Let's use I(n) for the integral. [In looks just like ln for natural logarithm]
And we'll just remember that the boundaries are 0, π/2 rather than writing them in each time.
I(n) = ∫(Cos^(n-2))(x) (cos^2)(x) dx
..... = ∫(Cos^(n-2))(x) (1 - (sin^2)(x) dx
..... = I(n-2) - ∫(Cos^(n-2))(x) (sin^2)(x) dx .......... (1)
(d/dx)[ (cos^(n-1))(x)] = (n-1)(cos^(n-2))(x) (-sinx)
hence if we write u = (cos^(n-1))(x) then the second term is
(1/(n-1) ∫ (du/dx) sinx dx
which becomes, using integration by parts,
(1/(n-1)([u sinx] - ∫ u cosx dx )
Now when x = 0, sinx = 0, and when x = π/2, u = 0
and u cosx = (cos^n)(x)
hence that term becomes
(1/(n-1))(0 - I(n) )
Thus equation (1) becomes
I(n) = I(n-2) - (1/(n-1)) I(n)
Add (1/(n-1)) I(n) to both sides:
(n/(n-1)) I(n) = I(n-2)
therefore
I(n) = ((n-1)/n) I(n-2)
so I don't know where the -1/(n^2) comes from.
EDIT: I've checked some values of I(n) using Wolframalpha.
I(1) = 1 (didn't need Wolfram for that!!!)
I(2) = π/4
I(3) = 2/3
I(4) = 3π/16
I(5) = 8/15
I(6) = 5π/32
which confirm the formula found above.
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Verified answer
Let's use I(n) for the integral. [In looks just like ln for natural logarithm]
And we'll just remember that the boundaries are 0, π/2 rather than writing them in each time.
I(n) = ∫(Cos^(n-2))(x) (cos^2)(x) dx
..... = ∫(Cos^(n-2))(x) (1 - (sin^2)(x) dx
..... = I(n-2) - ∫(Cos^(n-2))(x) (sin^2)(x) dx .......... (1)
(d/dx)[ (cos^(n-1))(x)] = (n-1)(cos^(n-2))(x) (-sinx)
hence if we write u = (cos^(n-1))(x) then the second term is
(1/(n-1) ∫ (du/dx) sinx dx
which becomes, using integration by parts,
(1/(n-1)([u sinx] - ∫ u cosx dx )
Now when x = 0, sinx = 0, and when x = π/2, u = 0
and u cosx = (cos^n)(x)
hence that term becomes
(1/(n-1))(0 - I(n) )
Thus equation (1) becomes
I(n) = I(n-2) - (1/(n-1)) I(n)
Add (1/(n-1)) I(n) to both sides:
(n/(n-1)) I(n) = I(n-2)
therefore
I(n) = ((n-1)/n) I(n-2)
so I don't know where the -1/(n^2) comes from.
EDIT: I've checked some values of I(n) using Wolframalpha.
I(1) = 1 (didn't need Wolfram for that!!!)
I(2) = π/4
I(3) = 2/3
I(4) = 3π/16
I(5) = 8/15
I(6) = 5π/32
which confirm the formula found above.