Integral completing the square

[karush]

$\int\frac{1}{\sqrt{16+4x-2{x}^{2}}}dx$
factored out 2 then completed the square
$\int\frac{1}{\sqrt{2\left[9-{(x-1)}^{2}\right]}}dx$
next?

 

[MarkFL]

I would then write:

\(\displaystyle I=\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{3^2-(x-1)^2}}\,dx\)

I would then consider the trigonometric substitution:

\(\displaystyle x-1=3\sin(\theta)\)

Can you proceed?

 

[karush]

$$I=\frac{1}{3\sqrt{2}}\int \frac{1}{\cos\left({x}\right)}dx $$
What about dx?

 

[MarkFL]

Your denominator should have $\cos(\theta)$ (as long as it is non-negative), to express $dx$ as a function of $\theta$, implicitly differentiate the substitution and leave in differential form...:D

 

[karush]

as a function of θ, implicitly differentiate the substitution and leave in differential form...

How do you do this?

 

[MarkFL]

Since we have:

\(\displaystyle x-1=3\sin(\theta)\)

We may then state:

\(\displaystyle dx=3\cos(\theta)\,d\theta\)

 

[karush]

$\frac{1}{\sqrt{2}}\int \frac{\cos\left({\theta}\right)}{\cos\left({\theta}\right)}d\theta $

I got to this but can't be right

 

[MarkFL]

Yes, now reduce the integrand, compute the anti-derivative and then back-substitute for $\theta$. :D

 

[karush]

Reducing the $\int$ is $\frac{\sqrt{2}}{2}\theta$
From $x-1=3\sin\left({\theta}\right)$ and$\theta=\sin^{-1}\left({\frac{x-1}{3}}\right)$
So $\frac{\sqrt{2}}{2}\sin^{-1}\left({\frac{x-1}{3}}\right)+C$
However the TI gave $x+1$ in the numerator

 

[Prove It]

Reducing the $\int$ is $\frac{\sqrt{2}}{2}\theta$
From $x-1=3\sin\left({\theta}\right)$ and$\theta=\sin^{-1}\left({\frac{x-1}{3}}\right)$
So $\frac{\sqrt{2}}{2}\sin^{-1}\left({\frac{x-1}{3}}\right)+C$
However the TI gave $x+1$ in the numerator

Then you must have either written the question down incorrectly or input it incorrectly into the calculator, because what you have given is the correct answer to the problem as written.

 

[karush]

Your right had sign error on input