integration?

Question

please someone do tell me what is definite integral (cosx)^2dx with limits pi & 0.
this is a part of a bigger question.
i proceeded (1-(sinx)^2)dx
also integral cos x* cos x... the product rule...
yet i didn't get correct answer. please help me...
beleive me i tried it lots of times. so u wouln't be simply answering my HW but really helping me...

Answer 1

cos 2x = 2 cos²x - 1
cos ² x = (1/2) (cos 2x + 1)

I = (1/2) ∫ (cos 2x + 1) dx
I = (1/2) [ (sin 2x) / 2 + x ] lims 0 to π
I = (1/2) [ π ]
I = π / 2

Answer 2

Integral (cos^2 x dx)

cos2x = 2cos^2 x - 1
1/2 (cos2x + 1) = cos^2 x
now substitute cos^2(x) in terms of cos2x in the integral.

Integral ( 1/2 (cos2x + 1) dx)
= [1/4 sin2x + 1/2 x ]
substitute the limits (0 to pi),
= [1/4 sin(2pi) + pi/2] - [1/4 sin0 + 0]
= [ pi/2 ] - [0]
= pi/2

Answer 3

Mental math approach:
sin^2 x + cos^2 x = 1
∫sin^2 x + cos^2 x dx, x from 0 to pi = pi
By symmetry,
∫cos^2 x dx, x from 0 to pi = pi/2
-------
Ideas:
∫cos^2 x dx has the same area as ∫sin^2 x dx, if x is from 0 to pi.

Answer 4

The trick is to use the double angle formula:
cos2x = 2cos(x)^2 - 1
cos(x)^2 = (cos2x+1)/2

So now you can integrate...
1/2 * int (cos2x + 1) from 0 to pi =>
1/2 * (1/2 sin 2x + x) from 0 to pi =>
1/4 sin2x + x/2 from 0 to pi =>
1/4 sin(2pi) + pi/2 - [1/4 sin0 + 0/2] = pi/2

Good luck!

Answer 5

Because, is is a definite integarl from 0 to pi

and cos(pi/2 + x) = cos(pi/2)sin(x) + sin(pi/2)sin(x) =
0*sin(x) + 1*cos(x) = sin(x)

integral from 0 to pi (sin x)^2 dx =
integral from pi/2 to pi (sin x)^2 dx +
+ integral from pi to 1.5pi (-sin x)^2 dx =
= integral from 0 to pi/2 (cos x)^2 dx+
+ integral from pi/2 to pi (cos x)^2 dx
= integral from 0 to pi (sin x)^2 dx

pi = integral from 0 to pi [(sin x)^2 + (cos x)^2] dx =
= integral from 0 to pi (sin x)^2 dx +
+ integral from 0 to pi (cos x)^2 dx =
= 2integral from 0 to pi (cos x)^2 dx

Divide LHS and RHS by 2 and get
integral from 0 to pi (cos x)^2 dx = pi/2