MHB 206.8.4.35 integral complete the square

[karush]

$\tiny{206.8.4.35} \\
\text{given }$
$$\displaystyle
I_{35}=\int \frac{1}{\sqrt{x^2+2x+65}} \, dx = $$
$\text{complete the square} \\
x^2+2x+65 \implies x^2+2x+64+1
\implies \left[x+1\right]^2+8^2 \\$
$\text{u substitution } \\
\displaystyle x+1= 8 \tan\left({u}\right)
\therefore du=8\sec{u} \, du \\
u=\arctan{\left(\frac{x+1}{8}\right)}$
$$I_{35}=\int\frac{2\sec^2{u}}{8\sec{u}}\, du
= \frac{1}{4}u+C$$
$\text{back substitution } \\
I_{35}=\frac{1}{4} \arctan{\left(\frac{x+1}{8}\right)} + C$

mistake somewhere!

 

[MarkFL]

After your substitution, you should get:

$$I_{35}=\int \sec(u)\,du$$

 

[karush]

$\tiny{206.8.4.35} \\
\text{given }$
$$\displaystyle
I_{35}=\int \frac{1}{\sqrt{x^2+2x+65}} \, dx = $$
$\text{complete the square}$
$$ \\
x^2+2x+65 \implies x^2+2x+64+1
\implies \left[x+1\right]^2+8^2 \\$$
$\text{u substitution }$
$$
\displaystyle x+1= 8 \tan\left({u}\right)
\therefore du=8\sec^2{u} \, du \\
u=\arctan{\left(\frac{x+1}{8}\right)}$$
$$I_{35}=\int\frac{8\sec^2{u}}{8\sec{u}}\, du
\implies \int \sec{u} \, du =- \ln\left({\cos{u}}\right)-\ln\left({\sin\left({u}\right)-1}\right)+C$$
$\text{how do you back substitute into this } $
$$I_{35}=$$

 

[MarkFL]

Your integration is incorrect...I would suggest using this:

$$I_{35}=\int \sec(u)\,du=\ln\left|\sec(u)+\tan(u)\right|+C$$

And then you can use the fact that if:

$$\tan(\theta)=\frac{a}{b}$$

then:

$$\sec(u)=\frac{\sqrt{a^2+b^2}}{b}$$

 

[karush]

$$u= \arctan\left(\frac{x+1}{8}\right)$$
Then
$$\tan\left({u}\right)=\frac{x+1}{8}=\frac{a}{b}$$
$$\displaystyle \sec(u)=\frac{\sqrt{a^2+b^2}}{b}=
\frac{\sqrt{(x+1)^2+8^2}}{8}$$
So then
$$I_{35}=\ln{\left[\frac{\sqrt{(x+1)^2+8^2}}{8}
+\frac{x+1}{8}\right]}+C$$
Hopefully. ...bar simplification

 

[MarkFL]

Ok but my TI returned
$$\int\sec{u} \, du
= \ln\left[{\frac{{-\cos\left({u}\right)}}
{\sin\left({u}\right)-1}}\right]+C$$
which may be the same thing

$$\frac{-\cos(u)}{\sin(u)-1}=\frac{\cos(u)}{1-\sin(u)}=\frac{\cos(u)}{1-\sin(u)}\cdot\frac{1+\sin(u)}{1+\sin(u)}=\frac{1+\sin(u)}{\cos(u)}=\sec(u)+\tan(u)$$

What you posted before though was not equivalent:

$$\ln\left|\frac{\cos(u)}{1-\sin(u)}\right|\ne-\ln\left|\cos(u)\right|-\ln\left|\sin(u)-1\right|$$

 

[karush]

I updated post 5 hope its correct

 

[MarkFL]

I updated post 5 hope its correct

Looks good, except use absolute value instead of brackets (since the log argument could be negative otherwise). Then combine terms, and then use a property of logs and the fact that a constant added/subtracted to/from an arbitrary constant is still an arbitrary constant. :D