $\begingroup$ I think our answers might actually be the same: the differences lie in the arcsin vs arg sinh expressions. I may have a sign error somewhere, though. $\endgroup$
    – Chappers
    Jul 21 '20 at 2:47
  • $\begingroup$ It not my answer; it is yours ! You are missing a factor $2$ (typo for sure when looking at the end results). $dx$ is not correct. Cheers :-) $\endgroup$ Jul 21 '20 at 2:57
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First, a small historical precision : most books call the tangent half-angle substitution the Weierstrass $(1815-1897)$ substitution while in fact the technique appears in Euler's work $-1707-1783).

Concerning the possible substitutions, what @Chappers proposed is probably the best since leading to a real closed form expression in terms of elliptic integrals.

Just to continue with what I early wrote in comments, letting $x=\sin ^{-1}(u)$, we end with $$I=\int \frac{\tan^{\frac{3}{2}}(x)}{1-\sin(x)} \,dx=\int \frac{u^{3/2}}{(1-u)^{9/4} \,(u+1)^{5/4}}\,du$$ $$I=\frac{2 u^{3/2} (3 u-2)}{5 (1-u)^{5/4}\, (u+1)^{1/4}}+\frac{4}{5} u^{3/2} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{7}{4};u^2\right)$$ which I have not been able to simplify further.

Remarks

Looking again at @Chappers's answer, I have the filling that there are a few minot typo's.

$$x= \tan ^{-1}\left(u^2\right)\implies dx=\frac{2 u}{u^4+1}\,du \qquad \text{and} \quad \sin(x)=\frac{u^2}{\sqrt{u^4+1}}$$ making $$I =2\int \left(u^4+\frac{u^6}{\sqrt{u^4+1}}\right)\,du$$ making $$I=\frac{2}{5} u^3 \left(u^2+\sqrt{u^4+1}\right)+$$ $$\frac{6}{5} (-1)^{3/4} \left(E\left(\left.i \sinh ^{-1}\left((-1)^{1/4} u\right)\right|-1\right)-F\left(\left.i \sinh ^{-1}\left((-1)^{1/4} u\right)\right|-1\right)\right)$$

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  • $\begingroup$ While both answers are very informative, I accepted @Chappers because he answered earlier. Thanks to both of you! $\endgroup$
    – Dimitris
    Jul 21 '20 at 3:07
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    $\begingroup$ @Veriun. I should have been very unhappy to see you not accepting Chappers's answer (which is really great !). $\endgroup$ Jul 21 '20 at 3:10
  • $\begingroup$ @Claude Leibovici: Do you know examples of the use of this change of variable in Euler's writings? $\endgroup$
    – FDP
    Jul 21 '20 at 7:47

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