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?

  • $\begingroup$ after that i'm doing integration by part then i got stuck. look i edit again. $\endgroup$ – Siddhant Trivedi Jul 22 '12 at 13:08
  • $\begingroup$ NCERT problem @SiddhantTrivedi $\endgroup$ – Hyperbola Jul 22 '12 at 14:39
  • $\begingroup$ One way to handle $\int(\sqrt x/\sqrt{1-x})\,dx$: multiply top and bottom by $\sqrt{1-x}$ and manipulate to $(1/2)\int(\sqrt{1-(2x-1)^2}/(1-x))\,dx$; substitute $2x-1=\sin u$ and use trig identities to bring it to the form $(1/2)\int(1+\sin u)\,du$. $\endgroup$ – Gerry Myerson Jul 22 '12 at 23:31
  • $\begingroup$ Rather than multiplying top and bottom by $\sqrt{1-x}$ I would prefer multiplying by $\sqrt{x}$ to get $\int \dfrac{x}{\sqrt{x-x^2}}dx$ which is a standard form in our textbook (NCERT). $\endgroup$ – Chaitanya Tappu Oct 6 '13 at 14:07
  • $\begingroup$ Have you tried the Weierstrass trig substitutions? $\endgroup$ – CivilSigma May 16 '15 at 17:30

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 $$ I_1 = 2zcos^{-1}z+ \int \frac{z^2}{\sqrt{1-z^2}}dz$$ $$I_1 = 2zcos^{-1}z+ \int \frac{1}{\sqrt{1-z^2}}dz-\int\sqrt{1-z^2}dz$$ $$\int \frac{1}{\sqrt{1-z^2}}dz=-\cos^{-1}z$$ $$\int\sqrt{1-z^2}dz=\frac{z\sqrt{1-z^2}}{2}+\frac{1}{2}\sin^{-1}z$$

  • 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.

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. )

  • $\begingroup$ but i can't get easily ans. in this i was tried this way. $\endgroup$ – Siddhant Trivedi Jul 22 '12 at 12:51
  • 1
    $\begingroup$ You tried that substitution? It's not clear to me where you got stuck. $\endgroup$ – Gerry Myerson Jul 22 '12 at 12:57
  • $\begingroup$ i edit my que. where i got stuck.check it.help me $\endgroup$ – Siddhant Trivedi Jul 22 '12 at 13:15

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