integration (1 Viewer)

elv09 · Aug 26, 2009

find attached, two integration questions that i can't do.

thanks

Trebla · Aug 27, 2009

For a), there is a problem at x = π/2 in the integrated interval.

For b)
$\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\\\\b)I=\int_{0}^{2\pi}\dfrac{x}{2+\cos x}\,dx\\\\a = 2\pi\\ f(x)=\dfrac{x}{2+\cos x} \\\\\int_{0}^{2\pi}\dfrac{x}{2+\cos x}\,dx=\int_{0}^{2\pi}\dfrac{2\pi-x}{2+\cos(2\pi-x)}\,dx\\\\=\int_{0}^{2\pi}\dfrac{2\pi-x}{2+\cos x}\,dx\:\\\\$ as $\cos (2\pi-x)=\cos x \\\\=\displaystyle\int_{0}^{2\pi}\dfrac{2\pi}{2+\cos x}\,dx-\displaystyle\int_{0}^{2\pi}\dfrac{x}{2+\cos x}\,dx\\\\\Rightarrow \int_{0}^{2\pi}\dfrac{x}{2+\cos x}\,dx=\displaystyle\int_{0}^{2\pi}\dfrac{\pi}{2+\cos x}\,dx\\\\$

Use the substitution t = tan x/2 to evaluate the integral...

elv09 · Aug 27, 2009

the problem is, is that if you use the substitution t = tan(x/2), when changing the limits of integration, it will go from 0 --> 0

i.e. when x = 2pi --> t = 0
x = 0 --> t = 0
the substitution fails i think?

(that integration, after the manipulation using f(a-x) is the part im having trouble with)

untouchablecuz · Aug 27, 2009

intuitively i would say that S [0->0] (something) dx = 0, but is the S [0->0] (something) dx defined? can you integrate from 1->1 or from 2->2?

however, graphing y = 1/(2+cosx), i dont see how the area under the graph is 0. enlighten us trebla

Trebla · Aug 28, 2009

Ah yes, my bad. The substitution won't work because the interval contains x = π which makes the substitution tan x/2 undefined...

Trebla · Aug 28, 2009

$$In that case, try using the t-formula without formal substitution$\\\\\displaystyle\int_0^{a}\dfrac{1}{2+\cos x}\,dx=\displaystyle\int_0^{a}\dfrac{1}{2+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}\,dx\\\\=\displaystyle\int_0^{a}\dfrac{1+\tan^2\frac{x}{2}}{2+\tan^2\frac{x}{2}+1-\tan^2\frac{x}{2}}\,dx\\\\=\displaystyle\int_0^{a}\dfrac{1+\tan^2\frac{x}{2}}{3+\tan^2\frac{x}{2}}\,dx\\\\=2\displaystyle\int_0^{a}\dfrac{\frac{1}{2}\sec^2\frac{x}{2}}{3+\tan^2\frac{x}{2}}\,dx\\\\=\dfrac{2}{\sqrt{3}}\left [\tan^{-1}\left( \dfrac{\tan \frac{x}{2}}{\sqrt{3}} \right ) \right ]_0^{a}\: $ by reverse chain rule$\\\\= \dfrac{2}{\sqrt{3}}\tan^{-1}\left ( \dfrac{\tan \frac{a}{2}}{\sqrt{3}} \right)\\\\$ As $a \to \pi,\Rightarrow \dfrac{\tan \frac{a}{2}}{\sqrt{3} }\to \infty \Rightarrow \tan^{-1}\left ( \dfrac{\tan \frac{a}{2}}{\sqrt{3}} \right) \to \dfrac{\pi}{2}\\\\\therefore \displaystyle\int_0^{a}\dfrac{\pi}{2+\cos x}\,dx \to \dfrac{\pi^2}{\sqrt{3}}\:$ as $ a \to \pi$

Then redo integral from $\pi$ to $2\pi$ using similar method....

lolokay · Aug 28, 2009

tan pi -> infinity :O

what values of x is inverse tan defined for?

untouchablecuz · Aug 28, 2009

Trebla said:
$$In that case, try using the t-formula without formal substitution$\\\\\displaystyle\int_0^{a}\dfrac{1}{2+\cos x}\,dx=\displaystyle\int_0^{a}\dfrac{1}{2+\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}\,dx\\\\=\displaystyle\int_0^{a}\dfrac{1+\tan^2\frac{x}{2}}{2+\tan^2\frac{x}{2}+1-\tan^2\frac{x}{2}}\,dx\\\\=\displaystyle\int_0^{a}\dfrac{1+\tan^2\frac{x}{2}}{3+\tan^2\frac{x}{2}}\,dx\\\\=2\displaystyle\int_0^{a}\dfrac{\frac{1}{2}\sec^2\frac{x}{2}}{3+\tan^2\frac{x}{2}}\,dx\\\\=\dfrac{2}{\sqrt{3}}\left [\tan^{-1}\left( \dfrac{\tan \frac{x}{2}}{\sqrt{3}} \right ) \right ]_0^{a}\: $ by reverse chain rule$\\\\= \dfrac{2}{\sqrt{3}}\tan^{-1}\left ( \dfrac{\tan \frac{a}{2}}{\sqrt{3}} \right)\\\\$ As $a \to 2\pi,\Rightarrow \dfrac{\tan \frac{a}{2}}{\sqrt{3} }\to \infty \Rightarrow \tan^{-1}\left ( \dfrac{\tan \frac{a}{2}}{\sqrt{3}} \right) \to \dfrac{\pi}{2}\\\\\therefore \displaystyle\int_0^{a}\dfrac{\pi}{2+\cos x}\,dx \to \dfrac{\pi^2}{\sqrt{3}}\:$ as $ a \to 2\pi$

doesnt tan(a/2) -> 0 as a -> 2pi

untouchablecuz · Aug 28, 2009

lolokay said:
tan pi -> infinity :O

what values of x is inverse tan defined for?

-inf < x < inf

untouchablecuz · Aug 28, 2009

3unitz said:
his idea is fine, just let a approach pi instead:

S[0 -> 2pi] 1/[2 + cos(x)] dx

= 2 * S[0 -> pi] 1/[2 + cos(x)] dx (even about pi)

= 2 * lim[a -> pi] S[0 -> a] 1/[2 + cos(x)] dx

= 2*(pi/sqrt3)

ah ok

just one more thing

with my integration of 1/(2+cosx)

i end up with

pi/sqrt3 [the expression involving the inverse tan's][from 2pi -> 0]=
2pi/sqrt3 [the expression involving the inverse tan's][from pi -> 0]

but [the expression involving the inverse tan's][from pi -> 0] gives 0 upon evaluating. do you know what the problem is? is my integration incorrect?

untouchablecuz · Aug 28, 2009

ah ok, thanks mate

to remedy this, would i integrate from [0->a] and [2pi->a] as a approaches pi/2 ?

lolokay · Aug 28, 2009

$\int_{0}^{\pi }\frac{1}{2+cosx}dx\\ let\ u=2\pi -x\\ \int_{2\pi}^{\pi}\frac{1}{2+cos(2\pi -u}-du\\ =\int_{\pi}^{2\pi}\frac{1}{2+cosu}du\\ =\int_{\pi}^{2\pi}\frac{1}{2+cosx}dx\\$
just integrate from 0 to pi and double it
(should give the same answer as trebla's)

Trebla · Aug 28, 2009

woops...made a mistake...hopefully edit is correct now....

integration (1 Viewer)

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