MHB Further applications of integration

[ineedhelpnow]

so i know I've asked this question before but id really like a step by step walk through with a few questions. starting with $\int_{\pi/3}^{\pi} \ \sqrt{1+\frac{4}{x^2}},dx$

i know I am not showing any work but id like to see how to this can properly be done. thanks

wait never mind i think i got it

 

[ineedhelpnow]

i didnt put the integral sign for each step
$\int_{\pi/3}^{\pi} \ \sqrt{1+\frac{4}{x^2}},dx = \sqrt{\frac{x^2+4}{x^2}} = \frac{\sqrt{x^2+4}}{x}$
let $x=2\tan\left({\theta}\right)$ and $dx=2(\sec\left({\theta}\right))^{2}$
$\frac{\sqrt{4(\tan\left({\theta}\right))^2+4}}{2\tan\left({\theta}\right)} \times (\sec\left({\theta}\right))^{2}$
all i did then was simplify that until i got to $\frac{1}{2} \int_{\pi/3}^{\pi} \ \frac{1}{\sin\left({\theta}\right)} d\theta + 2\int_{\pi/3}^{\pi} \ \tan\left({\theta}\right) \times \sec\left({\theta}\right) d\theta$

what do i do now? do i do u-sub?

 

[MarkFL]

This is how I would work the problem:

We are given to evaluate:

$$I=\int_{\frac{\pi}{3}}^{\pi}\sqrt{1+\left(\frac{2}{x}\right)^2}\,dx$$

Make the substitution:

$$\frac{2}{x}=\tan(\theta)\implies x=2\cot(\theta)\,\therefore\,dx=-2\csc^2(\theta)\,d\theta$$

To change our limits, consider:

$$\theta=\tan^{-1}\left(\frac{2}{x}\right)$$

And so we now have:

$$I=2\int_{\tan^{-1}\left(\frac{2}{\pi}\right)}^{\tan^{-1}\left(\frac{6}{\pi}\right)} \sec(\theta)\csc^2(\theta)\,d\theta$$

Using the Pythagorean identity:

$$\csc^2(\theta)=1+\cot^2(\theta)$$

We may now write:

$$I=2\int_{\tan^{-1}\left(\frac{2}{\pi}\right)}^{\tan^{-1}\left(\frac{6}{\pi}\right)} \sec(\theta)+\cot(\theta)\csc(\theta)\,d\theta$$

Applying the FTOC, we obtain:

$$I=2\left[\ln\left|\sec(\theta)+\tan(\theta)\right|-\csc(\theta)\right]_{\tan^{-1}\left(\frac{2}{\pi}\right)}^{\tan^{-1}\left(\frac{6}{\pi}\right)}=2\left(\left(\ln\left(\frac{6+\sqrt{\pi^2+36}}{\pi}\right)-\frac{\sqrt{\pi^2+36}}{6}\right)-\left(\ln\left(\frac{2+\sqrt{\pi^2+4}}{\pi}\right)-\frac{\sqrt{\pi^2+4}}{2}\right)\right)$$

Simplifying, we get:

$$I=2\left(\ln\left(\frac{6+\sqrt{\pi^2+36}}{2+\sqrt{\pi^2+4}}\right)+\frac{3\sqrt{\pi^2+4}-\sqrt{\pi^2+36}}{6}\right)$$

 

[Prove It]

i didnt put the integral sign for each step
$\int_{\pi/3}^{\pi} \ \sqrt{1+\frac{4}{x^2}},dx = \sqrt{\frac{x^2+4}{x^2}} = \frac{\sqrt{x^2+4}}{x}$
let $x=2\tan\left({\theta}\right)$ and $dx=2(\sec\left({\theta}\right))^{2}$
$\frac{\sqrt{4(\tan\left({\theta}\right))^2+4}}{2\tan\left({\theta}\right)} \times (\sec\left({\theta}\right))^{2}$
all i did then was simplify that until i got to $\frac{1}{2} \int_{\pi/3}^{\pi} \ \frac{1}{\sin\left({\theta}\right)} d\theta + 2\int_{\pi/3}^{\pi} \ \tan\left({\theta}\right) \times \sec\left({\theta}\right) d\theta$

what do i do now? do i do u-sub?

