integral with sqrt

**ZOOZ** · January 9th 2011, 01:06 AM

in the picture you can see the problem, i did a little bit of the way to solve it but i stuck now.

hope you can help

thanks.

**ZOOZ** · January 9th 2011, 01:07 AM

by the way 1 of the answer is between A-F.

thanks.

**TheCoffeeMachine** · January 9th 2011, 01:09 AM

For $\displaystyle \int_{\sqrt{2}}^{\sqrt{6}}\frac{1}{\left(\frac{x}{ \sqrt{2}}\right)^2+1}\;{dx}$ , let $y = \frac{x}{\sqrt{2}}$ , then use:

$\displaystyle \int\frac{1}{y^2+1}\;{dy} = \arctan{y}+k.$

**Jameson** · January 9th 2011, 01:30 AM

Another way to do it if you can't remember the formulas is through trig substitution. Draw a right triangle such that $\displaystyle \tan( \theta) = \left( \frac{x}{\sqrt{2}} \right)$ . Find $d \theta$ in terms of x then substitute everything. You'll be left with a nice cancellation and an easy integral. Then back substitute. You'll get the same answer that TheCoffeeMachine is leading you to.

**ZOOZ** · January 9th 2011, 03:15 AM

hi thanks to both of you, so after i Calculate i find the answer is E .

can you Please just tell me if i correct?

thanks.

**TheCoffeeMachine** · January 9th 2011, 04:25 AM

Originally Posted by ZOOZ

So after i Calculate i find the answer is E.

Correct!

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