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= 2çdt/4(1+t^2)+5(1-t^2)=2çdt/(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?

Answer 1

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.

Answer 2

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.