∫ (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, 06:22 PMtennis0216Evaluate 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 PMdwsmith
- Jan 17th 2011, 06:40 PMtennis0216
- Jan 17th 2011, 06:41 PMdwsmith
- Jan 17th 2011, 06:54 PMtennis0216
- Jan 17th 2011, 06:57 PMdwsmith
- Jan 18th 2011, 06:50 AMHallsofIvy
- Jan 18th 2011, 07:05 AMProve It
- Jan 18th 2011, 07:35 AMHallsofIvy
I think you intended $\displaystyle x= \sqrt{20}tan(\theta)$. That is, the tangent is not squared.

That way $\displaystyle x^2+ 20= 20tan^2(\theta)+ 20= 20sec^2(\theta)$

and $\displaystyle dx= \sqrt{20}sec^2(\theta)d\theta$

The integral becomes

$\displaystyle \int \frac{20 tan^2(\theta)}{\sqrt{20}sec(\theta)}(sec^2(theta)d \theta)$

$\displaystyle \sqrt{20}\int tan^2(\theta)sec(\theta)d\theta$

which, switching to sine and cosine, becomes

$\displaystyle \sqrt{20}\int \frac{sin^2(\theta)}{cos^2(\theta)}\frac{1}{cos(\t heta)}d\theta= \sqrt{20}\int \frac{sin^2(\theta)}{cos^3(\theta)}d\theta$

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

$\displaystyle \sqrt{20}\int \frac{sin^2(\theta)}{cos^4(\theta)}cos(theta(d\the ta= \sqrt{20}\int \frac{sin^2(\theta)}{(1- sin^2(\theta))^2}cos(\theta)d\theta$

$\displaystyle = \sqrt{20}\int \frac{u^2}{(1- u^2)^2}du$

a rational integral that can by done by "partial fractions". - Jan 18th 2011, 10:44 AMSammyS
Another approach to: $\displaystyle \displaystyle \int{{x^2}\over{\sqrt{x^2+20}}}\,dx$

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

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

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

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

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

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

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

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

Integration by parts:

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

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

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

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

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

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

Let $\displaystyle \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:

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

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