Question

Integral of the trigonometric functions:? B-) `int_0^pisin^2x*cos^4xdx`

Answer 1

`x/16+1/64sin(2x)-1/64sin(4x)-1/192sin(6x)+C`

Explanation:

Using the double angle formulas

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

we rewrite the integrand as

`sin^2 x cos^4x = (sin x cos x)^2cos^2x`
`qquad = (1/2sin(2x))^2 1/2(1+cos(2x))`
`qquad = 1/8 sin^2(2x) (1+cos(2x))`

Thus, the integral is

`int sin^2 x cos^4x = 1/8 int sin^2(2x) (1+cos(2x))dx`
`qquad = 1/8 int sin^2(2x)dx+1/8int sin^2(2x) cos(2x)dx`
`qquad = 1/16 int (1-cos(4x))dx`
`qquadquad +1/8 int sin^2(2x) 1/2d(sin(2x))`
`qquad = x/16-1/64sin(4x)+1/48sin^3(2x)+C`

Using the triple angle formula

`sin^3x = 1/4(3sinx-sin(3x))`

this result can be recast in the form

`x/16-1/64sin(4x)+1/48[1/4(3sin(2x)-sin(6x))]+C`
`qquad = x/16+1/64sin(2x)-1/64sin(4x)-1/192sin(6x)+C`