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

1. anonymous

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

2. anonymous

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

3. anonymous

1+sinh^2 I mean

4. anonymous

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

5. anonymous

should we use tangent bc there is a plus sign

6. anonymous

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

7. anonymous

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

8. anonymous

9. anonymous

yah

10. anonymous

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

11. anonymous

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

12. anonymous

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,$\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta -\int\limits_{}{}\sec \theta \tan^2 \theta d \theta$

13. anonymous

$=\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$

14. anonymous

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,$\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta-\int\limits_{}{}\sec^3 \theta d \theta + \ln|\sec \theta + \tan \theta| + c$

15. anonymous

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

16. anonymous

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

17. anonymous

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.

18. anonymous

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.

19. anonymous

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

20. anonymous

i became a fan

21. anonymous

Awesome...cheers ;)

22. anonymous

You fine to finish it from here?

23. anonymous

yah ty

24. anonymous

No worries.