1. ## Trigonometric Substitution

I am working on a problem that I have gone over and over and can't seem to find my mistake. The problem involves solving an integral using trigonometric substitution. My final answer is identical to the text's answer except my answer has a factor of 1/4 that is not in the book's answer and I cannot figure out why. Thanks in advance to anyone who takes the time to work it!

$\displaystyle \int \frac{x}{\sqrt{x^2+4x+8}} dx$

Book answer: $\displaystyle \sqrt{x^2+4x+8} - 2ln|\sqrt{x^2+4x+8} + (x+2)| + C$

My answer: $\displaystyle \frac{1}{4}\sqrt{x^2+4x+8} - 2ln|\sqrt{x^2+4x+8} + (x+2)| + C$

2. Hello, kaiser0792!

At the end, did you multiply by $\displaystyle \tfrac{1}{2}$ instead of dividing?

If you approached like I did, you had this at the end of them problem:

. . $\displaystyle \int (x^2+4x+8)^{-\frac{1}{2}}(x+2)\,dx \;-\; 2\int\frac{dx}{\sqrt{(x+4)^2 + 4}}$

For the first integral:

Let: .$\displaystyle u \:=\:x^2+4x+8 \quad\Rightarrow\quad dx \:=\:(2x+4)\,dx \quad\Rightarrow\quad(x+2)\,dx \,=\,\tfrac{1}{2}\,du$

Substitute: .$\displaystyle \int u^{-\frac{1}{2}}\left(\tfrac{1}{2}\,du\right) \;=\; \tfrac{1}{2}\int u^{-\frac{1}{2}}du \;=\;\tfrac{1}{2}\cdot{\color{red}\frac{1}{\frac{1 }{2}}}\,u^{\frac{1}{2}} + C \;=\;u^{\frac{1}{2}}+C$

3. Note that $\displaystyle \int{\frac{x}{\sqrt{x^2 + 4x + 8}}\,dx} = \int{\frac{x + 2 - 2}{\sqrt{(x + 2)^2 + 4}}\,dx}$

$\displaystyle = \int{\frac{x + 2}{\sqrt{(x + 2)^2 + 4}}\,dx} - \int{\frac{2}{\sqrt{(x + 2)^2 + 4}}\,dx}$.

Hyperbolic substitution is easier in this case.

Let $\displaystyle x + 2 = 2\sinh{t}$ so that $\displaystyle dx = 2\cosh{t}\,dt$.

The integral becomes:

$\displaystyle \int{\frac{2\sinh{t}}{\sqrt{(2\sinh{t})^2 + 4}}\,2\cosh{t}\,dt} - \int{\frac{2}{\sqrt{(2\sinh{t})^2 + 4}}\,2\cosh{t}\,dt}$

$\displaystyle = \int{\frac{4\sinh{t}\cosh{t}}{\sqrt{4(\sinh^2{t} + 1)}}\,dt} - \int{\frac{4\cosh{t}}{\sqrt{4(\sinh^2{t} + 1)}}\,dt}$

$\displaystyle = \int{\frac{4\sinh{t}\cosh{t}}{2\sqrt{\cosh^2{t}}}\ ,dt} - \int{\frac{4\cosh{t}}{2\sqrt{\cosh^2{t}}}\,dt}$

$\displaystyle = \int{2\sinh{t}\,dt} - \int{2\,dt}$

$\displaystyle = 2\cosh{t} - 2t + C$

Since $\displaystyle x + 2 = 2\sinh{t}$ that means $\displaystyle t = \textrm{arsinh}\,\left({\frac{x + 2}{2}}\right)$.

Also, since $\displaystyle \cosh^2{t} - \sinh^2{t} = 1$

that means $\displaystyle \cosh{t} = \sqrt{1 - \sinh^2{t}}$

$\displaystyle = \sqrt{1 + \left(\frac{x + 2}{2}\right)^2}$

$\displaystyle = \sqrt{\frac{(x + 2)^2 + 4}{4}}$

$\displaystyle = \frac{\sqrt{x^2 + 4x + 8}}{2}$.

$\displaystyle 2\cosh{t} - 2t + C = 2\left(\frac{x^2 + 4x + 8}{2}\right) - 2\,\textrm{arsinh}\,\left({\frac{x + 2}{2}}\right) + C$
$\displaystyle = \sqrt{x^2 + 4x + 8} - 2\ln{|\sqrt{x^2 + 4x + 8} + x + 2|} + C$ after converting to the arsinh logarithmic equivalent.