IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

divalli_oct07

any one pls integrate cos²

d

within the limits 0 to 2pi.

puneet

hey

I think the answer is alerady there .. it was a simple enuf question ..

cheers

waterdemon

The question is _{[ 0]}

^[2] Cos²x.dx

Let f(x) = Cos²x

f(2 - x) = [Cos(2-x)]² = Cos²x

Therefore ,

_{[0 ]}^{[2pi ]} Cos²x.dx = 2 _[0]^[pi] Cos²x.dx.......................1

This has been derived by a property such as

_[0]^[2a] f(x)dx = 2 _[0]^[a] f(x).dx , If f(2a-x) = f(x) And

_[0]^[2a] f(x).dx = 0 , If f(2a-x) = - f(x)

Now , We will get

f(-x) = [Cos(-x)]² = - Cos²x = f(x)

Therefore ,

Now applying the property again

_[0]^[pi] Cos²x.dx = 2 _[0]^[pi/2] Cos²x.dx = 2(L)

Where L = _[0]^[pi/2] Cos²x.dx ................2

L = _[0]^[pi/2] Cos² ( /2 - x).dx

L = _[0]^[pi/2] Sin²x.dx .............................3

Adding 2 and 3

2L = _[0]^[pi/2] (Sin²x + Cos²x)dx

2L = _[0]^[pi/2] dx

2L = /2

_[0]^[pi] Cos²x.dx = /2

Now from Equation 1 we will get

2 _[0]^[pi] Cos²x.dx = 2 * /2

Therefore the final answer is

_[0]^[2pi] Cos²x.dx =

Hope it helps you.Rate if it does help you,

Cheers !!!!!!!!!!!!!!!!!!

neeraj_agarwal_1990

I think it would be much easier if we use (cosx)^2 =(1+cos2x)/2...then u can do it in mind!

waterdemon

Yes wht not but i choosed to give the answer in the step by step

method and using the properties.

Cheers !!!!!!!!!!!!!!!!!!!!!