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How to evaluate

$ \int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $

I know that $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ but after that I have no idea, so please help me. Thanks in advance.

I tried this way,

$ \int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\frac{\pi}{2}}dx $ then I put the value $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ so $ \frac{2}{\pi}\int\left(\sin^{-1} \sqrt{x} -\left(\frac{\pi}{2}-\sin^{-1} \sqrt{x}\right)\right)dx $ Is this right?

after that I integrate by part and get,

$ \int \frac{\sqrt{x}}{\sqrt{1-x}}$ now,what can i do?

  • 0
    Have you tried the Weierstrass trig substitutions?2015-05-16

3 Answers 3

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Let $ I_0=\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $$\Rightarrow I_0=\int\frac{\frac{\pi}{2}-2\cos^{-1}\sqrt{x}}{\frac{\pi}{2}}dx$$\Rightarrow I_0=\int (1-\frac{4}{\pi}\cos^{-1}\sqrt{x})dx$ $\Rightarrow I_0=x-\frac{4}{\pi}\int\cos^{-1}\sqrt{x})dx$ Now Consider $I_1= \int\cos^{-1}\sqrt{x}dx$ $\Rightarrow I_1=\int 2z\cos^{-1} zdz$ Where $ x=z^2$ Hence Integrating by parts we get $ I_1 = 2zcos^{-1}z+ \int \frac{z^2}{\sqrt{1-z^2}}dz$ $I_1 = 2zcos^{-1}z+ \int \frac{1}{\sqrt{1-z^2}}dz-\int\sqrt{1-z^2}dz$ $\int \frac{1}{\sqrt{1-z^2}}dz=-\cos^{-1}z$ $\int\sqrt{1-z^2}dz=\frac{z\sqrt{1-z^2}}{2}+\frac{1}{2}\sin^{-1}z$

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  • Using the relation $\arcsin(\sqrt{x}) + \arccos(\sqrt{x}) = \frac{\pi}{2}$ (which is valid for $0 \leqslant x \leqslant 1$, so this must an implicit assumption in your problem) solve for $\arccos(\sqrt{x})$ and substitute that into the integrand.
  • After that make a $u$-substitution $u = \arcsin(\sqrt{x})$. This should lead to $\int \left( \frac{4 u}{\pi} - 1\right) \sin(2u) \mathrm{d} u$. This can be integrated by parts.
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Given what you know, you should be able to get the answer if you can get $\int\arcsin\sqrt x\,dx$ and you can get that starting with the substitution $u=\arcsin\sqrt x$ ($\sin u=\sqrt x$, $x=\sin^2u$, $dx=2\sin u\cos u$, etc. )

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    i edit my que. where i got stuck.check it.help me2012-07-22