WolframAlpha says this is a NASTY one - devolves into hyper-geometric functionI really cant see this working. I wondered about trying a different identity but I am not sure that would work either. Its got infinite recursive pattern.
That is not cheating - that is called "looking before leaping".haha. Lets not cheat. I think this one needs to be tried again the morning after sleep. I will try a different method.
cos(2*x) \(\displaystyle \to cos(2*0) = cos(0) =1\)Theres a clue in the Q. the limit is 1/4 pi and it is dealing with cos 2x. so that would make it cos pi/2 when you sub in the limit which will make the term 1. I think at some point we put in the limits. I am going to try IBP this morning but by rearranging differently.
Replace [MATH]\sin^2(2x)[/MATH] with [MATH]1 - \cos^2(2x)[/MATH] and continue. You should get an expression for [MATH]I_n[/MATH] in terms of [MATH]I_n[/MATH] and [MATH]I_{n-1}[/MATH].
But also check your final form of du/dx.
Remember: Whenever you write a line, check that it is correct compared with the previous line. This is the only way to avoid silly mistakes that waste your time.