Evaluate the integral by using substitution prior to integration by parts

**tennis0216** · January 17th 2011, 06:22 PM

∫ (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!

**dwsmith** · January 17th 2011, 06:24 PM

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.

**tennis0216** · January 17th 2011, 06:40 PM

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?

**dwsmith** · January 17th 2011, 06:41 PM

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.

**tennis0216** · January 17th 2011, 06:54 PM

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?

**dwsmith** · January 17th 2011, 06:57 PM

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.

**HallsofIvy** · January 18th 2011, 06:50 AM

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"?

**Prove It** · January 18th 2011, 07:05 AM

Originally Posted by HallsofIvy

How? What would you use for "u" and what for "dv"?

What about $\displaystyle u = \frac{1}{2}x$ and $\displaystyle dv = \frac{2x}{x^2 + 20}\,dx$ ?

**HallsofIvy** · January 18th 2011, 07:35 AM

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".

**SammyS** · January 18th 2011, 10:44 AM

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.

**SammyS** · January 18th 2011, 11:47 AM

$\displaystyle \int \sec^3(t)\,dt$

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

**TheCoffeeMachine** · January 18th 2011, 10:35 PM

This integral demonstrates the simplicity/efficiency of hyperbolic substitutions!

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

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

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