# Find the integral integrate from 1 to 0 of (sin(pi*x)^2)(cos(pi*x))^2 dx

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### 1 Answer

You should use half angle identities such that:

`sin^2 (pi*x) = (1 - cos 2pi*x)/2`

`cos^2 (pi*x) = (1 + cos 2pi*x)/2`

Substituting `(1 - cos 2pi*x)/2` for `sin^2 (pi*x)` and (1 + cos 2pi*x)/2 for `cos^2 (pi*x)` yields:

`int sin^2 (pi*x) cos^2 (pi*x) dx = (1/4)int (1 - cos 2pi*x)(1 + cos 2pi*x) dx`

You need to convert the product `(1 - cos 2pi*x)(1 + cos 2pi*x)` into a difference of squares such that:

`(1 - cos 2pi*x)(1 + cos 2pi*x) = 1 - cos^2 2pi*x`

`int sin^2 (pi*x) cos^2 (pi*x) dx = (1/4)int (1 - cos^2 2pi*x)dx`

Using the linearity yields:

`int sin^2 (pi*x) cos^2 (pi*x) dx = (1/4)int - (1/4)intcos^2 2pi*x dx`

Using the half angle identity yields:

`int cos^2 2pi*x dx = int (1 + cos 4pi*x)/2 dx`

`int cos^2 2pi*x dx = (1/2)(int dx + int cos 4pi*x dx)`

`int cos^2 2pi*x dx = (1/2)(x + (sin(4pi*x))/(4pi))`

`int sin^2 (pi*x) cos^2 (pi*x) dx = (1/4)x - (1/8)(x + (sin(4pi*x))/(4pi))`

You may evaluate the definite integral such that:

`int_0^1 sin^2 (pi*x) cos^2 (pi*x) dx = ((1/4)x -(1/8)(x + (sin(4pi*x))/(4pi)))|_0^1`

`int_0^1 sin^2 (pi*x) cos^2 (pi*x) dx = 1/4 - (1/8)(1 + (sin 4pi)/(4pi) - 0)`

`int_0^1 sin^2 (pi*x) cos^2 (pi*x) dx = 1/4 - (1/8)`

`int_0^1 sin^2 (pi*x) cos^2 (pi*x) dx = 1/8`

**Hence, evaluating the definite integral yields`int_0^1 sin^2 (pi*x) cos^2 (pi*x) dx = 1/8.` **