From the problem, we have :
[tex]\int -\frac{3}{\sqrt[]{64-9x^2}}dx[/tex]
We need to think of a square number that if we multiply it by 9, the result is 64.
That would be 64/9
[tex]9\times\frac{64}{9}=64[/tex]
The square root of 64/9 is 8/3.
Let u = 3x/8
8u = 3x
x = 8u/3 (This value is the same as we've got above 8/3)
du = 3/8 dx
dx = 8/3 du
Subsitute x = 8u/3 and dx = 8/3 du
[tex]\begin{gathered} \int -\frac{3}{\sqrt[]{64-9(\frac{8u}{3})^2}}\times\frac{8}{3}du \\ \Rightarrow\int -\frac{8}{\sqrt[]{64-9(\frac{64}{9}u^2)}}du \\ \Rightarrow\int -\frac{8}{\sqrt[]{64-64u^2}}du \end{gathered}[/tex]
Extract the square root of 64 which is equal to 8.
[tex]\begin{gathered} \Rightarrow\int -\frac{8}{\sqrt[]{64-64u^2}}du \\ \Rightarrow\int -\frac{8}{8\sqrt[]{1-u^2}}du \\ \Rightarrow\int -\frac{1}{\sqrt[]{1-u^2}}du \end{gathered}[/tex]
Note that the identity :
[tex]\int \frac{1}{\sqrt[]{1-u^2}}du=\sin ^{-1}(u)+C[/tex]
It follows that :
[tex]\int -\frac{1}{\sqrt[]{1-u^2}}du=-\sin ^{-1}(u)+C[/tex]
Bring back u = 3x/8, the answer is :
[tex]-\sin ^{-1}(\frac{3x}{8})+C[/tex]
