# Find $\int\arcsin(\sqrt{x})dx$

Find $\displaystyle\int\arcsin(\sqrt{x})dx$

My Attempt

Put $y=\sqrt{x}\implies dy=\frac{1}{2\sqrt{x}}dx\implies dx=2ydx$ $$\int\arcsin(\sqrt{x})dx=2\int \arcsin(y)\,y\,dy=2\bigg[\frac{y^2}{2}\arcsin(y)-\int\frac{1}{\sqrt{1-y^2}}\frac{y^2}{2}dy\bigg]\\ =y^2\arcsin(y)-\int\frac{y^2}{\sqrt{1-y^2}}dy.$$ How do I proceed further and find the solution or is there any easier way ?

• An upvote to this post, and to every answer below! $(+1)$ – Mr Pie May 22 '18 at 11:32

Hint. Note that $$-\int\frac{y^2}{\sqrt{1-y^2}}\,dy=\int\frac{1-y^2}{\sqrt{1-y^2}}\,dy-\int\frac{dy}{\sqrt{1-y^2}}=\int\sqrt{1-y^2}\,dy-\arcsin(y).$$

I would proceed as follows:\begin{align}\int\frac{y^2}{\sqrt{1-y^2}}\,\mathrm dy&=\int y\frac y{\sqrt{1-y^2}}\,\mathrm dy\\&=-y\sqrt{1-y^2}+\int\sqrt{1-y^2}\,\mathrm dy.\end{align}

$$\int\frac{-y^2}{\sqrt{1-y^2}}dy=\int\frac{1-y^2-1}{\sqrt{1-y^2}}dy =\int\frac{1-y^2}{\sqrt{1-y^2}}dy-\int\frac{1}{\sqrt{1-y^2}}dy=$$ $$= \int\sqrt{1-y^2}dy-\arcsin(y)=\left[y=\sin{t}, dy=\cos{t}\ dt\right]=\int\cos^2{t}dt-\arcsin(y)=$$ $$=\frac12\int(1+\cos{2t})dt-\arcsin(y)=\frac12\left[t+\frac12\sin{2t}\right]_{t=\arcsin{y}}-\arcsin(y)+C=$$ $$=\frac12\arcsin{y}+\frac14\sin{(2\arcsin{y})}-\arcsin(y)+C=$$ $$=-\frac12\arcsin{y}+\frac12\sin{(\arcsin{y})}\cos{(\arcsin{y})}+C=$$ $$=-\frac12\arcsin{y}+\frac12y\sqrt{1-y^2}+C$$

Hint. Notice that $\int\frac{1}{\sqrt{1-y^2}}\mathrm{d}y=\sin^{-1}(y)$, so you could try writing $y^2$ as $y^2-1+1$ and splitting into two integrals:

$$\int\frac{y^2}{\sqrt{1-y^2}}\mathrm{d}y=\int\left(\frac{y^2-1}{\sqrt{1-y^2}}+\frac{1}{\sqrt{1-y^2}}\right)\mathrm{d}y\\=\int\left (\frac{1}{\sqrt{1-y^2}}-\sqrt{1-y^2}\right )\mathrm{d}y=\int\frac{1}{\sqrt{1-y^2}} \mathrm{d}y-\int\sqrt{1-y^2} \mathrm{d}y$$ Solving $\int\sqrt{1-y^2} \mathrm{d}y$ can be made via trigonometric substitution, setting $y=\sin u$.

By parts directly:

$$\begin{cases}u=\arcsin\sqrt x,&u'=\frac1{2\sqrt x\sqrt{1-x}}\\{}\\ v'=1,&v=x\end{cases}\;\;\implies \int\arcsin\sqrt x\,dx=x\arcsin\sqrt x-\frac12\int\sqrt\frac x{1-x}\,dx$$

and now substitute

$$\;u^2=\frac x{1-x}\implies x(-u^2-1)=-u^2\implies x=\frac{u^2}{u^2+1}=1-\frac1{1+u^2}\implies$$

$$dx=\frac{2u}{(1+u^2)^2}\;$$

and from here your integrals equals

$$x\arcsin\sqrt x-\int\frac{u^2}{(1+u^2)^2}\,du=x\arcsin x-\frac12\left(\arctan u-\frac u{1+u^2}\right)$$

and now go back to $\;x\;$ and etc.

Note that $$d/dx((x - 1/2) \sin^{-1}(\sqrt x)) = \frac{x - 1/2}{2 \sqrt{x(1 - x)} } + \sin^{-1}(\sqrt x )$$ So you can easily integrate $$\int\sin^{-1}\sqrt{x}dx \\ = (x - 1/2) \sin^{-1}(\sqrt x) - \int \frac{x - 1/2}{2 \sqrt{x(1 - x)} } dx \\ = (x - 1/2) \sin^{-1}(\sqrt x) + \frac{ \sqrt{x(1 - x)}}{2}$$

• The outcome is correct (I checked it by differentiating...very nice!), yet the very first step seems like magic: where did it come from? Imo students trying to grasp advanced integration techniques won't like come up with things like this one. – DonAntonio May 22 '18 at 11:18
• @DonAntonio Yes, there is a common step and an unusual step to it. The common step is to differentiate $x \sin^{-1}(\sqrt x)$ since it is clear that $\sin^{-1}(\sqrt x)$ will arise, plus a second term from the product rule. The unusual step is to try and compensate for this second term by using the derivative of $\sin^{-1}(\sqrt x)$. It needs a little luck that this can indeed be used for compensation. – Andreas May 22 '18 at 14:10