Question

How do I calculate this integral?

Answer 1

\[\int\limits\limits_{2}^{5} \pi(\frac{ 1 }{ \sqrt{x^2 + 10} })^2\]

Answer 2

What's stopping you from canceling the square with the radical ?

Answer 3

Well I'm actually stuck on a certain step. \[\pi \int\limits_{2}^{5}\frac{ 1 }{ x^2 + 10 }\]

The bottom looks like something for the arctan thingy, but the top isn't du and I can't figure out how to make it du.

Answer 4

\[\Large \pi \int\limits\limits_{2}^{5} \frac{ dx }{ x^2 + 10} \]
Let \[\large x = \sqrt{10} \tan \theta\]

Answer 5

That \[\pi \int\limits_{2}^{5}\frac{ \sqrt{10}\sec^2 \theta }{ (\sqrt{10}\tan \theta)^2 + 10 } d \theta\]

Answer 6

Very nice, now just simplify (factor out the 10 in the denominator, then notice the trig identity). Then you'll see it's a very simple integral.

btw, you forgot to change the limits when you made the substitution.

Answer 7

\[\frac{ \pi }{ \sqrt{10} } [\arctan \frac{ \sqrt{10} }{ 2 } - \arctan \frac{ \sqrt{10} }{ 5 }]\]

Answer 8

That might be right, I'm too tired to check. You could always check on wolfram alpha or something.

Answer 9

Thank god you were not doing something related to the prime counting function...

Answer 10

that was my first thought haha!