Q on reduction formula.

[Sonal7]

I have tried to work this through a bit but I cant see a why through. I just need help getting the formula.


So here's my working:

 

[Sonal7]

I have included the choices in the attachment. Thanks for any help.

 

[HallsofIvy]

cos(2x) is NOT equal to \(\displaystyle 1- sin^2(x)\)! (\(\displaystyle 1- sin^2(x)\) is equal to \(\displaystyle cos^2(x)\).) \(\displaystyle cos(2x)= cos^2(x)- sin^2(x)\).

 

[Sonal7]

I will try that now

 

[Sonal7]

 

[Sonal7]

I am not sure if I am going it right or working in the right direction. Sorry I think i spot errors, give me a few more minutes. I have an online class so might be longer.

 

[Sonal7]

 

[Sonal7]

I 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.

 

[Deleted member 4993]

I 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.

WolframAlpha says this is a NASTY one - devolves into hyper-geometric function

 

[Sonal7]

haha. Lets not cheat. I think this one needs to be tried again the morning after sleep. I will try a different method.

 

[Deleted member 4993]

haha. Lets not cheat. I think this one needs to be tried again the morning after sleep. I will try a different method.

That is not cheating - that is called "looking before leaping".

 

[Sonal7]

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.

 

[Deleted member 4993]

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.

cos(2*x) \(\displaystyle \to cos(2*0) = cos(0) =1\)

cos(2*x) \(\displaystyle \to cos(\frac{2*\pi}{4}) = cos(\frac{\pi}{2}) =0\)

 

[Sonal7]

Yes true. Thanks

 

[Dr.Peterson]

Have you solved it yet? I finally got around to trying it; integration by parts using [MATH]u=\cos^{n-1}(2x)[/MATH] and [MATH]dv=\cos(2x)dx[/MATH] works.

 

[Sonal7]

thats what i have been trying but the du/dx is hard for me. give me a few min.

 

[Sonal7]

 

[Sonal7]

The last line should have an equal sign before it. It doesnt work out from my working.

 

[Dr.Peterson]

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.

 

[Sonal7]

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.