ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: yang_dong on August 22, 2014, 05:27:14 pm

Title: in need of help :(
Post by: yang_dong on August 22, 2014, 05:27:14 pm

how would i integrate:
(1/x)(sqrt((x^4+1)/x^4))

Title: Re: in need of help :(
Post by: tlbdrummerkid on August 28, 2014, 09:43:37 am

Let x^2 =tanu
2xdx=(secu)^2du
2x^2dx/x= (secu)^2du
(1/x)dx= (secu)^2du/(2x^2)
(1/x)dx= (secu)^2du/(2tanu)
(1/x)dx= 1/2cotudu

∫(1/x)(sqrt((x^4+1)/x^4))dx
=∫((1/2)cotu)(sqrt(((tanu)^2+1)/(tanu)^2))du
=∫((1/2)cotu)(sqrt(((secu)^2)/(tanu)^2))du
=∫((1/2)cotu)(sqrt((cosecu)^2))du
=∫((1/2)cotu)(cosecu)du
=∫((1/2)secu)du
=∫((1/2)secu(secu+tanu)/ (secu+tanu))du
=∫((1/2)((secu)^2+secutanu)/ (secu+tanu))du

Let w=secu+tanu
dw=(secutanu+(secu)^2)du

∫((1/2)((secu)^2+secutanu)/ (secu+tanu))du
=∫((1/2)1/w)dw
=(1/2)ln(w)+c
=(1/2)ln(secu+tanu)

x^2=tanu
x^4=(tanu)^2
x^4+1=(secu)^2
secu=sqrt(x^4+1)

(1/2)ln(secu+tanu)
=(1/2)ln(sqrt(x^4+1)+x^2)+c