# Math Help - Proofs: definite integrals

1. ## Proofs: definite integrals

May I have a little help here please?

a.) Prove that $\int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).$

b.) Show that $\int_{7}^{10} \frac{7}{x^2 - 5x - 6}\,dx = \ln \bigg(\frac{32}{11}\bigg).$

2. Originally Posted by Hellbent

a.) Prove that $\int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).$
You are going to need a substitution here me thinks. Maybe $x = \tan \theta$

Originally Posted by Hellbent

b.) Show that $\int_{7}^{10} \frac{7}{x^2 - 5x - 6}\,dx = \ln \bigg(\frac{32}{11}\bigg).$
With this guy make some partial fractions i.e. $x^2-5x-6 = (x-6)(x+1)$

3. Originally Posted by Hellbent
May I have a little help here please?

a.) Prove that $\int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).$
$\int_0^2 x^2(x^2 + 4)^\frac{1}{2}$

$\text {let u} = x^2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = (x^2 + 4)^\frac{1}{2}$

$v = \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = 2x$

$=\frac{2x^2}{3}(x^2 + 4)^\frac{3}{2} = \int \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,2x$

$\text{let u} = 2x\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = \frac{2}{3}(x^2 + 4)^\frac{3}{2}$

$v = \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 2$

$= \frac{2x^2}{15}(x^2 + 4)^\frac{3}{2} - \frac{8x}{15}(x^2 + 4)^\frac{5}{2} \int \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\dot\,\,2$

$\text{let u} = 2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = ....$

$v = \frac{8}{15.7}(x^2 + 4)^\frac{7}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 0$

Gave up here. I believe I'm going up a blind alley.

4. Originally Posted by pickslides
With this guy make some partial fractions i.e. $x^2-5x-6 = (x-6)(x+1)$
$\text{Show that} \int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \ln\bigg(\frac{32}{11}\bigg)$
$\int_7^{10} \frac{7}{x^2 - 5x - 6}\,\,dx = \frac{A}{(x - 6)} + \frac{B}{(x + 1)}$

$Ax + A + Bx - 6B$
$A + B = 0, A - 6b = 7$
$A = -B\,\,\,\,\,(-B) - 6B = 7$
$\therefore A = 1\,\,\,\,\,-7B = 7\,\,\,B = -1$

$\frac{1}{(x - 6)} + \frac{1}{x + 1}$
$
\bigg[\ln|x - 6| - \ln|x + 1|\bigg]_7^{10} + C$

$\ln |\frac{x - 6}{x + 1}| - \ln |\frac{x - 6}{x + 1}|$
$\ln |\frac{4}{11}| - \ln |\frac{1}{8}|$

$\ln \bigg(\frac{4}{11}\div{\frac{1}{8}}\bigg)$

$\ln\bigg(\frac{4}{11} \times 8\bigg) = \ln \bigg(\frac{32}{11}\bigg)$

5. Originally Posted by Hellbent
a.) Prove that $\int_0^2 x^2\, \sqrt{(x^2 + 4)}\,\,dx = 6\sqrt{2} - \ln(3 + 2\sqrt{2}).$
Let $x = 2\sinh{\varphi}$, then $dx = 2\cosh{\varphi}\;{d\varphi}$. When $x = 0, ~ \varphi = 0$, and when [LaTeX ERROR: Convert failed] .
Thus $I = \int_{0}^{2} x^2\sqrt{x^2+4}\;{dx} = \int_{\varphi = 0}^{\text{arcsinh(1)}}(2\sinh{\varphi})^2\sqrt{(2\ sinh{\varphi})^2+4}(2\cosh{\varphi})\;{d\varphi}$
[LaTeX ERROR: Convert failed] (by by-parts).
$= 6\sqrt{2} - \ln(3 + 2\sqrt{2})$ (by using the identity $\text{arcsinh(t)} = \ln\left\{t + \sqrt{t^2 + 1}\right\}$ etc). Done!

b.) Show that $\int_{7}^{10} \frac{7}{x^2 - 5x - 6}\,dx = \ln \bigg(\frac{32}{11}\bigg).$
Since $x^2-5x-6 = (x-6)(x+1)$, let $t = \frac{x-6}{x+1}$, then $\frac{dt}{dx} = \frac{7}{(x+1)^2} ~ \Rightarrow dx = \frac{(x+1)^2}{7}dt.$
When $x = 7, ~ t = \frac{1}{8}$. When $x = 10$, $t = \frac{4}{11}$. Thus $\int_{7}^{10}\frac{7}{(x-6)(x+1)}\;{dx} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{7(x+1)^2}{7 (x-6)(x+1)}\;{dt}$
$= \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{x+1}{x-6}\;{dt} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{1}{t}\;{dt} = [\ln{t}]_{\frac{1}{8}}^{\frac{4}{11}}$ $= \ln\left(\frac{4}{11}\right)-\ln\left(\frac{1}{8}\right) = \ln\left(\frac{4}{11}\times\frac{8}{1}\right) = \ln\left(\frac{32}{11}\right)$.

