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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?

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after that i'm doing integration by part then i got stuck. look i edit again. –  Siddhant Trivedi Jul 22 '12 at 13:08
    
NCERT problem @SiddhantTrivedi –  Hyperbola Jul 22 '12 at 14:39
    
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 $(1/2)\int(1+\sin u)\,du$. –  Gerry Myerson Jul 22 '12 at 23:31
    
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). –  cmtappu96 Oct 6 '13 at 14:07
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3 Answers 3

up vote 3 down vote accepted

Let $$ I_0=\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $$ $$ \Rightarrow I_0=\int\frac{\frac{\pi}{2}-2\cos^{-1}\sqrt{x}}{\frac{\pi}{2}}dx $$ $$ \Rightarrow I_0=\int \left(1-\frac{4}{\pi}\cos^{-1}\sqrt{x}\right)dx $$ $$ \Rightarrow I_0=x-\frac{4}{\pi}\int\cos^{-1}\sqrt{x}dx $$ 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
$$ I_1 = 2z\cos^{-1}z+ \int \frac{z^2}{\sqrt{1-z^2}}dz $$ $$I_1 = 2z\cos^{-1}z+ \int \frac{1}{\sqrt{1-z^2}}dz-\int\sqrt{1-z^2}dz$$ $$ \int \frac{1}{\sqrt{1-z^2}}dz=-\cos^{-1}z$$ $$\int\sqrt{1-z^2}dz=\frac{z\sqrt{1-z^2}}{2}+\frac{1}{2}\sin^{-1}z $$

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