Question

help with integral

Answer 1

\[\int\limits_{?}^{?} \sqrt{1+x^2}\]

Answer 2

using trigonometric substitution

Answer 3

try \(x=\tan u\)

Answer 4

so tan(theta)=x/sqrt(1)?

Answer 5

forgot the dx

Answer 6

I don't follow...
\(\tan\theta=x\) is the same thing as what I wrote, with \(\theta\) instead of \(u\)
you don't want to solve for \(\tan\theta\), you just want to substitute for \(x\) and \(dx\)

Answer 7

so substitute x^2 with tangent?

Answer 8

yes, well, substitute x with tangent, then square it
don't forget to sub dx too

Answer 9

okay and what about the 1

Answer 10

it stays a 1

Answer 11

i dont put a sqrt over it

Answer 12

well yes, you keep the expression as it was before
everything the same, except tan(u) where x was, and *something* where dx was (you still need to figure out what)

Answer 13

okay ill let you know if i ran into something confusing

Answer 14

cool, good luck

Answer 15

\[I=\int\limits \sqrt{1+x^2}*(1)dx=\sqrt{1+x^2}*x- \int\limits \frac{ 1 }{ 2 }\frac{ 2x }{ \sqrt{1+x^2} }*x ~dx+c\]
\[\int\limits \frac{ x^2 }{ \sqrt{1+x^2} }dx=\int\limits \frac{ 1+x^2-1 }{ \sqrt{1+x^2} }dx=\int\limits \sqrt{1+x^2}dx-\int\limits \frac{ 1 }{ \sqrt{1+x^2} }dx+c\]
put \(x=tan \theta,dx=sec^2 \theta d \theta\)
?

Answer 16

\[\int\limits \sec \theta~d \theta=\ln \left| \sec \theta+\tan \theta \right| \]