what's the integral of cosine square (x) sine to the forth power(x) dx?

Answer 1

Are you asking: [integral sign] (cos x)^2(sin x)^4dx ?
The general approach is to try to reduce the power/order of the trig functions:
[integral sign] (cos x)^2(sin x)^4dx
= 1/4 [integral sign] (2cos x sin x)^2 (sin x)^2dx
= 1/4 [integral sign] (sin 2x)^2 (1 - cos 2x)/2 dx
= 1/8 [integral sign] (sin 2x)^2 (1 - cos 2x) dx
= 1/8 {[integral sign] (sin 2x)^2 dx - [integral sign] (sin 2x)^2 cos 2x dx}
=1/8 [integral sign] (1 - cos 4x)/2 dx - 1/16 [integral sign] (sin 2x)^2 d(sin2x)
= x/16 - (1/64) sin 4x - (1/48) (sin 2x)^3 + C

Answer 2

Check whether the equation is exact: Let M(x, y) = x + y Let N(x, y) = x - y ∂M(x, y)/∂y = 1 ∂N(x, y)/∂x = 1 The equation is exact. Therefore, there exists a function, F(x, y) the partial derivative with respect to x is M(x, y) and the partial derivative with respect to y is N(x, y). We begin by integrating M(x, y) with respect to x: ∫(x + y)dx = (1/2)x² + yx + h(y) Where h(y) is a function that may disappear when we differentiate with respect to x. We differentiate this with respect to y: x + h'(y) This must equal N(x, y); x + h'(y) = x - y h'(y) = -y h(y) = (-1/2)y² + C The solution is: F(x, y) = (1/2)(x² - y²) + xy + C