My Math Forum Really need help with these two integration problems.

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October 9th, 2012, 02:34 PM   #1
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Joined: Oct 2012

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Really need help with these two integration problems.

Hello

I am really confused by these integrals.

I know that this one requires a u-substitution first.
u=tan(x)
You then must write everything in terms of tangent.
[attachment=3:2dyo6alo]gif.latex0001.gif[/attachment:2dyo6alo]

This requires integration by parts.
[attachment=2:2dyo6alo]gif.latex0000002.gif[/attachment:2dyo6alo]

To save you some time:
[attachment=1:2dyo6alo]gif.latexfsdfsdfsfs.gif[/attachment:2dyo6alo]
and
[attachment=0:2dyo6alo]gif.latexfsfsaaaaaaa.gif[/attachment:2dyo6alo]

Attached Images
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 October 9th, 2012, 06:21 PM #2 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Really need help with these two integration problems. 1. Let $t=\tan{x}$, then $x=\arctan {t}$, $\sec^4 {x}=(1+t^2)^2$ $dx=\frac{dt}{1+t^2}$ $I=\int\frac{1+t^2}{8sqrt{-t^2-6t+7}}dt=\int\frac{1+t^2}{8sqrt{16-(t+3)^2}}dt$ $=\frac{1}{8}\int\frac{1}{sqrt{16-(t+3)^2}}dt+\frac{1}{8}\int\frac{t^2}{sqrt{16-(t+3)^2}}dt$ $=\frac{1}{8}\{\arcsin{\frac{t+3}{4}}+$t^2\arcsin{\frac{t+3}{4}}-2\int t\arcsin{\frac{t+3}{4}}dt$\}$ $=\frac{1}{8}\{\arcsin{\frac{t+3}{4}}+$t^2\arcsin{\frac{t+3}{4}}-\frac{1}{2}$$(t-9)sqrt{16-(t+3)^2}+2(t^2-17)\arcsin{\frac{t+3}{4}}$$$\}+C$ $=\frac{1}{8}$18\arcsin{\frac{t+3}{4}}+\frac{9-t}{2}\sqrt{7-6t-t^2}$+C$ Then substitute $t=\tan x$ into something, I think you can figure it out.
 October 9th, 2012, 09:25 PM #3 Newbie   Joined: Oct 2012 Posts: 10 Thanks: 0 Re: Really need help with these two integration problems. Thank you thank you thank you! Still stuck on the other problem. I know you do integration by parts but I can't get the solution. I know it's a messy problem.
 October 9th, 2012, 09:34 PM #4 Newbie   Joined: Oct 2012 Posts: 10 Thanks: 0 Re: Really need help with these two integration problems. I love how you did a t-substitution there.
October 9th, 2012, 10:31 PM   #5
Senior Member

Joined: Oct 2010
From: Changchun, China

Posts: 492
Thanks: 14

Re: Really need help with these two integration problems.

Quote:
 Originally Posted by brh27 Thank you thank you thank you! Still stuck on the other problem. I know you do integration by parts but i can't get the solution. I know its a messy problem.
You are welcome!

Don't worry, your problems are really time-consuming. I'm working on it!

 October 9th, 2012, 11:10 PM #6 Senior Member     Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Re: Really need help with these two integration problems. 2. $I=2\int(x+1)\ln[(x+1)(x^2+4x+13)]dx$ $=2\{\int(x+1)\ln(x+1)dx+\int(x+1)\ln(x^2+4x+13)dx\ }$ $=2\{[\frac{1}{2}(x+1)^2\ln(x+1)-\frac{1}{2}\int\frac{(x+1)^2}{x+1}dx]+[\frac{1}{2}(x+1)^2\ln(x^2+4x+13)-\frac{1}{2}\int\frac{(2x+4)(x+1)^2}{x^2+4x+13}dx]\}$ $=(x+1)^2\ln[(x+1)(x^2+4x+13)]-\int{(x+1)}dx-\int\frac{2x^3+8x^2+10x+4}{x^2+4x+13}dx$ $=(x+1)^2\ln[(x+1)(x^2+4x+13)]-\frac{1}{2}(x+1)^2-\int(2x-\frac{8(2x+4)}{x^2+4x+13}+\frac{36}{x^2+4x+13})dx$ $=(x+1)^2\ln[(x+1)(x^2+4x+13)]-\frac{1}{2}(x+1)^2-x^2+8\ln(x^2+4x+13)-12\arctan\frac{x+2}{3}+C$
 October 10th, 2012, 06:54 AM #7 Newbie   Joined: Oct 2012 Posts: 10 Thanks: 0 Re: Really need help with these two integration problems. Thanks so much again! That must have taken a long time. If I knew you in real life, I'd give you a cookie.
October 10th, 2012, 05:31 PM   #8
Senior Member

Joined: Oct 2010
From: Changchun, China

Posts: 492
Thanks: 14

Re: Really need help with these two integration problems.

Quote:
 Originally Posted by brh27 Thanks so much again! That must have taken a long time. If I knew you in real life, I'd give you a cookie.
Good idea, Thanks~

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