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anonymous
 5 years ago
whats the integral of sqrt 1+ x^2
anonymous
 5 years ago
whats the integral of sqrt 1+ x^2

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to make a substitution. Set x=sinh(theta) and take it from there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}{}\sqrt{1+x^2}dx=\int\limits_{}{}\sqrt{1=\sinh^2 \theta}.\cosh \theta d \theta=\int\limits_{}{}\cosh^2 \theta d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\cosh^2 \theta = \frac{\cosh 2 \theta +1 }{2}\]where I've used\[\cosh^2 \theta + \sinh^2 \theta = \cosh2 \theta\]and \[\cosh^2 \theta  \sinh^2 \theta =1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0should we use tangent bc there is a plus sign

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then your integral is,\[\frac{1}{2}\int\limits_{}{}(\cosh2 \theta +1) d \theta=\frac{1}{2}[\frac{1}{2}\sinh 2 \theta + \theta]+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can use tangent if you want. That's another way of doing it. I just find hyperbolic trigs easier.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that your preferred method?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...lol...then set x=tan(theta) and go...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your integral would be, \[\int\limits_{}{}\sqrt{1+\tan^2 \theta}.\sec^2 \theta. d \theta\]\[=\int\limits_{}{}\sec^3 \theta d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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,\[\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta \int\limits_{}{}\sec \theta \tan^2 \theta d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[=\sec \theta \tan \theta  \int\limits_{}{}\sec \theta (\sec^2 \theta 1) d \theta \]\[=\sec \theta \tan \theta  \int\limits_{}{}\sec^3 \theta d \theta+ \int\limits_{}{}\sec \theta d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now \[\int\limits_{}{}\sec \theta d \theta = \ln\sec \theta + \tan \theta+c\] (I'm assuming you don't have to show that), so that your integral equation becomes,\[\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta\int\limits_{}{}\sec^3 \theta d \theta + \ln\sec \theta + \tan \theta + c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Add the sec^3(theta) integral to both sides, \[2\int\limits_{}{}\sec^3 \theta d \theta =\sec \theta \tan \theta + \ln\sec \theta + \tan \theta+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and divide by 2 to get,\[\int\limits_{}{}\sec^3 \theta d \theta =\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln\sec \theta + \tan \theta+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This site is causing me grief  won't post properly.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You fine to finish it from here?
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