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Integrate $\int_0^1 \sin^{-1}{\frac{x^2}{1+x^2}}dx$

I tried to put $x=\tan\theta$, which gives $\int_0^{\frac{\pi}{4}} {\sin^{-1}({\sin^2\theta}})\sec^2\theta d\theta$, but I don't know how to proceed after this. Is there something I am missing here?

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  • $\begingroup$ Did you miss a factor $\sec^2 (\theta)$? $\endgroup$ – xbh Aug 11 '18 at 8:29
  • $\begingroup$ @xbh yeah corrected that. $\endgroup$ – Happy Mittal Aug 11 '18 at 8:36
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    $\begingroup$ @HappyMittal Sorry, but should it be times $\sec^2(\theta)$ rather than divides it? $\endgroup$ – xbh Aug 11 '18 at 8:38
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    $\begingroup$ Again, why downvote this? This question seems to be hard to start. It is possible to be clueless about this. $\endgroup$ – xbh Aug 11 '18 at 8:44
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    $\begingroup$ Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. I've edited the question in now. $\endgroup$ – Simply Beautiful Art Aug 12 '18 at 23:14
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$$\int_{0}^{1}\arcsin\left(\frac{x^2}{1+x^2}\right)\,dx = \int_{0}^{1}\arcsin\left(\frac{x}{1+x}\right)\frac{dx}{2\sqrt{x}} $$ equals $$ \int_{0}^{1/2}\frac{\arcsin(u)}{2u^{1/2}(1-u)^{3/2}}\,du $$ which (pretty incredibly) can be managed by integration by parts. It boils down to $$ \frac{\pi}{6}-\int_{0}^{1/2}\sqrt{\frac{u}{1-u}}\cdot\frac{du}{\sqrt{1-u^2}}\,du $$ then to $$ \color{red}{\frac{\pi}{6}+\log(2+\sqrt{3})-\sqrt{2}\log(\sqrt{2}+\sqrt{3})}\approx 0.219563.$$

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  • $\begingroup$ How did you compute the last integral? $\endgroup$ – Szeto Aug 11 '18 at 9:19
  • $\begingroup$ @Szeto: by letting $u=\sinh^2 v$. $\endgroup$ – Jack D'Aurizio Aug 11 '18 at 9:25
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Hint:

You might know the property that $$\int_a^b f(x) dx+\int_{f(a)}^{f(b)} f^{-1} (x) dx=bf(b)-af(a)$$

Using this we get $$\int_0^1 \arcsin \left( \frac {x^2}{1+x^2}\right) dx+\int_0^{\frac {\pi}{6}} \sqrt {\frac {\sin x}{1-\sin x}} dx=\frac {\pi}{6}$$

Therefore $$\int_0^1 \arcsin \left( \frac {x^2}{1+x^2}\right) dx=\frac {\pi}{6}- \int_0^{\frac {\pi}{6}} \sqrt {\frac {\sin x}{1-\sin x}} dx$$

Now for the integral on right hand side use the substitution $u=\sin x$ which will lead you to a very simple integral $$\int_0^{\frac 12} \sqrt {\frac {u}{1-u}} \frac {du}{\sqrt {1-u^2}}$$ Which can be solved as stated by Jack D'Aurizio

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integration by parts $${\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}$$

$$f=\arcsin\left(\dfrac{x^2}{x^2+1}\right) \space and \space g'=1 $$

$$f'=\dfrac{\frac{2x}{x^2+1}-\frac{2x^3}{\left(x^2+1\right)^2}}{\sqrt{1-\frac{x^4}{\left(x^2+1\right)^2}}}$$ $$g=x$$

$$I=x\arcsin\left(\dfrac{x^2}{x^2+1}\right)-{\displaystyle\int}\dfrac{x\left(\frac{2x}{x^2+1}-\frac{2x^3}{\left(x^2+1\right)^2}\right)}{\sqrt{1-\frac{x^4}{\left(x^2+1\right)^2}}}\,\mathrm{d}x$$

$$I_1={\displaystyle\int}\dfrac{x\left(\frac{2x}{x^2+1}-\frac{2x^3}{\left(x^2+1\right)^2}\right)}{\sqrt{1-\frac{x^4}{\left(x^2+1\right)^2}}}\,\mathrm{d}x$$

$$I_1=2{\displaystyle\int}\dfrac{x\left(x\left(x^2+1\right)-x^3\right)}{\left(x^2+1\right)^2\sqrt{1-\frac{x^4}{\left(x^2+1\right)^2}}}\,\mathrm{d}x$$

$$I_1=-2{\displaystyle\int}\dfrac{x\left(x^3+x\left(-x^2-1\right)\right)}{\left(x^2+1\right)^2\sqrt{1-\frac{x^4}{\left(x^2+1\right)^2}}}\,\mathrm{d}x$$ $$I_1=2{\displaystyle\int}\dfrac{x^2}{\left(x^2+1\right)\sqrt{2x^2+1}}\,\mathrm{d}x$$ $$I_1=2{\displaystyle\int}\left(\dfrac{1}{\sqrt{2x^2+1}}-\dfrac{1}{\left(x^2+1\right)\sqrt{2x^2+1}}\right)\mathrm{d}x$$ $$I_1=2{\displaystyle\int}\dfrac{1}{\sqrt{2x^2+1}}\,\mathrm{d}x-2{\displaystyle\int}\dfrac{1}{\left(x^2+1\right)\sqrt{2x^2+1}}\,\mathrm{d}x$$

