We'll choose the function to be
integrated:
f(x) = (sin
x)^n
We'll integrate the
function:
Int (sin x)^n
dx
We'll re-write the given function as a product of (sin
x)*(sin x)^n-1.
We'll write (sin x)^(n-1) = [1 - *(cos
x)^2]^(n-1)/2.
We'll re-write the
integral:
Int (sin x)^n dx = Int [1 - (cos
x)^2]^(n-1)/2*(cos x)'dx
We'll give values to n. We'll
choose n = 5.
Int (sin x)^5 dx = Int [1 - (cos
x)^2]^(5-1)/2*(cos x)'dx
We'll substitute cos x =
t
(cos x)'dx = -dt
We'll
re-write the integral of the function in t:
Int [1 - (cos
x)^2]^(5-1)/2*(cos x)'dx = -Int (1 - t^2)^2dt
We'll expand
the square:
-Int (1 - t^2)^2dt = -Int (1 - 2t^2 + t^4)
dt
We'll apply the property of additivity of
integrals:
Int (1 - t^2)^2dt = -Int dt + 2Int t^2dt - Int
t^4dt
Int dt - 2Int t^2dt + Int t^4dt = -t + 2t^3/3 -
t^5/5
Int (sin x)^5 dx = -cos x + 2(cos
x)^3/3 - (cos x)^5/5 + C
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