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 October 27th, 2010, 06:15 PM #1 Senior Member   Joined: Oct 2010 Posts: 126 Thanks: 0 Integration Hi, I require help with this problem. I'm asking kindly for help. Hints... $\text{The integral I_n is defined by I_n= \int_0^{\frac{1}{2}\pi} \cos^n\,\theta\,d\theta\,(n \geq 2).}$ $\text{By writing \cos^n\,\theta = \cos^{n-1}\,\theta\,\cos\,\theta\, and using integration by parts, or otherwise show that I_n = \frac{n-1}{n}I_{n-2}. Hence, or otherwise, evaluate I_8.}$.
October 27th, 2010, 07:29 PM   #2
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Re: Integration

Quote:
 Originally Posted by MathematicallyObtuse Hi, I require help with this problem. I'm asking kindly for help. Hints... $\text{The integral I_n is defined by I_n= \int_0^{\frac{1}{2}\pi} \cos^n\,\theta\,d\theta\,(n \geq 2).}$ $\text{By writing \cos^n\,\theta = \cos^{n-1}\,\theta\,\cos\,\theta\, and using integration by parts, or otherwise show that I_n = \frac{n-1}{n}I_{n-2}. Hence, or otherwise, evaluate I_8.}$.
$I_n= \int_0^{\frac{1}{2}\pi} \cos^n\,\theta\,d\theta\,(n \geq 2).}$

$=sin{\theta}cos^{n-1}\theta \|_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}sin{\theta}\,d{cos^{n-1}\theta}$

$=(n-1)\int_0^{\frac{\pi}{2}}sin^{2}{\theta}\,{cos^{n-2}\theta}\,d\theta$

$=(n-1)\int_0^{\frac{\pi}{2}}(1-cos^{2}{\theta})\,{cos^{n-2}\theta}\,d\theta$

$=(n-1)\int_0^{\frac{\pi}{2}}{cos^{n-2}\theta}\,d\theta-(n-1)\int_0^{\frac{\pi}{2}}{cos^{n}\theta}\,d\theta$

$=(n-1)I_{n-2}-(n-1)I_n$

Therefor....

$I_n=(n-1)I_{n-2}-(n-1)I_n \longrightarrow I_{n}=\frac{n-1}{n}I_{n-2}$

So,

$I_8=\frac{8-1}{8}.I_6=\frac{8-1}{8}.\frac{6-1}{6}.I_4=......=\frac{8-1}{8}.\frac{6-1}{6}.\frac{4-1}{4}.\frac{2-1}{2}=\frac{105}{384}=0.2734....$

 October 27th, 2010, 07:51 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Integration We have: $I_n=\int^{\frac{\pi}{2}}_0 \cos^n\theta\,d\theta$ A definite integral can be evaluated using integration by parts in the following manner: $\int^b_a f(x)g'(x)\,dx=\left(f(b)g(b)-f(a)g(a)\right)-\int^b_a g(x)f#39;(x)\,dx$ We will let $g'(\theta)=\cos\theta$ and $f(\theta)=\cos^{n-1}\theta$, thus $g(\theta)=\sin\theta$ and $f'(\theta)=(n-1)(\cos\theta)^{n-2}(-\sin\theta)$ Putting it together, we have: $I_n=\left(\cos^{n-1}(\frac{\pi}{2})\sin(\frac{\pi}{2})-\cos^{n-1}(0)\sin(0)\right)+(n-1)\int^{\frac{\pi}{2}}_0 \cos^{n-2}\theta sin^2\theta \,d\theta$ The expression to the left of the integral is zero, and using the Pythagorean identity for $sin^2\theta$, we have: $I_n=(n-1)\int^{\frac{\pi}{2}}_0 \cos^{n-2}\theta(1-cos^2\theta)\,d\theta =(n-1)(I_{n-2}-I_n)$ Solve for $I_n$ $I_n+(n-1)I_n=(n-1)I_{n-2}$ $nI_n=(n-1)I_{n-2}$ $I_n=\dfrac{n-1}{n}I_{n-2}$ Thus, $I_8=\dfrac{7}{8}\cdot\dfrac{5}{6}\cdot\dfrac{3}{4} \cdot\dfrac{1}{2}=\dfrac{35}{128}$ edit: Sorry, stainburg, I didn't catch that you had already solved.
October 27th, 2010, 07:55 PM   #4
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Re: Integration

Quote:
 Originally Posted by MarkFL edit: Sorry, stainburg, I didn't catch that you had already solved.
Urh...haven't you ever slept? You are always online.

 October 27th, 2010, 08:04 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Integration I require very little sleep.
October 27th, 2010, 08:15 PM   #6
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Re: Integration

Quote:
 Originally Posted by MarkFL I require very little sleep.
You should take some physical exercise to help sleep

 November 2nd, 2010, 07:06 PM #7 Senior Member   Joined: Oct 2010 Posts: 126 Thanks: 0 Re: Integration Thanks. The next part may or not be a follow-up to the question. I worked it out. I just think it's unnecessarily tedious. Hmmm...ponders $\int_0^{\frac{\pi}{6}} cos^6\,\theta\,sin^2\,\theta\,\,d\theta$ $\text{I found the indefinite integral. Which gives:}\,\,\,\frac{120\,\theta\,+\,48\,\sin(2\thet a)\,-\,24\,\sin(4 \theta)\,-\,16\,\sin(6 \theta)\,-\,3\,\sin(8\theta)}{3072}+C$ Then for the definite integral: $\frac{120\,\frac{\pi}{6}\,+\,48\,\sin2(\frac{\pi}{ 6})\,-\,24\,\sin4(\frac{\pi}{6})\,-\,16\,\sin6(\frac{\pi}{6})\,-\,3\,\sin8(\frac{\pi}{6})}{3072}\,+\,C$ $= \frac{27\,\sqrt{3}\,+\,40\pi}{6144}$
 November 2nd, 2010, 07:42 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Integration You obtained the correct result (according to my TI-89) and in a much more elegant form. As a minor aside, you don't write the constant C after the result from a definite integral...it essentially gets subtracted away when you apply the F(b) - F(a) part of the anti-derivative form of the fundamental theorem of calculus...i.e. [F(b) + C] - [F(a) + C] = F(b) - F(a)
November 3rd, 2010, 01:26 AM   #9
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Re: Integration

Quote:
 Originally Posted by MarkFL You obtained the correct result (according to my TI-89) and in a much more elegant form. As a minor aside, you don't write the constant C after the result from a definite integral...it essentially gets subtracted away when you apply the F(b) - F(a) part of the anti-derivative form of the fundamental theorem of calculus...i.e. [F(b) + C] - [F(a) + C] = F(b) - F(a)
Thanks.

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