Respuesta :

W0lf93
To evaluate 17 int (sin^2 (x)  cos^3(x))
From Trig identity. Cos^2(x) + sin^2(x) =1. Cos^2(x) = 1 - sin^2 (x) 
Cos^3(x) = cosx * (1 - sin^2 (x)) = cosx - cosxsin^2x
So we have 17 int (sin^2x(cosx - cosxsin^2x))
int (sin^2x(cosx)dx - int (sin^4xcosx)dx. ----------(1)
Let u = sinx then du = cosxdx
Substituting into (1) we have
int (u^2du) - int (u^4du)
u^3/3 - u^5/5
Substitute value for u we have 
(sinx)^3/3 - (sinx)^5/5
Hence we have 17 [ sin^3x/3 - sin^5x/5]
lhmarianateixeira

In this exercise we have to use the knowledge of integral to calculate the value of the given function, so we have:

[tex]= [ sin^3x/3 - sin^5x/5][/tex]

The integral given in the exercise is:

[tex]\int\limits^{17}_0 {(sin^2 (x) cos^3(x))} \, dx[/tex]

we know that we will have to use the trigonometric identity as:

[tex]Cos^2(x) + sin^2(x) =1. Cos^2(x) = 1 - sin^2 (x) \\Cos^3(x) = cosx * (1 - sin^2 (x)) = cosx - cosxsin^2x[/tex]

Returning to the integral we have:

[tex]\int\limits {(sin^2x(cosx - cosxsin^2x))} \, dx \\\int\limits { (sin^2x(cosx)dx - int (sin^4xcosx)} \, dx[/tex]

Then using the replacement property we call:

[tex]u = sinx \\ du = cosxdx[/tex]

In this way we find that:

[tex]\int\limits { (u^2du)} \, dx - \int\limits{(xu^4du)} \, dx \\=u^3/3 - u^5/5=(sinx)^3/3 - (sinx)^5/5\\= [ sin^3x/3 - sin^5x/5][/tex]

See more about integrals at brainly.com/question/18651211