Indefinite integral complete square

[karush]

$$\int_{}^{} \frac{1}{\sqrt{16+4x-2x^2}}\,dx$$
$$\frac{\sqrt{2}} {2}\int_{}^{} \cos\left(\frac{x-1}{3}\right)\,dx$$

So far ? Not sure

 

[Dethrone]

Not quite, but close and I think you meant $\frac{\sqrt{2}} {2} \cos\left(\frac{x-1}{3}\right)$.

$$=\int \frac{1}{\sqrt{2[9-(x-1)^2]}}\,dx$$
$$=\frac{\sqrt{2}}{2}\int \frac{1}{\sqrt{9-u^2}}\,du$$
$$=\frac{\sqrt{2}}{2}\sin^{-1}\left({\frac{u}{3}}\right)+C$$
$$=\frac{\sqrt{2}}{2}\sin^{-1}\left({\frac{x-1}{3}}\right)+C$$