$$I_2={\displaystyle\int}\dfrac{1}{\sqrt{2x^2+1}}\,\mathrm{d}x$$ $$I_3={\displaystyle\int}\dfrac{1}{\left(x^2+1\right)\sqrt{2x^2+1}}\,\mathrm{d}x$$

now solving $I_2$ substitute $u=\sqrt{2} x$ thus $\mathrm{d}x=\dfrac{1}{\sqrt{2}}\,\mathrm{d}u$

$$I_2=\dfrac{1}{\sqrt{2}}{\displaystyle\int}\dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u$$ $$I_2=\dfrac{\ln\left(\sqrt{u^2+1}+u\right)}{\sqrt{2}}$$

$$I_2=\dfrac{\ln\left(\sqrt{2x^2+1}+\sqrt{2}x\right)}{\sqrt{2}}$$

now solve $$I_3={\displaystyle\int}\dfrac{1}{\left(x^2+1\right)\sqrt{2x^2+1}}\,\mathrm{d}x$$ $$x=\dfrac{\tan\left(u\right)}{\sqrt{2}}$$ $$\mathrm{d}x=\dfrac{\sec^2\left(u\right)}{\sqrt{2}}\,\mathrm{d}u$$ $$I_3={\displaystyle\int}\dfrac{\sec^2\left(u\right)}{\sqrt{2}\left(\frac{\tan^2\left(u\right)}{2}+1\right)\sqrt{\tan^2\left(u\right)+1}}\,\mathrm{d}u$$

$$I_3=\sqrt{2}{\displaystyle\int}\dfrac{\sec\left(u\right)}{\tan^2\left(u\right)+2}\,\mathrm{d}u$$ $$I_3=\sqrt{2}{\displaystyle\int}\class{steps-node}{\cssId{steps-node-5}{\cos\left(u\right)}}\class{steps-node}{\cssId{steps-node-6}{\left(-\dfrac{1}{\sin^2\left(u\right)-2}\right)}}\,\mathrm{d}u$$

substitute $$v=\sin\left(u\right)$$

$$I_3=-\sqrt{2}{\displaystyle\int}\dfrac{1}{v^2-2}\,\mathrm{d}v$$

$$I_3=-\sqrt{2}{\displaystyle\int}\dfrac{1}{\left(v-\sqrt{2}\right)\left(v+\sqrt{2}\right)}\,\mathrm{d}v$$

$$I_3=-\sqrt{2}{\displaystyle\int}\left(\dfrac{1}{2^\frac{3}{2}\left(v-\sqrt{2}\right)}-\dfrac{1}{2^\frac{3}{2}\left(v+\sqrt{2}\right)}\right)\mathrm{d}v$$

you can easily solve it $$I_3=-\sqrt{2}\big(\dfrac{\ln\left(v+\sqrt{2}\right)}{2^\frac{3}{2}}-\dfrac{\ln\left(v-\sqrt{2}\right)}{2^\frac{3}{2}}\big)$$

$$I_3=-\sqrt{2}\big(\dfrac{\ln\left(\sin\left(u\right)+\sqrt{2}\right)}{2^\frac{3}{2}}-\dfrac{\ln\left(\sin\left(u\right)-\sqrt{2}\right)}{2^\frac{3}{2}}\big)$$ $$I_3=\dfrac{\ln\left(\sin\left(u\right)+\sqrt{2}\right)}{2}-\dfrac{\ln\left(\sin\left(u\right)-\sqrt{2}\right)}{2}$$

undo the substituion $$u=\arctan\left(\sqrt{2}x\right)$$

$$I_3=\dfrac{\ln\left(\frac{\sqrt{2}x}{\sqrt{2x^2+1}}+\sqrt{2}\right)}{2}-\dfrac{\ln\left(\frac{\sqrt{2}x}{\sqrt{2x^2+1}}-\sqrt{2}\right)}{2}$$

plug back $I_2$ AND $I_3$

$$I_1=\sqrt{2}\ln\left(\sqrt{2x^2+1}+\sqrt{2}x\right)-\ln\left(\dfrac{\sqrt{2}x}{\sqrt{2x^2+1}}+\sqrt{2}\right)+\ln\left(\dfrac{\sqrt{2}x}{\sqrt{2x^2+1}}-\sqrt{2}\right)$$

$$I=-\sqrt{2}\ln\left(\sqrt{2x^2+1}+\sqrt{2}x\right)+\ln\left(\dfrac{x}{\sqrt{2x^2+1}}+1\right)-\ln\left(\dfrac{x}{\sqrt{2x^2+1}}-1\right)+x\arcsin\left(\dfrac{x^2}{x^2+1}\right)+C$$

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