# Find the integral of: `int sin^2xcos^4xdx`

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### 1 Answer

You should use half angle identity such that:

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

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

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

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

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

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

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

You should use the fundamental formula of trigonometry such that:

`1 - cos^2 2x = sin^2 2x`

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

Opening the brackets yields:

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

Using the linearity yields:

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

You need to solve the integral `int (sin^2 2x)dx` using half angle identity such that:

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

`int (sin^2 2x)dx = x/2 - (sin 4x)/8 + c`

You need to solve `int sin^2 2x*cos 2x dx` using the following substitution such that:

`sin 2x = t => 2 cos 2x dx = dt => cos 2x dx = (dt)/2`

`int sin^2 2x*cos 2x dx = int t^2 (dt)/2`

`int t^2 (dt)/2 = (1/2) (t^(2+1))/(2+1) + c`

`int t^2 (dt)/2 = t^3/6 + c`

Substituting back `sin 2x` for t yields:

`int sin^2 2x*cos 2x dx = (sin^3 2x)/6 + c`

`int sin^2 x*cos^4 x dx = (1/8)(x/2 - (sin 4x)/8) + (sin^3 2x)/48 + c`

`int sin^2 x*cos^4 x dx = x/16 - (sin 4x)/64 + (sin^3 2x)/48 + c`

**Hence, evaluating the given indefinite integral, using trigonometric identities yields `int sin^2 x*cos^4 x dx = x/16 - (sin 4x)/64 + (sin^3 2x)/48 + c.` **