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?
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.
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.
To show that the two results are the same, we can start by simplifying the expression obtained from the second procedure.
Using the trigonometric identity sec(u) = 1/cos(u), we can rewrite the expression as:
S = (2/3) * log [sec(arcsin(1/3 * tan(x/2))) + tan(arcsin(1/3 * tan(x/2)))]
Now, let's substitute back the values of arcsin(1/3 * tan(x/2)) using the equation t = tan(x/2):
S = (2/3) * log [sec(arcsin(t/3)) + tan(arcsin(t/3))]
We can use the Pythagorean identity sin^2(u) + cos^2(u) = 1 to solve for cos(u):
cos(u) = sqrt(1 - sin^2(u))
Substituting this back into our expression:
S = (2/3) * log [sec(arcsin(t/3)) + tan(arcsin(t/3))]
= (2/3) * log [1/cos(arcsin(t/3)) + sin(arcsin(t/3))/cos(arcsin(t/3))]
Using the definitions of sine and cosine in terms of t:
S = (2/3) * log [1/sqrt(1 - (t/3)^2) + (t/3)/sqrt(1 - (t/3)^2)]
Now, let's simplify the expression inside the logarithm:
S = (2/3) * log [(3 + t) / sqrt(9 - t^2)]
Using the logarithmic identity log(a/b) = log(a) - log(b):
S = (2/3) * [log(3 + t) - log(sqrt(9 - t^2))]
Finally, using the logarithmic identity log(sqrt(x)) = (1/2)log(x):
S = (2/3) * [log(3 + t) - (1/2)log(9 - t^2)]
= (2/3) * [log(3 + t) - (1/2)log((3 + t)(3 - t))]
Combining the logarithms:
S = (2/3) * [log(3 + t) - (1/2)[log(3 + t) + log(3 - t)]]
= (2/3) * [(1/2)log(3 + t) - (1/2)log(3 - t)]
Simplifying further:
S = (2/3) * (1/2) * log[(3 + t)/(3 - t)]
= (1/3) * log[(3 + t)/(3 - t)]
Now, we can see that the expression obtained from the second procedure is equal to the expression obtained from the first procedure:
S = (1/3) * log[(3 + t)/(3 - t)]
which matches the result obtained using the first procedure:
S = (2/3) * arctanh(1/3 * tan(x/2)) + C1
Therefore, both procedures yield the same result.