Question

can someone show me the easiest way to integrate

Answer 1

\[\int\limits_{5}^{10}\sqrt{10-x^2}dx\]

\[\int\limits 1/\sqrt{x^2-81} dx x>9\]

Answer 2

trig substitution ?? have you learned it yt ?

Answer 3

familiar with these
when \[\large\rm a^2+x^2 \rightarrow x=a \tan \theta\]\[\large \rm a^2-x^2 \rightarrow x=a\sin \theta \]
\[\large \rm x^2 -a^2 \rightarrow x=a sec theta \theta \]

Answer 4

\[\int\limits_{5}^{10} \sqrt{10 +x^2}\]
which can be written as
\[\int\limits_{5}^{10} \sqrt{ (\sqrt{10})^2 -x^2}\]
now under the square root you have a^2-x^2 form
so therefore
let \[x=\sqrt{10} \sin \theta\]

Answer 5

any question so far?
btw i forgot dx there \[\large\rm \int\limits_{5}^{10} \sqrt{(\sqrt{10})^2-x^2} ~~dx\]
take the derivative of
x=sqrt{10}sin theta
and then replace x with sqrt{10}sin theta and dx with the derivative of x