Question

whats the integral of sqrt 1+ x^2

Answer 1

You have to make a substitution. Set x=sinh(theta) and take it from there.

Answer 2

\[\int\limits_{}{}\sqrt{1+x^2}dx=\int\limits_{}{}\sqrt{1=\sinh^2 \theta}.\cosh \theta d \theta=\int\limits_{}{}\cosh^2 \theta d \theta\]

Answer 3

1+sinh^2 I mean

Answer 4

\[\cosh^2 \theta = \frac{\cosh 2 \theta +1 }{2}\]where I've used\[\cosh^2 \theta + \sinh^2 \theta = \cosh2 \theta\]and \[\cosh^2 \theta - \sinh^2 \theta =1\]

Answer 5

should we use tangent bc there is a plus sign

Answer 6

Then your integral is,\[\frac{1}{2}\int\limits_{}{}(\cosh2 \theta +1) d \theta=\frac{1}{2}[\frac{1}{2}\sinh 2 \theta + \theta]+c\]

Answer 7

You can use tangent if you want. That's another way of doing it. I just find hyperbolic trigs easier.

Answer 8

Is that your preferred method?

Answer 9

yah

Answer 10

Okay...lol...then set x=tan(theta) and go...

Answer 11

Your integral would be, \[\int\limits_{}{}\sqrt{1+\tan^2 \theta}.\sec^2 \theta. d \theta\]\[=\int\limits_{}{}\sec^3 \theta d \theta\]

Answer 12

You can integrate this by parts. Set u = sec(theta) and dv = sec^2(theta). Then u = sec(theta)tan(theta) and v = tan(theta). Your integral becomes,\[\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta -\int\limits_{}{}\sec \theta \tan^2 \theta d \theta\]

Answer 13

\[=\sec \theta \tan \theta - \int\limits_{}{}\sec \theta (\sec^2 \theta -1) d \theta \]\[=\sec \theta \tan \theta - \int\limits_{}{}\sec^3 \theta d \theta+ \int\limits_{}{}\sec \theta d \theta\]

Answer 14

Now \[\int\limits_{}{}\sec \theta d \theta = \ln|\sec \theta + \tan \theta|+c\] (I'm assuming you don't have to show that), so that your integral equation becomes,\[\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta-\int\limits_{}{}\sec^3 \theta d \theta + \ln|\sec \theta + \tan \theta| + c\]

Answer 15

Add the sec^3(theta) integral to both sides, \[2\int\limits_{}{}\sec^3 \theta d \theta =\sec \theta \tan \theta + \ln|\sec \theta + \tan \theta|+c\]

Answer 16

and divide by 2 to get,\[\int\limits_{}{}\sec^3 \theta d \theta =\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln|\sec \theta + \tan \theta|+c\]

Answer 17

Now you have to use your relationship for x=tan(theta) and the relationship betwen sec(theta) and tan(theta) to find your result in terms of x.

Answer 18

Now you have to use your relationship for x=tan(theta) and the relationship betwen sec(theta) and tan(theta) to find your result in terms of x.

Answer 19

This site is causing me grief - won't post properly.

Answer 20

i became a fan

Answer 21

Awesome...cheers ;)

Answer 22

You fine to finish it from here?

Answer 23

yah ty

Answer 24

No worries.