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How I can evaluate the following integrals? I tried too much but no luck.

First one:

$$\int \frac{\sqrt{\tan(x)}}{\cos(x)} dx$$

Second one:

$$\int \frac{x^2 \arcsin(x)}{1+x^6} dx$$

My try:

For first one, I tried to multiply by $\tan(x)/\tan(x)$, I got $\int \frac{\sqrt{\tan(x)}}{\tan(x)}\sec^2(x).\sin(x) dx$. I used by parts with differentiating the $\sin(x)$, I got stuck in integrating $\int \cos(x) \ln(\tan(x))dx$.

Second, nothing good.

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  • $\begingroup$ Both this indefinite integrals seem to be terrible. The first one has a solution with elliptic integrals of the first and second one, but the second one is a riddle (for me, I mean), eventhough that it has a very easy factor: $\;\int\frac{x^2}{1+x^6}dx=\frac13\arctan x^3+C\;$ . What are these things for? Where did you get them? Are you sure of the functions in the integrals? $\endgroup$ – Timbuc Mar 18 '15 at 12:57
  • $\begingroup$ In fact, one of my friend asked me to solve them for him, I doubt about them if there is a solution of any of them, but I assumed I am in wrong way ... $\endgroup$ – Leonardo Mar 18 '15 at 16:45
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    $\begingroup$ I see: change friends (just joking). $\endgroup$ – Timbuc Mar 18 '15 at 16:47
  • $\begingroup$ By the way, if you want, write \mathrm dx to generate $\mathrm dx$ as opposed to $dx$. $\endgroup$ – Mr Pie Apr 25 '18 at 5:54
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Both of them are a dead end. The latter is equivalent to evaluating $\displaystyle\int\dfrac{\arctan x^3}{\sqrt{1-x^2}}~dx$. The former either was meant to have $\cos^2x$ in the denominator, or lacks some limits, like $0$ and $\dfrac\pi4$ , for which the answer is $\sqrt2-\dfrac{2\pi\sqrt\pi}{\Gamma\bigg(\dfrac14\bigg)^2}$.

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  • $\begingroup$ Also, $\displaystyle\int_0^1\dfrac{\arctan x^3}{\sqrt{1-x^2}}~dx ~=~ \dfrac23\cdot~_5F_4\bigg(\bigg\{\dfrac12~,~\dfrac23~,~\dfrac43~,~1~,~1\bigg\}~,~\bigg\{\dfrac32~,~\dfrac32~,~\dfrac56~,~\dfrac76\bigg\}~,~-1\bigg).$ $\endgroup$ – Lucian Mar 18 '15 at 13:16
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$\bf{My\; Solution::}$ For $\displaystyle \int\frac{\sqrt{\tan x}}{\cos x}dx$

Let $\displaystyle \tan x=t^2\;,$ Then $\displaystyle \sec^2 xdx = 2tdt\Rightarrow dx = \frac{2t}{\sec^2 x}dt$

So Integral Convert into $\displaystyle \int\frac{2t^2}{\sec^2 x\cdot \cos x}dt = 2\int t^2\cdot \cos xdx = 2\int t^2\cdot \frac{1}{\sqrt{1+t^4}}dt$

$\displaystyle =2\int t^2\cdot \left(1+t^4\right)^{-\frac{1}{2}}dt.$

Campare with $\displaystyle \int x^{m}\cdot (a+bx^n)^{p}dx$

Which is Integrable in elementry form If $\displaystyle p\;,\frac{m+1}{n}\;,\frac{m+1}{n}+p$ is an Integer.

So Here $\displaystyle m=2\;,n=4\;,p=-\frac{1}{2}.$

So Here $\displaystyle p\notin \mathbb{Z}$ and $\displaystyle \frac{m+1}{n}\notin \mathbb{Z}$ and $\displaystyle \frac{m+1}{n}+p\notin \mathbb{Z}$.

So We can Not express Given Integral $\displaystyle \int\frac{\sqrt{\tan x}}{\cos x}dx = 2\int t^2\cdot \left(1+t^4\right)^{-\frac{1}{2}}dt$ in elementry form.

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  • $\begingroup$ Do $p, \frac{m+1}{n}, \frac{m+1}{n}+p $ all have to be integers or does only one of them have to be an integer? $\endgroup$ – 5o3x Mar 18 '15 at 13:17
  • $\begingroup$ To 503x anyone is integer. $\endgroup$ – juantheron Mar 18 '15 at 13:18
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    $\begingroup$ where I can find the proof of the $p$ argument about all the numbers should be integer $\endgroup$ – Leonardo Mar 18 '15 at 16:43

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