6. Originally Posted by Hellbent
$\int_0^2 x^2(x^2 + 4)^\frac{1}{2}$

$\text {let u} = x^2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = (x^2 + 4)^\frac{1}{2}$

$v = \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = 2x$

$=\frac{2x^2}{3}(x^2 + 4)^\frac{3}{2} = \int \frac{2}{3}(x^2 + 4)^\frac{3}{2}\,\,2x$

$\text{let u} = 2x\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = \frac{2}{3}(x^2 + 4)^\frac{3}{2}$

$v = \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 2$

$= \frac{2x^2}{15}(x^2 + 4)^\frac{3}{2} - \frac{8x}{15}(x^2 + 4)^\frac{5}{2} \int \frac{4}{15}(x^2 + 4)^\frac{5}{2}\,\,\dot\,\,2$

$\text{let u} = 2\,\,\,\,\,\,\,\,\,\frac{dv}{dx} = ....$

$v = \frac{8}{15.7}(x^2 + 4)^\frac{7}{2}\,\,\,\,\,\,\,\,\,\frac{du}{dx} = 0$

Gave up here. I believe I'm going up a blind alley.
pickslides suggested a trig substitution, the appropriate one is $x = 2 \tan \theta$, so do it that way. Or if you've learned about hyperbolic trig substitution, TheCoffeeMachine gave you the solution.

7. Originally Posted by Jhevon
pickslides suggested a trig substitution, the appropriate one is $x = 2 \tan \theta$.
Yes, I saw it. I can't do it his way. What's the reasoning behind giving me $x = 2 \tan \theta$ instead of using the already given $x = \tan \theta$?

8. Originally Posted by TheCoffeeMachine
Since $x^2-5x-6 = (x-6)(x+1)$, let $t = \frac{x-6}{x+1}$, then $\frac{dt}{dx} = \frac{7}{(x+1)^2} ~ \Rightarrow dx = \frac{(x+1)^2}{7}dt.$
When $x = 7, ~ t = \frac{1}{8}$. When $x = 10$, $t = \frac{4}{11}$. Thus $\int_{7}^{10}\frac{7}{(x-6)(x+1)}\;{dx} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{7(x+1)^2}{7 (x-6)(x+1)}\;{dt}$
$= \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{x+1}{x-6}\;{dt} = \int_{\frac{1}{8}}^{\frac{4}{11}}\frac{1}{t}\;{dt} = [\ln{t}]_{\frac{1}{8}}^{\frac{4}{11}}$ $= \ln\left(\frac{4}{11}\right)-\ln\left(\frac{1}{8}\right) = \ln\left(\frac{4}{11}\times\frac{8}{1}\right) = \ln\left(\frac{32}{11}\right)$.
Why did you substitute $\frac{x-6}{x+1}$ for t?
Neither do I get this part: $\frac{dt}{dx} = \frac{7}{(x+1)^2}$?!
It seems like you pulled it out of thin air. Where did the 7 and (x + 1)^2 come from?

Thanks.

9. Originally Posted by Hellbent
Yes, I saw it. I can't do it his way. What's the reasoning behind giving me $x = 2 \tan \theta$ instead of using the already given $x = \tan \theta$?
Because the objective here is to get rid of the radical, you need to find a function that makes the
expression under the radical a square. If you put $x = \tan{t}$, then the expression under the radical
becomes $(\tan{t})^2+4 = \tan^2{t}+4$, but $x = 2\tan{t}$ gives $4\tan^2{t}+4 = 4(\tan^2{t}+1)$
which is equal to $2^2\sec^2{t}$ and $\sqrt{2^2\sec^2{t}} = 2\sec{t}$... it gets rid of the $\sqrt$ (the other one doesn't).

Originally Posted by Hellbent
Why did you substitute $\frac{x-6}{x+1}$ for t?
It's just a substitution that works and makes an alternative to partial fractions. Whenever you have
integral of the form $\int\frac{k}{(ax+b)(cx+d)}\;{dx}$, you can let $t = \frac{ax+b}{cx+d}$ or $t = \frac{cx+d}{ax+b}$ (doesn't matter which).

Neither do I get this part: $\frac{dt}{dx} = \frac{7}{(x+1)^2}$?!
It seems like you pulled it out of thin air. Where did the 7 and (x + 1)^2 come from?
You have $t = \frac{x-6}{x+1}$. Differentiating it with respect to $x$ using the quotient rule, we have:

$\displaystyle \frac{dt}{dx} = \frac{(x-6)'(x+1)-(x-6)(x+1)'}{(x+1)^2} = \frac{x+1-(x-6)}{(x+1)^2} = \frac{7}{(x+1)^2}$.

10. Thank you, devourer of coffee