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

Question

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?

Answer 1

$$\int_{0}^{1}\arcsin\left(\frac{x^2}{1+x^2}\right)\,dx = \int_{0}^{1}\arcsin\left(\frac{x}{1+x}\right)\frac{dx}{2\sqrt{x}} $$ 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 $$ \color{red}{\frac{\pi}{6}+\log(2+\sqrt{3})-\sqrt{2}\log(\sqrt{2}+\sqrt{3})}\approx 0.219563.$$

Answer 2

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 $I_3$

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

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

Answer 3

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