In the other thread I told you step by step what to do, which was essentially identical to what you just posted here. I also went further and told you how to continue.

 

[ineedhelpnow]

@Prove It
I tried it this way to see if it would work because it seemed simple then I went back and did it your way again and I got stuck.

 

[Prove It]

@Prove It
I tried it this way to see if it would work because it seemed simple then I went back and did it your way again and I got stuck.

I told you the very next step already. Go back and re-read it.

 

[ineedhelpnow]

the substitution? I did that but I ended up getting lost. I found another way to do and it's pretty simple, the only problem I'm having with it is trying to simplify the final answer.

 

[ineedhelpnow]

$y=2\ln\left({x}\right) from \pi/3 to \pi$

$\int_{\pi/3}^{\pi} \ \sqrt{x+(\frac{2}{x})^2}dx = \int_{\pi/3}^{\pi} \ \sqrt{x+\frac{4}{x^2}}dx
= \int_{\pi/3}^{\pi} \ \sqrt{\frac{x^2+4}{x^2}}dx = \int_{\pi/3}^{\pi} \ \frac{\sqrt{x^2+2^2}}{x}dx
= [\sqrt{4+x^2}-2\ln\left({\frac{2+\sqrt{4+x^2}}{2}}\right)]_{\pi/3}^{\pi} = [\sqrt{4+\pi^2}-2\ln\left({\frac{2+\sqrt{4+\pi^2}}{2}}\right)]-[\sqrt{4+\pi^2/9}-2\ln\left({\frac{2+\sqrt{4+\pi^2/9}}{2}}\right)]$

is that correct?

 

[MarkFL]

Your final value for the definite integral is slightly different from the result I posted above, and I think I see why. According to the table of integrals in my old calculus textbook, there results:

$$\int\frac{\sqrt{a^2+u^2}}{u}\,du=\sqrt{a^2+u^2}-a\ln\left|\frac{a+\sqrt{a^2+u^2}}{u}\right|+C$$

It appears that you put $a$ in the denominator of the natural log function instead of $u$. Correcting this will give you the same result I gave above.

 

[ineedhelpnow]

oh that totally slipped past me. I didn't even notice, I'll fix it now.

 

[ineedhelpnow]

$\int_{\pi/3}^{\pi} \ \sqrt{x+(\frac{2}{x})^2}dx = \int_{\pi/3}^{\pi} \ \sqrt{x+\frac{4}{x^2}}dx
= \int_{\pi/3}^{\pi} \ \sqrt{\frac{x^2+4}{x^2}}dx = \int_{\pi/3}^{\pi} \ \frac{\sqrt{x^2+2^2}}{x}dx
= [\sqrt{4+x^2}-2\ln\left({\frac{2+\sqrt{4+x^2}}{x}}\right)]_{\pi/3}^{\pi} = [\sqrt{4+\pi^2}-2\ln\left({\frac{2+\sqrt{4+\pi^2}}{\pi}}\right)]-[\sqrt{4+\pi^2/9}-2\ln\left({\frac{2+\sqrt{4+\pi^2/9}}{\pi/3}}\right)]$

now is it right?

 

[MarkFL]

$\int_{\pi/3}^{\pi} \ \sqrt{x+(\frac{2}{x})^2}dx = \int_{\pi/3}^{\pi} \ \sqrt{x+\frac{4}{x^2}}dx
= \int_{\pi/3}^{\pi} \ \sqrt{\frac{x^2+4}{x^2}}dx = \int_{\pi/3}^{\pi} \ \frac{\sqrt{x^2+2^2}}{x}dx
= [\sqrt{4+x^2}-2\ln\left({\frac{2+\sqrt{4+x^2}}{x}}\right)]_{\pi/3}^{\pi} = [\sqrt{4+\pi^2}-2\ln\left({\frac{2+\sqrt{4+\pi^2}}{\pi}}\right)]-[\sqrt{4+\pi^2/9}-2\ln\left({\frac{2+\sqrt{4+\pi^2/9}}{\pi/3}}\right)]$

now is it right?

Yes, now it is equivalent to the value I gave, which I verified with W|A before posting.