Question
s=çdx/(4+5cos x). By using t-substitution, i.e. t=tan(x/2) we get cos x=(1-t^2)/(1+t^2) and dx=2dt/(1+t^2). Substituting in s and simplifying, we get
s= 2çdt/4(1+t^2)+5(1-t^2)=2çdt/(9-t^2). Using standard result çdx/(a^2-x^2)=1/a*arctanh (x/a) we get
s=2/3*arctanh {1/3*tan(x/2)} +C1, which is correct as given in the book.
However, if we use further substitution of t=3 sin u, we get du=3 cos u du and u=arcsin (t/3)
S=2/3*çcos u du/(1-sin^2 u)=2/3*çsec u du= 2/3*log[sec u+tan u)
=2/3*log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] +C2.
Though second procedure gives complicated result, is it correct?
If so, how can we show that the two results are same?
s= 2çdt/4(1+t^2)+5(1-t^2)=2çdt/(9-t^2). Using standard result çdx/(a^2-x^2)=1/a*arctanh (x/a) we get
s=2/3*arctanh {1/3*tan(x/2)} +C1, which is correct as given in the book.
However, if we use further substitution of t=3 sin u, we get du=3 cos u du and u=arcsin (t/3)
S=2/3*çcos u du/(1-sin^2 u)=2/3*çsec u du= 2/3*log[sec u+tan u)
=2/3*log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] +C2.
Though second procedure gives complicated result, is it correct?
If so, how can we show that the two results are same?
Answers
Steve
Wow. Looks good, but I'm stumped on the transformations to use. If you plot them at wolframalpha.com via
plot log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] , arctanh {1/3*tan(x/2)}
The two graphs overlap completely, so they must be equal.
plot log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] , arctanh {1/3*tan(x/2)}
The two graphs overlap completely, so they must be equal.
MS
Thanks. I checked and really got them as same. But can we analytically show them to be so? I tried but could not make it. Please guide.