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$\int_0^\frac{1}{\sqrt2}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$

I tried using substitute, $x=\sin\theta$

But I ended up with $\int_0^\frac{\pi}{4}\frac{d\theta}{1+\sin^2\theta}$

Is my substitution correct? Please give me a hint to work this out! Thank you very much.

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hint: $$\frac{1}{1+\sin^2 x}=\frac{\sec^2 x}{1+2\tan^2 x}$$

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We see that $$\frac1{1+\sin(x)^2}=\frac{\sec(x)^2}{1+2\tan(x)^2}$$ as was hinted by @E.H.E. We then preform the sub $u=\tan(x)$ to get $$\int_0^1 \frac{du}{1+2u^2}$$ Then we may in fact compute the integral $$I(x;a,b,c)=\int\frac{ dx}{ax^2+bx+c}=\int\frac{dx}{a(x+\frac{b}{2a})^2+g}$$ Here $g=c-\frac{b^2}{4a}$. If we assume that $4ac>b^2$, then we may make the substitution $x+\frac{b}{2a}=\sqrt{\frac{g}{a}}\tan u$ which gives $$I(x;a,b,c)=\sqrt{\frac{g}{a}}\int\frac{\sec^2u\, du}{g\tan^2u+g}$$ $$I(x;a,b,c)=\frac{u}{\sqrt{ag}}$$ $$I(x;a,b,c)=\frac2{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}+C$$ And by noting that your integral is given by $I(1;2,0,1)-I(0;2,0,1)$ we have your integral at the value $$\frac{\arctan\sqrt2}{\sqrt2}\approx 0.675510858856$$

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  • $\begingroup$ Why was the hint @EHE gave not enough? Why do the calculation for the OP? $\endgroup$ – JavaMan Apr 1 at 2:43
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    $\begingroup$ @JavaMan Because it was an opportunity to demonstrate something more than the OP asked, namely $I(x;a,b,c)$. If I were a beginning calc. student I would find it very helpful, even interesting, to see such an answer. IMO, a good answer helps you in more ways than $1$, and helps you connect concepts/ideas (and in this case integrals). $\endgroup$ – clathratus Apr 1 at 3:31
  • $\begingroup$ Of course I found it very useful. Using a second substitution is somewhat challenging task. Thank you for pointing it out! $\endgroup$ – emil Apr 1 at 15:26
  • $\begingroup$ @emil You are very welcome :) $\endgroup$ – clathratus Apr 1 at 15:45
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Hint

Divide numerator & denominator by $\sin^2\theta$

and set $\cot\theta =u$

Or divide numerator & denominator by $\cos^2\theta$

and set $\tan\theta=v$

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