# Find `int (9-4x^2)^(1/2) dx` .

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### 2 Answers

`int(9-4x^2)^(1/2)dx = int sqrt(9-4x^2) dx`

Note that the integrand has a form `sqrt(a^2-u^2)` . Base on this, let's consider the formula,

`int 1/sqrt(a^2-u^2)du = sin^(-1) u/a + C` .

To be able to apply this, multiply the integrand by `sqrt(9-4x^2)/sqrt(9-4x^2)` .

`=int sqrt(9-4x^2)*(sqrt(9-4x^2)/sqrt(9-4x^2))dx =int(9-4x^2)/sqrt(9-4x^2)dx`

Then, express the integrand as two fractions.

`=int (9/sqrt(9-4x^2)dx - (4x^2)/sqrt(9-4x^2))dx = int 9/sqrt(9-4x^2)dx - int (4x^2)/sqrt(9-4x^2)dx`

Next, let's determine the integral of each separately.

> For the first integral we have,

`int9/sqrt(9-4x^2)dx = 9 int1/sqrt((9-4x^2))dx`

Applying the mentioned formula above, we have a=3 and u=2x and du =2dx.

`= 9/2int 1/sqrt(3^2-(2x)^2)*2dx`

`=9/2 sin^(-1) ((2x)/3)`

> For the second integral, we have,

`int(4x^2)sqrt(9-4x^2)=4intx^2/sqrt(9-4x^2)dx`

Then, let's use integration by parts. The formula is `intudv=uv - intvdu`

Let ,

`u=x ` and `dv=x/sqrt(9-4x^2)dx `

`du=dx` `v=int x/sqrt(9-4x^2)dx`

To determine v, use substitution method. Let,

`y=9-4x^2` `dy=-8xdx` `-dy/8=xdx`

Replacing the variable x with y yields,

`v=int 1/sqrty *(-dy)/8 = -1/8 int y^(-1/2)dy = -1/4y^(1/2)=-sqrty/4`

Susbtitute back `y=9-4x^2` .

`v= -sqrt(9-4x^2)/4`

Then, susbtitute u,v, and du to the formula of integration by parts.

`int(4x^2)/sqrt(9-4x^2)dx= 4[x*(-sqrt(9-4x^2)/4) -int-sqrt(9-4x^2)/4dx]`

`=-xsqrt(9-4x^2)+intsqrt(9-4x^2)dx`

So we have,

`int sqrt(9-4x^2)dx = int9/sqrt(9-4x^2)dx- int(4x^2)/sqrt(9-4x^2)dx`

`int sqrt(9-4x^2)dx=9/2sin^(-1)((2x)/3) +xsqrt(9-4x^2) - int sqrt(9-4x^2) dx`

Note that both sides have the same integral. To simplify, add both sides by `int sqrt(9-4x^2)dx` .

`2int sqrt(9-4x^2)dx = 9/2sin^(-1)((2x)/3) + xsqrt(9-4x^2)`

To solve for `int sqrt(9-4x^2)dx` , divide both sides by 2.

`intsqrt(9-4x^2)dx = (9sin^(-1)((2x)/3))/4+ (xsqrt(9-4x^2))/2`

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**Since we have an indefinite integral, therefore:**

**`int (9-4x^2)^(1/2)dx = (9sin^(-1)((2x)/3))/4 + (xsqrt(9-4x^2))/2 + C`**

**Sources:**

When an integral has an expression `sqrt{a^2-x^2}` in it, we often need to use the trigonometric substitution `x=sin theta` to be able to use trig identities like `sin^2 theta+cos^2 theta=1` to simplify the integral.

In this case, let `2/3x=sin theta` which means that `dx=3/2 costheta d theta`. Also, using some simple algebra, we get `sqrt{1-4/9x^2}=costheta` which means the integral becomes.

`int sqrt{9-4x^2}dx`

`=3 int sqrt{1-4/9x^2}dx`

`=3 int cos theta 3/2 cos theta d theta`

`=9/2 int cos^2 theta d theta` now use identity `cos 2theta=2cos^2 theta-1`

`=9/4int d theta + 9/4 int cos 2theta d theta`

`=9/4 theta+9/8 sin 2 theta +C` where C is a constant

`=9/4 sin^{-1}({2x}/3)+9/4 sin theta cos theta +C` back-substitute for `theta` using Pythagorean Theorem

`=9/4sin^{-1}({2x}/3)+3/2 xsqrt{1-4/9x^2}+C`

`=9/4sin^{-1}({2x}/3)+1/2xsqrt{9-4x^2}+C`

**The integral evaluates to `9/4sin^{-1}({2x}/3)+1/2xsqrt{9-4x^2}+C` . **

**Sources:**