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| − | I am having a little trouble figuring out what to do here. I know that <math> sqrt | + | I am having a little trouble figuring out what to do here. I know that <math> \sqrt{y} </math> can be the u value for arctan but what do you do with the other <math> \sqrt{y} </math>? --[[User:Klosekam|Klosekam]] 15:39, 8 October 2008 (UTC) |
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| + | * i had a little trouble with this one at first too. Think about it this way: if<math>u=\sqrt{y}</math> then <math>du=\frac{dx}{2\sqrt{y}}</math> which can also be written as <math>2du=\frac{dx}{\sqrt{y}}</math>. You can then substitute 2du for the dy over the square root of y. Then, you have to notice that if <math>u=\sqrt{y}</math> then <math>u^2=y</math> Use that to substitute for the other y. Now you should get something along the lines of: | ||
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| + | <math>\int\frac{6*2du}{1+u^2}</math> an easily-integrated integral. :) [[User:Jhunsber|Jhunsber]] | ||
Latest revision as of 13:30, 8 October 2008
I am having a little trouble figuring out what to do here. I know that $ \sqrt{y} $ can be the u value for arctan but what do you do with the other $ \sqrt{y} $? --Klosekam 15:39, 8 October 2008 (UTC)
- i had a little trouble with this one at first too. Think about it this way: if$ u=\sqrt{y} $ then $ du=\frac{dx}{2\sqrt{y}} $ which can also be written as $ 2du=\frac{dx}{\sqrt{y}} $. You can then substitute 2du for the dy over the square root of y. Then, you have to notice that if $ u=\sqrt{y} $ then $ u^2=y $ Use that to substitute for the other y. Now you should get something along the lines of:
$ \int\frac{6*2du}{1+u^2} $ an easily-integrated integral. :) Jhunsber
