# How do I solve this integration problem?

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 $\tan x = t$.

$$\int\frac{\sec^{x-2}x}{\sqrt{t\,}}dt$$

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

• it is $$\int\frac{\sec(x)}{\sqrt{\tan(x)}}dx?$$ – Dr. Sonnhard Graubner Mar 16 '18 at 16:04
• 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 $x \not\in\Bbb Z$, $\sec(x)<0$? – John Doe Mar 16 '18 at 16:05
• I have checked and my question is correct. – Abhishekstudent Mar 16 '18 at 16:11
• I do not understand the reason for the down votes. My question is legit. – Abhishekstudent Mar 16 '18 at 16:12
• 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. – 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 $1+\epsilon$.

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