# Integral Question

• Oct 9th 2010, 12:41 PM
VkL
Integral Question
How do you do this integral without a table?

Integral [ Sqrt(4x^2 + 20)]

I got it down to

2* integral [ sqrt(x^2 + 5) ]

Now I can use the table and get a answer that has a natural logarithm in it, But how do you do it without a table?? I know that's a stupid question to ask, But this does not look complicated to do!!! I just don't know where to start. U-sub does not work! ...by parts? I can't see how that would work..I was maybe thinking Switching it to polar coordinates?...Or maybe a trig substitution?

Any help would be much appreciated!
• Oct 9th 2010, 01:28 PM
e^(i*pi)
From what I can see on Wolfram you should use a trig subsitution.

Unfortunately I don't know anything about them :(
• Oct 9th 2010, 01:42 PM
yeKciM
Quote:

Originally Posted by VkL
How do you do this integral without a table?

Integral [ Sqrt(4x^2 + 20)]

I got it down to

2* integral [ sqrt(x^2 + 5) ]

Now I can use the table and get a answer that has a natural logarithm in it, But how do you do it without a table?? I know that's a stupid question to ask, But this does not look complicated to do!!! I just don't know where to start. U-sub does not work! ...by parts? I can't see how that would work..I was maybe thinking Switching it to polar coordinates?...Or maybe a trig substitution?

Any help would be much appreciated!

integral &#40;1&#47;&#40;x&#94;2&#43;5&#41;&#94;&#40;1&#47; 2&#41;&#41; dx - Wolfram|Alpha

hit show steps :D

or just remember those 12 or so table integrals and you will not have any problem :D:D:D you don't even have to learn them as must, because you will while practice learn them without knowing that you know them :P hehehehe

$\displaystyle \int \frac {dx}{\sqrt{x^2\pm a^2 } } = \ln |x+\sqrt{x^2 \pm a^2 }| + C$
• Oct 9th 2010, 02:39 PM
Soroban
Hello, VkL!

Trig Substitution is the way to go . . . and it's still very messy!

Quote:

$\displaystyle \int\!\sqrt{4x^2 + 20}\:dx \;=\;2\!\!\int\!\sqrt{x^2+5}\:dx$

Let: . $x \,=\,\sqrt{5}\,\tan\theta \quad\Rightarrow\quad dx \:=\:\sqrt{5}\sec^2\!\theta\,d\theta$

And: . $\sqrt{x^2+5} \:=\:\sqrt{5\tan^2\!\theta + 5} \:=\:\sqrt{5(\tan^2\!\theta+1)} \:=\:\sqrt{5\sec^2\!\theta} \:=\:\sqrt{5}\sec\theta$

Substitute: . $\displaystyle 2\!\1\int(\sqrt{5}\sec\theta)(\sqrt{5}\sec^2\!\the ta\,d\theta) \;=\;10\!\!\int\!\sec^3\!\theta\,d\theta$

This is the dreaded secant-cubed integral.
. . It requires some tricky integration-by-parts.
But I memorized the formula years ago . . .

We have: . $10\cdot\frac{1}{2}\bigg[\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg] + C$

Back-substitute: . $\tan\theta \:=\:\dfrac{x}{\sqrt{5}} \:=\:\dfrac{opp}{adj}$

From Pythagorus: . $opp = x,\;adj = \sqrt{5} \quad\Rightarrow\quad hyp = \sqrt{x^2+5}$

. . Hence: . $\sec\theta = \dfrac{\sart{x^2+5}}{\sqrt{5}},\;\tan\theta = \dfrac{x}{\sqrt{5}}$

And we have: . $5\left[\dfrac{\sqrt{x^2+5}}{\sqrt{5}}\cdot\dfrac{x}{\sqrt {5}} + \ln\left|\dfrac{\sqrt{x^2+5}}{\sqrt{5}} + \dfrac{x}{\sqrt{5}}\right|\, \right] + C$

. . . . . . . . $=\;5\left[\dfrac{x\sqrt{x^2+5}}{5} + \ln\left|\dfrac{x+\sqrt{x^2+5}}{\sqrt{5}}\right|\, \right] + C$

. . . . . . . . $=\;x\sqrt{x^2+5} + 5\ln\left(\dfrac{x+\sqrt{x^2+5}}{\sqrt{5}}\right) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The above answer is absolutely correct,
. . but many textbooks simplify it beyond all recognition.

We have:

. . $x\sqrt{x^2+5} + 5\bigg[\ln\left(x + \sqrt{x^2+5}\,\right) - \ln\left(\sqrt{5}\right)\,\bigg] + C$

. . $=\;x\sqrt{x^2+5} + 5\ln\left(x+\sqrt{x^2+5}\,\right) - \underbrace{5\ln\left(\sqrt{5}\right)}_{\text{This is a constant!}} +\: C$

. . $=\;x\sqrt{x^2+5} + 5\ln\left(x+\sqrt{x^2+5}\,\right) + C$
• Oct 10th 2010, 05:12 AM
HallsofIvy
I would look at $\sqrt{4x^2+ 20}= 2\sqrt{x^2+ 5}$ and be reminded of " $a^2+ b^2= c^2$", the Pythagorean theorem. Or perhaps you will remember it better as " $sin^2(\theta)+ cos^2(\theta)= 1$", the trig identity derived from the Pythagorean theorem. Since that "5" is a constant and not a variable, divide the identity by $cos^2(\theta)$ to get $\frac{sin^2(\theta)}{cos^2(\theta)}+ 1= \frac{1}{cos^2(\theta)}$ or $tan^2(\theta)+ 1= sec^2(\theta)$.

That makes me think of the substitution $x= \sqrt{5}tan(\theta)$ so that $x^2+ 5= 5 tan^2(\theta)+ 5= 5(tan^2(\theta)+ 1)= 5 sec^2(\theta)$ and $\sqrt{x^2 +5}= \sqrt{5} sec(\theta)$.

Of course, if $x= \sqrt{5}tan(\theta)$ then $dx= \sqrt{5}sec^2(\theta)d\theta$ so we have
$2\int \sqrt{x^2+ 5} dx= 2\int \left(\sqrt{5}sec(\theta)\right)\left(\sqrt{5}sec^ 2(\theta) d\theta= 10\int sec^3(\theta) d\theta$ $= 10\int \frac{d\theta}{cos^3(\theta)}$.

Now that has $cos(\theta)$ to an odd power so we can use a standard "trick"- multiply both numerator and denominator inside the integral by cos(x) to get
$10\int \frac{ cos(\theta) d\theta}{cos^4(\theta)}= 10\int \frac{cos(\theta)d\theta}{(1- sin^2(\theta))^2}$

Now let $u= sin(\theta)$ so that $du= cos(\theta)d\theta$ and the integral becomes
$10\int \frac{du}{(1- u^2)^2}$

The denominator factors as $(1- u)^2(1+ u)^2$ so "partial fractions" can be used to integrate it.