arctan/arcsin and integration by substitution

**terrorsquid** · October 20th 2011, 11:51 PM

$\int \frac{1}{\sqrt{81-x^2}}~dx$

I know I need to use substitution and I am aware of the fact that $\frac{1}{\sqrt{1-x^2}}$ = derivative of arcsin(x), but I am unsure how I apply these to the above problem.

I have the answer $arcsin \left(\frac{x}{9}\right)+C$ but I don't follow the steps to get to it.

**Quacky** · October 21st 2011, 12:26 AM

Let $I=\int\frac{1}{\sqrt{81-x^2}}dx$

We can do it by substitution, but I wouldn't personally. However, it's good to start from there.

So I'm going to let $x=9sin(u)$

This means that $\frac{dx}{du}=9cos(u)$ so $dx=9cos(u)~du$

$I=\int\frac{9cos(u)}{\sqrt{81-[9sin(u)]^2}}du$

$=\int\frac{9cos(u)}{\sqrt{81-81sin^2(u)}}du$

Factoring the $81$ :

$=\int\frac{9cos(u)}{\sqrt{81[1-sin^2(u)]}}du$

Applying $\sqrt{a\cdot{b}}=\sqrt{a}\sqrt{b}$ for $a,b\geq{0}$ :

$I=\int\frac{9cos(u)}{9\sqrt{1-sin^2(u)}}du$

But as $sin^2(u)+cos^2(u)=1$ ,

$1-sin^2(u)=cos^2(u)$

So $I=\int\frac{cos(u)}{\sqrt{cos^2(u)}}du$

$=\int\frac{cos(u)}{cos(u)}du$

$=\int~du$

$=u+C$

Now, I let $x=9sin(u)$ right at the start of my substitution. This means that $\frac{x}{9}=sin(u)$ , so $u=arcsin(\frac{x}{9})$

Giving us the required final answer of $I=arcsin(\frac{x}{9})+C$

Is this clear? Usually I'd ask you to do part of the problem yourself but I don't think there's really a good place to stop with this question.

**salim** · October 21st 2011, 01:47 AM

hi,
the problem is easy
let y=x/9 => x = 9y and dx = 9dy
$\int \frac{9dy}{\sqrt{81-81y^{2}}}$
$\int \frac{9dy}{9\sqrt{1-y^{2}}}$
$\int \frac{dy}{\sqrt{1-y^{2}}}$
then
the solution is : arcsin(y) + c
i.e. arcsin(x/9) + c

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