Author Topic: most awesome question thread :) (Read 9140 times) Tweet Share

superflya · « **on:** January 02, 2010, 10:12:53 pm »

$\int_{-2}^{-1}\frac {e^x}{1-e^x}\, dx$

Mod Edit: Fixed up incorrect LaTeX

GerrySly · « **Reply #1 on:** January 02, 2010, 10:21:53 pm »

Recognition...

$\begin{align*} \int_{-2}^{-1} \frac{e^x}{1-e^x}.dx&=-\int_{-2}^{-1} \frac{-e^x}{1-e^x}.dx\\ &=\left [-\log_e{(1-e^x)}\right ]_{-2}^{-1}\\ &=\log_e{(1-e^{-2})}-\log_e{(1-e^{-1})}\\ &=\log_e{\left ( \frac{1-e^{-2}}{1-e^{-1}}\right )}\\ &=\log_e{\left ( \frac{1+e}{e}\right )}\\ &=\log_e{(1+e)}-\log_e{e}\\ &=\log_e{(1+e)}-1 \end{align*}$

TrueTears · « **Reply #2 on:** January 02, 2010, 10:23:04 pm »

If inspection fails.

Let $u = e^x$

Then how do you integrate $\frac{u}{1-u}$ (long divide)

superflya · « **Reply #3 on:** January 02, 2010, 10:41:24 pm »

thanks gerry

getting stumped at the after the long division part TT :S

TrueTears · « **Reply #4 on:** January 02, 2010, 10:45:42 pm »

$\frac{u}{1-u} = \frac{-1}{u-1}-1$

The antiderivative of $\frac{-1}{u-1}$ is $-\log_e(u-1)$

superflya · « **Reply #5 on:** January 02, 2010, 10:52:16 pm »

legend

superflya · « **Reply #6 on:** January 02, 2010, 11:11:28 pm »

$U_{n}=\int_{0}^{\frac{\pi}{4}}\tan ^{n} x dx$

express $U_{n}+U_{n-2}$ in terms of n

hence, show $U_{6}=\frac{13}{15}-\frac{\pi}{4}$

Mod Edit: Merged triple post into one.

TrueTears · « **Reply #7 on:** January 02, 2010, 11:40:40 pm »

Quote from: superflya on January 02, 2010, 11:11:28 pm

$U_{n}=\int_{0}^{\frac{\pi}{4}}\tan ^{n} x dx$

express $U_{n}+U_{n-2}$ in terms of n

hence, show $U_{6}=\frac{13}{15}-\frac{\pi}{4}$

Mod Edit: Merged triple post into one.

$U_n+U_{n-2} = \int_0^{\frac{\pi}{4}} \tan^n(x)+\tan^{n-2}(x)\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)(1+\tan^2(x))\,dx$

$=\int_0^\frac{\pi}{4} \tan^{n-2}(x)\sec^2(x)\,dx$

Let $u=\tan(x)\Rightarrow \frac{du}{dx} = \sec^2(x)$

$dx = \frac{du}{\sec^2(x)}$

$\Rightarrow U_n+U_{n-2} = \int_0^{1} tan^{n-2}(x) du=\int_0^1 u^{n-2}du$

$\therefore U_n+U_{n-2} = \left[ \frac{u^{n-1}}{n-1}\right]_0^1 = \frac{1}{n-1}$

Hopefully I didn't make any mistakes, stupid computer keeps on freezing for like 10-15 seconds each min or so.

superflya · « **Reply #8 on:** January 02, 2010, 11:45:43 pm »

thats right

ur too quick for it TT

TrueTears · « **Reply #9 on:** January 02, 2010, 11:54:13 pm »

cool, do you still need help on the U_6, i think it should be fairly straightforward if you understood how to do the other part

superflya · « **Reply #10 on:** January 03, 2010, 12:00:31 am »

i got the answer but cood u show me ur workings, i probs used triple the amount of necessary steps

the essentials explanations are really getting to me, they show u s***t all
$\int \sin ^{4}x dx$
u use the double angle formulae for that right? i can do it using the books jibbed methods but how wood u do it?

TrueTears · « **Reply #11 on:** January 03, 2010, 12:12:16 am »

$\int \sin^4(x) dx = \int \left(-\frac{1}{2}\left(\cos(2x)-1\right)\right)^2 dx$

Is that how you did it? I also used double angle formula, it is the only spesh method.

~~Or you can use complex numbers... but that's outside the spesh course.~~

won't work for this one

superflya · « **Reply #12 on:** January 03, 2010, 12:20:12 am »

yea i ended up with that integration but took a while.
wanted to finish the 2 chapters on antidifferentiation and its apps in 2 days but doesnt look lyk happening

TrueTears · « **Reply #13 on:** January 03, 2010, 12:22:41 am »

Quote from: superflya on January 03, 2010, 12:20:12 am

yea i ended up with that integration but took a while.
wanted to finish the 2 chapters on antidifferentiation and its apps in 2 days but doesnt look lyk happening

Yeah it sure does, but it is right

Good work!

superflya · « **Reply #14 on:** January 03, 2010, 12:29:14 am »

partial fractions look so tempting in the next exercise but ceebs finishing this lol.

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