3
$\begingroup$

$ \int _{0}^1 \frac{1}{{\sqrt{1+x^2}}(1+x^2)} dx $

The answer is $\frac{\sqrt{2}}{2}$, we've tried many ways to find it, and we wish to NOT use sech.. so looking for another way to do it like with substitution method or by parts.. we've also noticed that $\frac{1}{\sqrt{1+x^2}}$ is the derivative of $arcsinh(x)$, but we haven't found how to use it properly with the "by parts" method

Thanks

  • 2
    Hint: $x=tg(u)$ – 2017-01-05

1 Answers 1

5

With $x=\tan t$, you have $dx=(1+\tan^2t)\,dt$, so the integral becomes $$ \int_0^{\pi/4}\frac{1}{\sqrt{1+\tan^2t}(1+\tan^2t)}(1+\tan^2t)\,dt= \int_0^{\pi/4}\cos t\,dt $$

Alternatively, with $x=\sinh t$, we have $$ \int_0^{\operatorname{arsinh}1} \frac{1}{\cosh^2t}\,dt=\Bigl[\tanh t\Bigr]_0^{\operatorname{arsinh}1} $$

  • 0
    Damn.. you made my day dude, thank you!! – 2017-01-05
  • 0
    @JohnAdams Finding the right substitution is also a question of being lucky, sometimes. `;-)` – 2017-01-05
  • 0
    yea... I spent like 2 hours on this to find a legit way and was going deeper and deeper on the wrong path.. – 2017-01-05