# Find the error

1. Nov 17, 2013

### Saq_Lfc

∫√(16-9χ²) dx

this is what I tried

let 9/16 χ² = sin²

then x=4/3(sinu) and u=arcsin(3/4x) and dx=4/3(cosu) du

∫√(16(1-sin²u) )x 4/3(cosu) du

∫16/3 cos²u du

∫8/3(cos2u + 1) du

4/3(sin2u) +8/3 u +k

4/3(2sinucosu) +8/3 u +k

8/3 (sin(arcsin(3/4χ))xcos(arcsin(3/4χ))) +8/3(arcsin(3/4χ))+k

2χ√(1-16/9x²) +8/3arcsin(3/4χ) +k

2. Nov 18, 2013

### brmath

From the way you wrote it, it's a little hard for me to tell exactly what you did. Suppose I start the same way you did:

$\int \sqrt{16-9x^2} dx = 4\int \sqrt{1-9x/16}dx \hspace{50px}$ (1)

Let sin u = 3x/4; cos u du =3/4 dx; So (1) becomes

$3\int \sqrt {1-sin^2}(u) cos(u) du = 3\int cos^2(u) du$

You can finish this up by using the half angle formula on the cos.

3. Nov 18, 2013

### Saq_Lfc

can u solve it for me I wanna check plus how did that 4 outta the integral turned 3

Last edited: Nov 18, 2013
4. Nov 18, 2013

### dirk_mec1

We do not solve anything for you. Moreover, you've failed to apply the standard PF format.

5. Nov 18, 2013

### brmath

Hi Saq,

du = 3/4 dx.

As Dirk says, we don't do your homework -- we provide hints, clues, get you started etc. I think I did enough that you should be able to finish up.