Nov 11, 2009 #1 Roshdy Member level 3 Joined Nov 23, 2005 Messages 57 Helped 2 Reputation 4 Reaction score 0 Trophy points 1,286 Location Egypt Activity points 1,738 integrations how those integrals can be performed ∫ x (inv cosh x) dx ∫ (x^8 ) ln (x^5) dx ∫ (1/x) (x^2 -25)^1.5 dx
integrations how those integrals can be performed ∫ x (inv cosh x) dx ∫ (x^8 ) ln (x^5) dx ∫ (1/x) (x^2 -25)^1.5 dx
Nov 12, 2009 #2 K katko Member level 4 Joined Feb 22, 2009 Messages 73 Helped 5 Reputation 10 Reaction score 0 Trophy points 1,286 Location Durham, USA Activity points 1,750 I would start with integration by parts. See: https://en.wikipedia.org/wiki/List_of_integrals_of_inverse_hyperbolic_functions for inverse cosh, then you can do the first. same with the second ln(a^b)=b*ln(a). Same with the third.
I would start with integration by parts. See: https://en.wikipedia.org/wiki/List_of_integrals_of_inverse_hyperbolic_functions for inverse cosh, then you can do the first. same with the second ln(a^b)=b*ln(a). Same with the third.
Nov 13, 2009 #3 P Peter Nachtwey Newbie level 4 Joined Nov 10, 2009 Messages 6 Helped 2 Reputation 4 Reaction score 0 Trophy points 1,281 Location Vancouver, WA Activity points 1,335 The easy way is to download wxMaxmia from Sourceforge. wxMaxima can solve these easily. integrate(x*acosh(x),x)$ grind(%); x^2*acosh(x)/2-(log(2*sqrt(x^2-1)+2*x)/2+x*sqrt(x^2-1)/2)/2$
The easy way is to download wxMaxmia from Sourceforge. wxMaxima can solve these easily. integrate(x*acosh(x),x)$ grind(%); x^2*acosh(x)/2-(log(2*sqrt(x^2-1)+2*x)/2+x*sqrt(x^2-1)/2)/2$
Nov 14, 2009 #4 B bednyk Newbie level 4 Joined Mar 24, 2008 Messages 5 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,281 Activity points 1,309 You still need some help with this...just let me know. I can explain you how to solve it - in mathematical way.
You still need some help with this...just let me know. I can explain you how to solve it - in mathematical way.