The integral is $$\int\frac{\arcsin{\sqrt{x}}-\arccos{\sqrt{x}}}{\arcsin{\sqrt{x}}+\arccos{\sqrt{x}}}\cdot dx$$

What I did was to sum up the denominator to equal $\frac\pi2$ and then applied integration by parts on the remaining $\arccos{\sqrt{x}}$ and $\arcsin{\sqrt{x}}$ integrals but this process was very lengthy.

Can someone suggest a shorter way?


$\arcsin{\sqrt{x}}+\arccos{\sqrt{x}}=\frac{\pi}{2}$ and $$\int\arcsin{\sqrt{x}}=x\arcsin{\sqrt{x}}-\int\frac{x}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}dx$$ Can you end it now?

  • $\begingroup$ That's what I did as I've already mentioned. Is this the only feasible way? $\endgroup$ – infinite-blank- Jul 10 at 4:13
  • 1
    $\begingroup$ @infinite-blank- I think, this is the way. Use the substitution $\sqrt{\frac{x}{1-x}}=t$. $\endgroup$ – Michael Rozenberg Jul 10 at 4:15
  • $\begingroup$ No I meant as in- is applying integration by parts the only possible way? $\endgroup$ – infinite-blank- Jul 10 at 4:17
  • 1
    $\begingroup$ Yes, I think it's an unique way here. By the way, $\int\arccos\sqrt{x}dx=\frac{\pi}{2}x-\int\arcsin\sqrt{x}dx.$ $\endgroup$ – Michael Rozenberg Jul 10 at 4:18
  • $\begingroup$ Alternatively $$\arcsin\sqrt x=y\implies x=\sin^2y$$ $\endgroup$ – lab bhattacharjee Jul 10 at 4:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.