# Integral - trigonometric substitution?

• Dec 17th 2010, 08:54 AM
drewbear
Integral - trigonometric substitution?
Solve the following problem using trig substitution:

$\int{\frac{x}{\sqrt{x^2+x+1}}}$
• Dec 17th 2010, 09:07 AM
snowtea
I think you forgot a dx.

$\frac{x}{\sqrt{x^2+x+1}} = \frac{x}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}$

Let $y = x + \frac{1}{2}$
$\frac{y - \frac{1}{2}}{\sqrt{y^2+\frac{3}{4}}} = \frac{y}{\sqrt{y^2+\frac{3}{4}}} - \frac{\frac{1}{2}}{\sqrt{y^2+\frac{3}{4}}}$

Now, its split into one simple integral $\frac{y}{\sqrt{y^2+\frac{3}{4}}}$ and a trig sub $\frac{\frac{1}{2}}{\sqrt{y^2+\frac{3}{4}}}$.

Use $y^2 = \frac{3}{4}tan^2(\theta)$
• Dec 17th 2010, 09:14 AM
drewbear
i did forget the dx, sorry.
but wouldn't $(x+\frac{1}{2})^2+\frac{1}{2}=x^2+x+\frac{3}{4}$ and not $x^2+x+1$
• Dec 17th 2010, 09:16 AM
snowtea
Yes, you are absolutely right :D
Let me fix that.
But you get the idea.
• Dec 17th 2010, 09:17 AM
TheCoffeeMachine
I solved this integral in here, and here, and here. (Giggle)
• Dec 17th 2010, 09:19 AM
drewbear
thank you, but is there any other way of doing this problem without setting y= to some number?
• Dec 17th 2010, 10:52 AM
Soroban
Hello, drewbear!

This problem is quite awful!

Quote:

$\displaystyle \text{Solve using trig substitution: }\:\int \frac{x\,dx}{\sqrt{x^2+x+1}}$

Under the radical, complete the square:

. . $x^2 + x + 1 \;=\;x^2 + x + \frac{1}{4} + \frac{3}{4} \;=\;\left(x+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$

The integral becomes: . $\displaystyle \int \frac{x\,dx} {\sqrt{(x + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}}$

Let $x + \frac{1}{2} \:=\:\frac{\sqrt{3}}{2}\tan\theta \quad\Rightarrow\quad x \:=\:\frac{\sqrt{3}}{2}\tan\theta - \frac{1}{2} \quad\Rightarrow\quad dx \:=\:\frac{\sqrt{3}}{2}\sec^2\!\theta\,d\theta$

Substitute: . $\displaystyle \int\frac{\left(\frac{\sqrt{3}}{2}\tan\theta - \frac{1}{2}\right)
\left(\frac{\sqrt{3}}{2}\sec^2\theta\,d\theta)}{\f rac{\sqrt{3}}{2}\sec\theta}$

. . . . . $\displaystyle =\; \tfrac{1}{2}\int(\sqrt{3}\tan\theta - 1)\,\sec\theta\,d\theta$

. . . . . $\displaystyle =\; \tfrac{1}{2}\int (\sqrt{3}\sec\theta\tan\theta - \sec\theta)\,d\theta$

. . . . . $=\;\frac{1}{2}\bigg[\sqrt{3}\sec\theta + \ln|\sec\theta + \tan\theta|\bigg] + C$

Back-substitute: . $\displaystyle \tan\thet \:=\:\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \;=\;\frac{2x+1}{\sqrt{3}}$

. . $\displaystyle \sec^2\!\theta \:=\:\tan^2\!\theta + 1 \:=\:\left(\frac{2x+1}{\sqrt{3}}\right)^2 + 1 \;=\; \frac{4(x^2+x+1)}{3}$

. . $\sec\theta \;=\;\frac{2}{\sqrt{3}}\sqrt{x^2+x+1}$

We have: . $\frac{1}{2}\left[\sqrt{3}\cdot\frac{2}{\sqrt{3}}\sqrt{x^2+x+1} - \ln\left|\frac{2\sqrt{x^2+x+1}}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}\right|\, \right] + C$

. . . . . . $=\;\sqrt{x^2+x+1} - \frac{1}{2}\ln\left|\frac{2\sqrt{x^2+x+1} + 2x + 1}{\sqrt{3}}\right| + C$

This can be simplified further, but I won't bother . . .
• Dec 17th 2010, 06:58 PM
Prove It
Quote:

Originally Posted by drewbear
Solve the following problem using trig substitution:

$\int{\frac{x}{\sqrt{x^2+x+1}}}$

Maybe an alcoholic substitution would work well here...

$\displaystyle \int{\frac{x}{\sqrt{x^2+x+1}}\,dx} = \int{\frac{x}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}}}\,dx}$

Make the substitution $\displaystyle x + \frac{1}{2} = \frac{\sqrt{3}}{2}\sinh{t}$ so that $\displaystyle dx = \frac{\sqrt{3}}{2}\cosh{t}\,dt$ and the integral becomes

$\displaystyle \int{\frac{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}}{\sqrt{\frac{3}{4}\sinh^2{t} + \frac{3}{4}}}\left(\frac{\sqrt{3}}{2}\cosh{t}\righ t)\,dt}$

$\displaystyle =\int{\frac{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}}{\sqrt{\frac{3}{4}}\sqrt{\sinh^2{t} + 1}}\left(\frac{\sqrt{3}}{2}\cosh{t}\right)\,dt}$

$\displaystyle =\int{\frac{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}}{\frac{\sqrt{3}}{2}\cosh{t}}\left(\fra c{\sqrt{3}}{2}\cosh{t}\right)\,dt}$

$\displaystyle =\int{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}\,dt}$

$\displaystyle = \frac{\sqrt{3}}{2}\cosh{t} - \frac{t}{2} + C$

$\displaystyle = \frac{\sqrt{3}}{2}\sqrt{\sinh^2{t} + 1} - \frac{t}{2} + C$

$\displaystyle = \frac{\sqrt{3}}{2}\sqrt{\left(\frac{2x+1}{\sqrt{3} }\right)^2 + 1}- \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C$

$\displaystyle = \frac{\sqrt{3}}{2}\sqrt{\frac{4x^2 + 4x + 1}{3} + \frac{3}{3}} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C$

$\displaystyle = \frac{\sqrt{3}}{2}\sqrt{\frac{4x^2 + 4x + 4}{3}} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C$

$\displaystyle = \frac{\sqrt{3}}{2}\sqrt{\frac{4(x^2+x+1)}{3}} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C$

$\displaystyle = \frac{\sqrt{3}}{2}\frac{2}{\sqrt{3}}\sqrt{x^2+x+1} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C$

$\displaystyle = \sqrt{x^2+x+1} -\frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C$