Problem with ln expression and integral

Question

For $\displaystyle\int{\frac{1}{x\ln x\sqrt{\ln^2x-1}}}dx$,

• I first applied a $u$-sub. $u = \ln x$ -> $\displaystyle du = \frac{1}{x}dx$
• the integral then become $\displaystyle\int{\frac{1}{u\sqrt{u^2-1}}}du$
• This is the directly integral formula or $arcsec(x)$, as a result, the result is $\text{arcsec}(u)+C$ -> $\text{arcsec}(\ln x)+C$

However, I used the online calculator, they found the result of $\arctan(\sqrt{\ln^2x-1})+C$. And I used the function "Check my answer", they say what I found is not correct!

Can anybody tell me where I made a mistake? I couldn't see the mistake even trying to take the derivate of $\text{arcsec}(\ln x)$!

Answer 1

Inverse trig functions can always be thought of as angles. Let $\theta = \mbox{arcsec } x.$ Then $\sec \theta = x.$ If you draw a triangle showing this it would have hypotenuse of length $x$ and adjacent leg of length $1$. So the opposite side is $\sqrt{x^2-1}$. Then

$$\tan \theta = \frac{\sqrt{x^2-1}}{1}$$

or

$$\theta = \arctan \sqrt{x^2-1}.$$

So your answer is correct. One trick, as long as you have your online calculator up and running, is that you could plot both answers together and see if they differ only by a constant.

Answer 2

Hint: $$ \tan \alpha=\sqrt{\sec^2 \alpha -1} $$

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this means that the two solutions seems to be the same, but there is difference in the domain of the two functions:

$arcsec(\log x)$ is defined for $\log x \ge 1$

$\arctan\sqrt{\log^2 x -1}$ is defined for $\log^2 x-1 \ge 0$

since your starting function is defined for $\log^2 x-1 > 0$, the correct integral, on the almost full domain of the starting function is the second one.