Trigonometric Substitution

**kaiser0792** · March 30th 2010, 06:18 PM

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!

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

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

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

**Soroban** · March 30th 2010, 07:45 PM

Hello, kaiser0792!

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

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

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

For the first integral:

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

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

**Prove It** · March 30th 2010, 08:01 PM

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

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

Hyperbolic substitution is easier in this case.

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

The integral becomes:

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

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

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

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

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

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

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

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

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

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

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

So your integral is

$2\cosh{t} - 2t + C = 2\left(\frac{x^2 + 4x + 8}{2}\right) - 2\,\textrm{arsinh}\,\left({\frac{x + 2}{2}}\right) + C$

$= \sqrt{x^2 + 4x + 8} - 2\ln{|\sqrt{x^2 + 4x + 8} + x + 2|} + C$ after converting to the arsinh logarithmic equivalent.

**kaiser0792** · March 30th 2010, 08:08 PM

Soroban, thank you for the help. I approached it quite differently. I'm about to call it a night. I'm going to study your solution in the morning and see if I can follow it. I used a right triangle approach, with the legs labeled x + 2 = 2 cot (theta) and 2. The hypotenuse is [(x + 2) ^2 + 4]^1/2 = 2 csc (theta). From there I rewrote and solved the integral and then back-substituted in terms of x. I'm going to have to analyze your solution and see if I can make sense of it tomorrow.

Thanks for your help.

**kaiser0792** · March 30th 2010, 08:12 PM

Thanks Prove It, you are always a great help!

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