1. Differentiate $\arcsin\left(\dfrac{x}{a}\right)$ with respect to x.
    1. Integrate by parts: $\int\sqrt{a^2-x^2}dx$

The answer to part one of the question is $\dfrac{1}{\sqrt{a^2-{x^2}}}$

  • 2
    $\begingroup$ Your answer to part one is incorrect. Hint: The derivative with respect to $a$ is different than the derivative with respect to $x$. $\endgroup$ – John Barber May 10 at 15:56
  • $\begingroup$ Oops, Im sorry. I meant to say differentiate with respect to x. $\endgroup$ – NoLand'sMan May 10 at 16:04
  • $\begingroup$ The other point is that integration is a very restrictive process you have to learn when to do what. If you do this one, by parts you need to know another integral so on and so forth. The usual method to do this one is by the substitution $x= a \sin t.$ Cheers. $\endgroup$ – Dr Zafar Ahmed DSc May 10 at 16:04
  • $\begingroup$ @DrZafarAhmedDSc Although it may be easier by substitution, in the book it says that integration by parts needs to be used $\endgroup$ – NoLand'sMan May 10 at 16:12
  1. Since IBP needs to be use, take the integral and anti-rationalize it.

$$I=\int \sqrt{a^2-x^2}dx=\int \frac{a^2-x^2}{\sqrt{a^2-x^2}}dx=a^2\int \frac{1}{\sqrt{a^2-x^2}}dx-\int x\frac{x}{\sqrt{a^2-x^2}}dx$$ Integration by parts for the second integral with $\int \frac{x}{\sqrt{a^2-x^2}}dx=-\sqrt{a^2-x^2}+c$ $$=a^2\arcsin\left(\frac{x}{a}\right) + x\sqrt{a^2-x^2}-\int \sqrt{a^2-x^2}dx$$ $$2I=a^2\arcsin\left(\frac{x}{a}\right) + x\sqrt{a^2-x^2}+C$$ Division by $2$ on both sides now.


Answer to 2)

Let $I = \int \sqrt{a^2 - x^2}dx = \int udv$

where $u = \sqrt{a^2-x^2}, dv = dx$

and $ du = -\frac{2x}{2\sqrt{a^2-x^2}} = - \frac{x}{\sqrt{a^2-x^2}}, v = x$


$I = \int udv = uv - \int{vdu}$

$I = \sqrt{a^2-x^2}.x - \int{-x.\frac{x}{\sqrt{a^2-x^2}}}dx = x\sqrt{a^2-x^2} + \int{\frac{a^2 + x^2 - a^2}{\sqrt{a^2-x^2}}}dx +c$

$I = x\sqrt{a^2-x^2} + a^2\int{\frac{1}{\sqrt{a^2-x^2}}}dx - \int\frac{a^2-x^2}{\sqrt{a^2- x^2}}dx + c$

$I = x\sqrt{a^2-x^2} + a^2 sin^{-1}\frac{x}{a} - \int{\sqrt{a^2-x^2}dx} +c $

$I = x\sqrt{a^2-x^2} + a^2 sin^{-1}\frac{x}{a} - I + c$

$2I = x\sqrt{a^2-x^2} + a^2 sin^{-1}\frac{x}{a} + c$

$$I = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2} sin^{-1}\frac{x}{a} + C$$ where $C = \frac{c}{2}$


We know that integration by parts formula is $$\int\sqrt{a^2-x^2} dx=\int u dv=uv-\int v du$$ Comparing these two we can choose that $u=\sqrt{a^2-x^2}$, $dv=dx$, then we have $du=\frac{-x}{\sqrt{a^2-x^2}}$ and $v=x$, using all these in above integration by parts formula we have $$I=\int\sqrt{a^2-x^2} dx=x\sqrt{a^2-x^2}+\int \frac{x^2}{\sqrt{a^2-x^2}}dx$$. Now let $$I_{1}=\int \frac{x^2}{\sqrt{a^2-x^2}}dx$$ We substitute $x=a\cos\theta$, then we have $a^2-x^2=a^2 \cos^2\theta$ and $dx=a\cos\theta d\theta$, inserting all these in $I_{1}$, we have that $$I_{1}=a^2\int \sin^2\theta d\theta=\frac{a^2}{2}\int (1-\cos 2\theta)d\theta=\frac{a^2}{2}\theta-\frac{a^2}{4}\sin2\theta $$ Now $x=a\sin\theta$, so $\theta=\sin^{-1}\frac{x}{a}$ and $\cos\theta=\frac{\sqrt{a^2-x^2}}{a}$. Hence $$I_{1}=\frac{a^2}{2}\sin^{-1}\frac{x}{a}-\frac{\sqrt{a^2-x^2}}{2} +C$$. Putting $I_{1}$ in the first one we get that $$\int\sqrt{a^2-x^2} dx=\frac{a^2}{2}\sin^{-1}\frac{x}{a}+\frac{x\sqrt{a^2-x^2}}{2}+C$$


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