# Evaluate the integral by using substitution prior to integration by parts

• Jan 17th 2011, 06:22 PM
tennis0216
Evaluate the integral by using substitution prior to integration by parts
∫ (x^2)/ [square root (x^2 +20)] dx

this is another problem assigned in my homework and im just really confused on what to do.. on how to use both.. help appreciated!
• Jan 17th 2011, 06:24 PM
dwsmith
Quote:

Originally Posted by tennis0216
∫ (x^2)/ [square root (x^2 +20)] dx

this is another problem assigned in my homework and im just really confused on what to do.. on how to use both.. help appreciated!

Trig sub

$x=\sqrt{20}\tan^2(\theta)$

That is the only substitution I can think that will be useful.
• Jan 17th 2011, 06:40 PM
tennis0216
Quote:

Originally Posted by dwsmith
Trig sub

$x=\sqrt{20}\tan^2(\theta)$

That is the only substitution I can think that will be useful.

i dont even know how substitution would be usefull.. why cant we go directly to integration by parts in this problem?
• Jan 17th 2011, 06:41 PM
dwsmith
Quote:

Originally Posted by tennis0216
i dont even know how substitution would be usefull.. why cant we go directly to integration by parts in this problem?

I am just going by your directions.
• Jan 17th 2011, 06:54 PM
tennis0216
Quote:

Originally Posted by dwsmith
I am just going by your directions.

no no i am not saying you're wrong.. thank you for the advice by the way! .. im just saying in general.. cant this problem be solved by integration by parts only? why is substitution needed?
• Jan 17th 2011, 06:57 PM
dwsmith
Quote:

Originally Posted by tennis0216
no no i am not saying you're wrong.. thank you for the advice by the way! .. im just saying in general.. cant this problem be solved by integration by parts only? why is substitution needed?

You don't need integration by parts. My trig sub will work.
• Jan 18th 2011, 06:50 AM
HallsofIvy
Quote:

Originally Posted by tennis0216
no no i am not saying you're wrong.. thank you for the advice by the way! .. im just saying in general.. cant this problem be solved by integration by parts only? why is substitution needed?

How? What would you use for "u" and what for "dv"?
• Jan 18th 2011, 07:05 AM
Prove It
Quote:

Originally Posted by HallsofIvy
How? What would you use for "u" and what for "dv"?

What about $\displaystyle u = \frac{1}{2}x$ and $\displaystyle dv = \frac{2x}{x^2 + 20}\,dx$?
• Jan 18th 2011, 07:35 AM
HallsofIvy
Quote:

Originally Posted by dwsmith
Trig sub

$x=\sqrt{20}\tan^2(\theta)$

That is the only substitution I can think that will be useful.

I think you intended $x= \sqrt{20}tan(\theta)$. That is, the tangent is not squared.

That way $x^2+ 20= 20tan^2(\theta)+ 20= 20sec^2(\theta)$
and $dx= \sqrt{20}sec^2(\theta)d\theta$

The integral becomes
$\int \frac{20 tan^2(\theta)}{\sqrt{20}sec(\theta)}(sec^2(theta)d \theta)$
$\sqrt{20}\int tan^2(\theta)sec(\theta)d\theta$
which, switching to sine and cosine, becomes
$\sqrt{20}\int \frac{sin^2(\theta)}{cos^2(\theta)}\frac{1}{cos(\t heta)}d\theta= \sqrt{20}\int \frac{sin^2(\theta)}{cos^3(\theta)}d\theta$

Since that has an odd power of cosine, I would suggest multiplying both numerator and denominator by $cos(\theta)$, and then letting $u= sin(\theta)$ so that $du= cos(\theta)d\theta$.

$\sqrt{20}\int \frac{sin^2(\theta)}{cos^4(\theta)}cos(theta(d\the ta= \sqrt{20}\int \frac{sin^2(\theta)}{(1- sin^2(\theta))^2}cos(\theta)d\theta$
$= \sqrt{20}\int \frac{u^2}{(1- u^2)^2}du$
a rational integral that can by done by "partial fractions".
• Jan 18th 2011, 10:44 AM
SammyS
Quote:

Originally Posted by tennis0216
∫ (x^2)/ [square root (x^2 +20)] dx

this is another problem assigned in my homework and I'm just really confused on what to do.. on how to use both.. help appreciated!

Another approach to: $\displaystyle \int{{x^2}\over{\sqrt{x^2+20}}}\,dx$

I tried the substitution: $\displaystyle v={\sqrt{x^2+20} \quad \to \quad dv={{x}\over{\sqrt{x^2+20}}}\,dx\,,$ which left an x in the integrand, but it did give me an idea for integration by parts.

$\displaystyle \text{Let } dv={{x}\over{\sqrt{x^2+20}}}\,dx \quad \to \quad v={\sqrt{x^2+20} \,,$

$\displaystyle \text{and let } u=x\quad \to \quad du=dx\,.$

Then, $\displaystyle \int{{x^2}\over{\sqrt{x^2+20}}}\,dx$
$\displaystyle = x{\sqrt{x^2+20}-\int{\sqrt{x^2+20}\,dx$

The substitution: $\displaystyle x=\sqrt{20}\,\tan(t)\quad\to\quad dx=\sqrt{20}\,\sec^2(t)\,dt\,,$

gives: $\displaystyle \int{\sqrt{x^2+20}\,dx=20\int \sec^3(t)\,dt\,.$

This still leaves plenty to do.
• Jan 18th 2011, 11:47 AM
SammyS
$\displaystyle \int \sec^3(t)\,dt$

Integration by parts:

$\displaystyle \text{Let }\ u=\sec(t)\quad\to\quad du=\tan(t)\sec(t)\,dt\,,$

$\displaystyle \text{and let }\ dv=\sec^2(t)\,dt\quad\to\quad v=\tan(t)$

$\displaystyle \int \sec^3(t)\,dt$
$\displaystyle =\tan(t)\sec(t)-\int \tan^2(t)\sec(t)\,dt$

$\displaystyle =\tan(t)\sec(t)+\int \sec(t)\,dt-\int \sec^3(t)\,dt$

Therefore,     $\displaystyle 2\int \sec^3(t)\,dt=\tan(t)\sec(t)+\int \sec(t)\,dt$
• Jan 18th 2011, 10:35 PM
TheCoffeeMachine
This integral demonstrates the simplicity/efficiency of hyperbolic substitutions! (Cool)

Let $\displaystyle \varphi := \sinh^{-1}{\frac{x}{\sqrt{20}} \Rightarrow \frac{dx}{d\varphi} = \sqrt{20}\cosh{\varphi} \Rightarrow dx = \sqrt{20}\cosh{\varphi}\;{d\varphi}
$
, then:

$\displaystyle \rhd \int \frac{x^2}{\sqrt{x^2+20}}\;{dx} = \int \frac{\left(\sqrt{20}\sinh{\varphi}\right)^2\left( \sqrt{20}\cosh{\varphi}\right)}{\sqrt{\left(\sqrt{ 20}\sinh{\varphi}\right)^2+20}}\;{d\varphi} = \int \frac{\left({20}\sinh^2{\varphi}\right)\left(\sqrt {20}\cosh{\varphi}\right)}{\left(\sqrt{20}\cosh{\v arphi}\right)}\;{d\varphi}$
$\displaystyle = \int \left(10\cosh{2\varphi}-10\right) \;{d\varphi} = 5\sinh{2\varphi}-10{\varphi}+k$ $\displaystyle = 5\sinh\left(2\sinh^{-1}{\frac{x}{\sqrt{20}}\right)-10\sinh^{-1}{\frac{x}{\sqrt{20}}+k.$