Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
after that i'm doing integration by part then i got stuck. look i edit again. –  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
add comment

3 Answers

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

share|improve this answer
add comment

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. )

share|improve this answer
but i can't get easily ans. in this i was tried this way. –  Trivedi Jul 22 '12 at 12:51
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 –  Trivedi Jul 22 '12 at 13:15
add comment
  • 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.
share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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