2
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ An upvote to this post, and to every answer below! $(+1)$ $\endgroup$ – Mr Pie May 22 '18 at 11:32
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $y=\sin u$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$ \int\sin^{-1}\sqrt{x}dx \\ = (x - 1/2) \sin^{-1}(\sqrt x) - \int \frac{x - 1/2}{2 \sqrt{x(1 - x)} } dx \\ = (x - 1/2) \sin^{-1}(\sqrt x) + \frac{ \sqrt{x(1 - x)}}{2} $$

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – DonAntonio May 22 '18 at 11:18
  • $\begingroup$ @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. $\endgroup$ – Andreas May 22 '18 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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