Another approach Start with the integral I(a) = 0 to π intg (1/(a-cosx)) . Apply king's rule now I(a) = (1/(a+cosx)). Add the 2 versions of it 2I(a) = 2a/a²-cos²x . Write cosx as 1/secx . Now I(a) = asec²x/a²sec²x-1 . Put tanx = t . Sec²xdx becomes dt . I(a) = adt/a²(1+t²)-1 = adt/a²t² + a²-1 . Apply the formula of integral of dx/x²+a² . Back substitute the limits t = tanx and the limits of integral . Now we have I(a) in terms of a . Differentiate it twice wrt a to get (a-cosx)³ in the denominator. Finally put a = √5 and we're done ✅
Another marvel with the Weierstrass substitution!
Another approach
Start with the integral
I(a) = 0 to π intg (1/(a-cosx)) . Apply king's rule now I(a) = (1/(a+cosx)). Add the 2 versions of it 2I(a) = 2a/a²-cos²x . Write cosx as 1/secx . Now I(a) = asec²x/a²sec²x-1 . Put tanx = t . Sec²xdx becomes dt . I(a) = adt/a²(1+t²)-1 = adt/a²t² + a²-1 . Apply the formula of integral of dx/x²+a² . Back substitute the limits t = tanx and the limits of integral . Now we have I(a) in terms of a .
Differentiate it twice wrt a to get (a-cosx)³ in the denominator. Finally put a = √5 and we're done ✅
That's smart!
I shall go through your approach. Thanks
I just posted a new video for this integral based on your approach in your comment
Wow...
Nice
Thanks
Please how do I make a video like this...
I'll really be grateful if someone responds
I use Microsoft Word and power point to make my videos. The music is from TH-cam library
@@cipherunity Thanks man.
I really appreciate.
not very sick