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<math> 48: x - \arctan{x} + c </math> | <math> 48: x - \arctan{x} + c </math> | ||
| − | <math> 50: - | + | <math> 50: ln(9) - 4</math> |
| − | <math> 56: \frac{\pi}{4} | + | <math> 56: \frac{\pi}{4} - \ln{2} </math> |
<math> 84 a) : -\cos(\theta)+\frac{1}{3}\cos^3(\theta)+c</math> | <math> 84 a) : -\cos(\theta)+\frac{1}{3}\cos^3(\theta)+c</math> | ||
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---[[User:Gbrizend|Gary Brizendine II]] 14:51, 7 October 2008 (UTC) | ---[[User:Gbrizend|Gary Brizendine II]] 14:51, 7 October 2008 (UTC) | ||
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| + | I have corrected the answer to numbers 50 and 56. | ||
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| + | --[[User:Jmason|John Mason]] 15:41, 7 October 2008 (UTC) | ||
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| + | - You guys are life-savers. I was doing them right, but on 50 and 56 I made stupid simple math errors like saying a half times two was a fourth...Thanks guys. | ||
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| + | --[[User:Klosekam|Klosekam]] 15:26, 8 October 2008 (UTC)Shouldn't the answer to number 56 be pi/2 instead of pi/4 because you have to bring out a 2 before you take the integral meaning that you have to multiply the first part of the answer from above by 2. | ||
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| + | * No, note that after you separate the fracion you should get: | ||
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| + | <math>\int_{0}^{1/2}\frac{2}{1+4x^2}dx-\int_0^{1/2}\frac{8x}{1+4x^2}dx</math> | ||
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| + | Now in the first integral substitute <math>v=2x</math> Therefore <math>dv=2dx</math> and when x=0, v=0 and x=1/2, v=1. | ||
| + | In the second integral substitute <math>u=1+4x^2</math> therefore <math>du=8xdx</math> and when x=0, u=1 and x=1/2, u=2 | ||
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| + | Now after substitution we have: | ||
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| + | <math>\int_0^1\frac{dv}{1+v^2}-\int_0^2\frac{du}{u}</math> | ||
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| + | This eliminates the 2 you might have factored out. From this it is easy to see that the integrals will involve inverse tangetn and natural logs.[[User:Jhunsber|Jhunsber]] | ||
Latest revision as of 13:23, 8 October 2008
I'm going to give my even answers so far just to compare and see if everyone else is getting the same. Some are a little weird in my opinion. I used a calculator on a few too. -By the way, go here and make sure your preferences are set right so it makes these easier to read.
$ 38: \pi $
$ 48: x - \arctan{x} + c $
$ 50: ln(9) - 4 $
$ 56: \frac{\pi}{4} - \ln{2} $
$ 84 a) : -\cos(\theta)+\frac{1}{3}\cos^3(\theta)+c $
$ 84b) : -\cos(\theta)+\frac{2}{3}\cos^3(\theta)-\frac{1}{5}\cos^5(\theta)+c $
---Gary Brizendine II 14:51, 7 October 2008 (UTC)
I have corrected the answer to numbers 50 and 56.
--John Mason 15:41, 7 October 2008 (UTC)
- You guys are life-savers. I was doing them right, but on 50 and 56 I made stupid simple math errors like saying a half times two was a fourth...Thanks guys.
--Klosekam 15:26, 8 October 2008 (UTC)Shouldn't the answer to number 56 be pi/2 instead of pi/4 because you have to bring out a 2 before you take the integral meaning that you have to multiply the first part of the answer from above by 2.
- No, note that after you separate the fracion you should get:
$ \int_{0}^{1/2}\frac{2}{1+4x^2}dx-\int_0^{1/2}\frac{8x}{1+4x^2}dx $
Now in the first integral substitute $ v=2x $ Therefore $ dv=2dx $ and when x=0, v=0 and x=1/2, v=1. In the second integral substitute $ u=1+4x^2 $ therefore $ du=8xdx $ and when x=0, u=1 and x=1/2, u=2
Now after substitution we have:
$ \int_0^1\frac{dv}{1+v^2}-\int_0^2\frac{du}{u} $
This eliminates the 2 you might have factored out. From this it is easy to see that the integrals will involve inverse tangetn and natural logs.Jhunsber
