# How to evaluate this integral $\int\frac{\arcsin{\sqrt{x}}-\arccos{\sqrt{x}}}{\arcsin{\sqrt{x}}+\arccos{\sqrt{x}}}\cdot dx$?

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?

## 1 Answer

$$\arcsin{\sqrt{x}}+\arccos{\sqrt{x}}=\frac{\pi}{2}$$ 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?

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