integration (1 Viewer)

[azureus88]

Why is int [1/sqrt(x^2 +a^2)]dx = ln |x+sqrt(x^2 +a^2)| a standard integral. Is there like a proper way to derive it other than differentiating the RHS.

 

[shaon0]

Why is int [1/sqrt(x^2 +a^2)]dx = ln |x+sqrt(x^2 +a^2)| a standard integral. Is there like a proper way to derive it other than differentiating the RHS.

Yeah you can show this result by differentiating both sides. But i don't know the proof of it.

d/dx([1/sqrt(x^2 +a^2)] dx)) = d/dx (ln |x+sqrt(x^2 +a^2)|)
LHS= d/dx (ln |x+sqrt(x^2 +a^2)|)
Let u= x+sqrt(a^2+x^2)
LHS= d/dx(ln(u)
= [1/(2sqrt(a^2+x^2))*d/dx(a^2+x^2)+1]/(x+sqrt(a^2+x^2))
= ((x/sqrt(a^2+x^2)+1)/(x+sqrt(a^2+x^2))
= 1/sqrt(a^2+x^2)
= LHS

This is the only way i know how to do it.

 

[Iruka]

You can derive it using hyperbolic trig functions, which are not studied at HSC level. If you take maths at uni you should see it in first year.

http://en.wikipedia.org/wiki/Hyperbolic_trig_functions

 

[Iruka]

You can also derive it using the substitution x=a tan u, but you will need to know how to integrate sec u.

View attachment 17899

 

[azureus88]

thanks for the replies, they were all very helpful

 

[azureus88]

oh yeh, one more thing. is it important to put in the domains for the substitutions. like if u substitue x=asin@, is it really necessary to write
-pi/2 < @ < pi/2

 

[Trebla]

oh yeh, one more thing. is it important to put in the domains for the substitutions. like if u substitue x=asin@, is it really necessary to write
-pi/2 < @ < pi/2

If you have a definite integral, the restriction on the substitution is defined within the limits of the integral itself. If it is an indefinite integral, only natural restrictions need apply according to the original integrand and substitution. Normally, you don't need to actually state the restrictions of the substitutions.