The question is- Find $\int\frac{\sec^xx}{\sqrt{\tan x}}dx$

What I've tried: $$\int\frac{\sec^{x-2}x\sec^2x}{\sqrt{\tan x}}dx$$

Let $\tan x = t$.


What should I do next? Any sort of help is appreciated.

  • $\begingroup$ it is $$\int\frac{\sec(x)}{\sqrt{\tan(x)}}dx?$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 16 '18 at 16:04
  • 4
    $\begingroup$ Are you sure that this should be $\sec ^{\color{red}{x}} x$? Perhaps you mean $\sec ^{\color{red}{n}} x$? How would you even define $\sec^x(x)$ in regions where $x \not\in\Bbb Z$, $\sec(x)<0$? $\endgroup$ – John Doe Mar 16 '18 at 16:05
  • $\begingroup$ I have checked and my question is correct. $\endgroup$ – Abhishekstudent Mar 16 '18 at 16:11
  • $\begingroup$ I do not understand the reason for the down votes. My question is legit. $\endgroup$ – Abhishekstudent Mar 16 '18 at 16:12
  • $\begingroup$ It is not possible to evaluate this integral in terms of standard mathematical functions, according to wolfram alpha. Also your substitution doesn't make much sense, since you still have $x$'s in your integral, which should have been changed out for things in terms of $t$ (which would give an unpleasant $\arctan (t)$ in an exponent). Do you not even have any limits? Even then I don't see there being much hope. $\endgroup$ – John Doe Mar 16 '18 at 16:21

You can't. Integrands of the form $f^x(x)$ usually have no closed-form antiderivative, by the Liouville theorem.

The probability that your problem statement is wrong equals $1-\epsilon$.

And IMO the probability that the true question is

$$\int\frac{\sec^2x}{\sqrt{\tan x}}dx$$

equals $1+\epsilon$.

  • 2
    $\begingroup$ Wow! A probability greater than $1$ sounds interesting. $\endgroup$ – Mark Viola Mar 16 '18 at 16:36
  • $\begingroup$ @MarkViola: they call it "more than certainly". $\endgroup$ – Yves Daoust Mar 16 '18 at 16:37
  • $\begingroup$ Indeed. But are you certain about what "they call it?" $\endgroup$ – Mark Viola Mar 16 '18 at 16:38
  • $\begingroup$ Yves, in all seriousness, negative probabilities (and ones greater than 1) have applicability (See This Article). $\endgroup$ – Mark Viola Mar 16 '18 at 16:43
  • $\begingroup$ @MarkViola: less than impossibly ? $\endgroup$ – Yves Daoust Mar 16 '18 at 16:53

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