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February 22nd, 2014, 07:58 AM   #1
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Tricky integrals

I need to compute the following integrals. They are really tricky ones? I have solved the first one so far (lol). I tried wolframalpha but you need a pro account to see the full solution. I have looked into different books but i have not found similar ones yet . Can anybody help me?
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 February 22nd, 2014, 08:26 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Tricky integrals Most of them can be done by substitution or "integration by parts". These are the main two integration techniques, so I'm not sure where you've been looking if you haven't found these.
 February 22nd, 2014, 11:58 AM #3 Newbie   Joined: Feb 2014 Posts: 7 Thanks: 0 Re: Tricky integrals 8. The definite integral from 0 to infinity of -xe^x-e^-x+constant What do i need to with infinity ?
 February 22nd, 2014, 01:00 PM #4 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Tricky integrals You can integrate from 0 to a and take the limit as a -> infinity.
February 23rd, 2014, 12:04 AM   #5
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Re: Tricky integrals

So is the nr 8 right?
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 February 23rd, 2014, 04:23 AM #6 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Re: Tricky integrals The correct answer is 1. It seems that you have errors in taking the limit.
February 23rd, 2014, 06:03 AM   #7
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Re: Tricky integrals

I have solved 1,2 and 8.
I found the solution for the 6th. But i don't understand it
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 February 23rd, 2014, 06:27 AM #8 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Tricky integrals You can integrate cos(x)^3 using a substitution u = sin(x): $\int cos^3(x)dx= \int cos^2(x)cos(x)dx = \int (1- sin^2(x))cos(x) dx$ $= \int (1 - u^2)du = u - \frac13 u^3 + C = sin(x) - \frac13 sin^3(x) + C$ And, you can use the same technique for higher odd powers: $\int cos^{2n+1}(x)dx= \int (cos^2(x))^ncos(x)dx = \int (1- sin^2(x))^n cos(x) dx$ $= \int (1 - u^2)^n du = ...$ Where you can finish things off using the binomial expansion.
 February 23rd, 2014, 06:58 AM #9 Newbie   Joined: Feb 2014 Posts: 7 Thanks: 0 Re: Tricky integrals I had the same solution already on my paper but i thought it could be wrong (lol). Isn't the 8th one right? Don't look at the last line. The arrow is showing how i opened the brackets.
 February 23rd, 2014, 07:17 AM #10 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Tricky integrals The answer to 8 is 1. It's never a good idea to put in $\infty$ as though it is a number. The correct approach is: $\int_0^{\infty} x exp {-x} dx= \lim_{a \rightarrow \infty} \int_0^{a} x exp {-x} dx$ $= \lim_{a \rightarrow \infty} [-x exp {-x}]_0^{a} + \int_0^a exp {-x} dx = \lim_{a \rightarrow \infty} [-a exp {-a}] - [exp {-x} ]_0^a$ $= \lim_{a \rightarrow \infty} [-a exp {-a} - exp {-a} + 1] = 1 \ (as \ \lim_{a \rightarrow \infty} -a exp {-a} - exp {-a} = 0)$ $\frac{1}{\infty} \ and \ exp {\infty}$ are meaningless.

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