G
Guest
Guest
how do i do the integral (2, sqrt 2) of dx/x^2 times sqrt x^2-1
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Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
Re: Integral help!
Partial integration. Integrate the 1/x^2 factor. You then have to find the integral of 1/sqrt[x^2-1] which gives an arccosh(x).
Brit412 said:how do i do the integral (2, sqrt 2) of dx/x^2 times sqrt x^2-1
Partial integration. Integrate the 1/x^2 factor. You then have to find the integral of 1/sqrt[x^2-1] which gives an arccosh(x).
Hello, Brit412!
Trig Substitution . . .
Let \(\displaystyle x\,=\,\sec\theta\;\;\Rightarrow\;\;dx\,=\,\sec\theta\cdot\tan\theta\cdot d\theta\)
. . and: \(\displaystyle \:\sqrt{x^2\,-\,1}\:=\:\sqrt{\sec^2\theta\,-\,1}\:=\:\sqrt{\tan^2\theta} \:=\:\tan\theta\)
Substitute: \(\displaystyle \L\:\int\frac{\sec\theta\cdot\tan\theta\cdot d\theta}{\sec^2\theta\cdot\tan\theta} \;=\;\int\frac{d\theta}{\sec\theta} \;=\;\int\)\(\displaystyle \cos\theta\,d\theta\)
Can you finish it now?
\(\displaystyle \L\int^{\;\;\;2}_{\sqrt{2}}\frac{dx}{x^2\sqr{x^2\,-\,1}}\)
Trig Substitution . . .
Let \(\displaystyle x\,=\,\sec\theta\;\;\Rightarrow\;\;dx\,=\,\sec\theta\cdot\tan\theta\cdot d\theta\)
. . and: \(\displaystyle \:\sqrt{x^2\,-\,1}\:=\:\sqrt{\sec^2\theta\,-\,1}\:=\:\sqrt{\tan^2\theta} \:=\:\tan\theta\)
Substitute: \(\displaystyle \L\:\int\frac{\sec\theta\cdot\tan\theta\cdot d\theta}{\sec^2\theta\cdot\tan\theta} \;=\;\int\frac{d\theta}{\sec\theta} \;=\;\int\)\(\displaystyle \cos\theta\,d\theta\)
Can you finish it now?
G
Guest
Guest
yes i can...thank u much!!