Integration by substitution (1 Viewer)

steve001 · Apr 27, 2009

hey, does anybody know how to do this question
Its: by using the substitution of $x=sin\theta, show \int_{\frac{1}{root 2}}^{\frac{root 3 }{2}}\frac{1-2x^{2}}{x^{5}-x^{3}}dx$

shuning · Apr 27, 2009

回复: Integration by substitution

where is the question? xD

lychnobity · Apr 27, 2009

Yeah, kind of need a question to help with.

steve001 · Apr 27, 2009

umm do you know how to copy it from the latex thing?

shuning · Apr 27, 2009

回复: Re: Integration by substitution

draw it with MS paint and host it with image shack. that's what i always do b4 i knew how to use LaTeX xD

steve001 · Apr 27, 2009

by using the substitution of x= $sin\theta, show \int_{\frac{1}{root 2}}^{\frac{root 3 }{2}}\frac{1-2x^{2}}{x^{5}-x^{3}}dx$

there we go.

lychnobity · Apr 27, 2009

Alternatively, write "integrate [insert expression here]" using the substitution u = [blah], limits from _ to _

shuning · Apr 27, 2009

回复: Re: Integration by substitution

u sure the question is right?

annabackwards · Apr 27, 2009

Re: 回复: Re: Integration by substitution

I haven't used that latex thingo before and i'm not bothered trying.

I've uploaded what i've managed to do so far here

No idea where to go next.

By the way, i used $\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$ in one of my steps. (aha i used latex... so magical)

shuning · Apr 27, 2009

回复: Re: 回复: Re: Integration by substitution

lol @ anna's pro solution xDDDD

ROFL'ing

annabackwards · Apr 27, 2009

Re: 回复: Re: 回复: Re: Integration by substitution

Why thank you, shuning XD

Can someone tell me if i'm right? I haven't done 3U math in ~2 weeks aha.

Dumbledore · Apr 27, 2009

how did u get from integral[(cos(2@)/(sin^2(@)*sin(2@))] to the next step? u just substituted without integrating... i think

shuning · Apr 27, 2009

回复: Re: Integration by substitution

Dumbledore said:
how did u get from integral[(cos(2@)/(sin^2(@)*sin(2@))] to the next step? u just substituted without integrating... i think

see anna this person thinks the same as i do ^^

annabackwards · Apr 27, 2009

Dumbledore said:
how did u get from integral[(cos(2@)/(sin^2(@)*sin(2@))] to the next step? u just substituted without integrating... i think

Oh yeah i did too =="

Edited my original post so as to not confuse people. Hmm not sure where to go next but i'll continue to try.

@ Shuning - i know what you mean now. LOl aren't you supposed to be asleep? oO

Dumbledore · Apr 27, 2009

its acutally much easier to solve using partial fractions

annabackwards · Apr 27, 2009

Dumbledore said:
its acutally much easier to solve using partial fractions

Probably. I end up with 1/sinx^3.cosx - 2 tanx. So i can integrate the 2nd partial fraction but not the 1st. Do you have a solution?

Dumbledore · Apr 27, 2009

i'm bad at typing so i'll do the 1st partial fraction

S(dx/[x^5-x^3])
=S(dx/((x^3)(x-1)(x+1)))

= a/(x^3) + b/(x^2) + c/x + d/(x-1) + e/(x+1)

d = 1/2, e = 1/2, a = -1 [heavyside cover method]
a(x-1)(x+1)+b(x-1)(x+1)x+c(x-1)(x+1)x^2 + dx^3(x+1) + ex^3(x-1)
equating d3; 0=c+d+e -> c=-1
equating d2; 0=b+d-e -> b=0

and u can integrate from there

//hard to show working for heaviside cover method

shaon0 · Apr 27, 2009

steve001 said:
hey, does anybody know how to do this question
Its: by using the substitution of $x=sin\theta, show \int_{\frac{1}{root 2}}^{\frac{root 3 }{2}}\frac{1-2x^{2}}{x^{5}-x^{3}}dx$

Is the answer; 5/3 - ln(3)?

GUSSSSSSSSSSSSS · Apr 27, 2009

hmmm yes i got up to a stage to where it has integral of (cosecx)^3 - cosecx which i really cannot be bothered doing at the moment lol xD

Trebla · Apr 28, 2009

$x = \sin \theta \Rightarrow dx = \cos \theta d\theta\\x = \frac{\sqrt{3}}{2}\Rightarrow\theta = \frac{\pi}{3}\\x = \frac{1}{\sqrt{2}}\Rightarrow\theta = \frac{\pi}{4}\\\displaystyle\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \dfrac{1 - 2x^2}{x^5 - x^3}\,dx= \displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{1 - 2\sin^2\theta}{\sin^3\theta(\sin^2\theta - 1)}\cos\theta\,d\theta\\=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{1 - 2\sin^2\theta}{\sin^3\theta(-\cos^2\theta)}\cos\theta\,d\theta\\=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{2\sin^2\theta-1}{\sin^3\theta\cos\theta}\,d\theta$

$=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{1-2\cos^2\theta}{\sin^3\theta\cos\theta}\,d\theta\\=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{cosec^2\theta}{\sin\theta\cos\theta}\,d\theta-\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{2\cos^2\theta}{\sin^3\theta\cos\theta}\,d\theta\\=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{1+\cot^2\theta}{\sin\theta\cos\theta}\,d\theta-\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{2\cos\theta}{\sin^3\theta}\,d\theta\\ =\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{1}{\sin\theta\cos\theta}\,d\theta+\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\cos^2\theta}{\sin^3\theta\cos\theta}\,d\theta-\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{2\cos\theta}{\sin^3\theta}\,d\theta\\$

$=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\cos\theta}{\sin\theta\cos^2\theta}\,d\theta+\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\cos\theta}{\sin^3\theta}\,d\theta-\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{2\cos\theta}{\sin^3\theta}\,d\theta\\=\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\cos\theta}{\sin\theta\cos^2\theta}\,d\theta-\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\cos\theta}{\sin^3\theta}\,d\theta\\ =\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \dfrac{\sec^2\theta}{\tan\theta}\,d\theta-\displaystyle\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \cot\theta\,cosec^2\theta\,d\theta\\= \left [\ln(\tan\theta) \right ]_{\frac{\pi}{4}}^{\frac{\pi}{3}} + \left [\dfrac{\cot^2\theta}{2} \right ]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\\=\dfrac{1}{2}\ln 3-\dfrac{1}{3}$

That was hell...lol check my working...

Integration by substitution (1 Viewer)

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