Integrate the function y = x*sin^3x

We'll integrate the given function by parts, using the
formula:

Int f*g'  = f*g - Int
f'*g

We'll put f = x => f' =
1

We'll put g' = (sinx)^3 and we'll have to integrate
(sinx)^3 to find the function g.

We'll write the function
as a product:

(sinx)^3 = (sinx)^2*sin
x

We'll integrate both
sides:

Int (sinx)^3dx = Int [(sinx)^2*sin
x]dx

We'll write (sinx)^2  = 1 -
(cosx)^2

Int [(sinx)^2*sin x]dx = Int [(1 - (cosx)^2)*sin
x]dx

We'll remove the
brackets:

Int [(1 - (cosx)^2)*sin x]dx  = Int sin xdx - Int
(cosx)^2*sin xdx

We'll solve Int (cosx)^2*sin xdx using
substitution technique:

cos x =
t

We'll differentiate both
sides:

cos xdx = dt

We'll
re-write the integral, changing the variable:

Int
(cosx)^2*sin xdx = Int t^2dt

Int t^2dt = t^3/3 +
C

Int (cosx)^2*sin xdx = (cos x)^3/3 +
C

Int (sinx)^3dx = Int sin xdx - Int (cosx)^2*sin
xdx

Int (sinx)^3dx = -cos x - (cos x)^3/3 +
C

So, if g' = (sinx)^3 => g = -cos x - (cos
x)^3/3

Now, we can integrate by
parts:

Int x*(sinx)^3dx = x*[-cos x - (cos x)^3/3] - Int
[ -cos x - (cos x)^3/3] dx

We'll apply the additive
property of integrals:

Int x*(sinx)^3dx = x*[-cos x - (cos
x)^3/3] + Int cos x dx + (1/3)Int (cos x)^3 dx

We'll write
the function as a product:

(cos x)^3 = (cos x)^2*cos
x

We'll integrate both
sides:

Int (cos x)^3 dx = Int (cos x)^2*cos x
dx

We'll write (cos x)^2  = 1 - (sin
x)^2

Int (cos x)^2*cos x dx = Int [(1 - (sin x)^2)*cos
x]dx

We'll remove the
brackets:

Int [(1 - (sin x)^2)*cos x]dx  = Int cos xdx -
Int (sinx)^2*cos xdx

We'll solve Int (sinx)^2*cos xdx using
substitution technique:

sin x =
t

We'll differentiate both
sides:

cos xdx = dt

We'll
re-write the integral, changing the variable:

Int
(sinx)^2*cos xdx = Int t^2dt

Int t^2dt = t^3/3 +
C

Int [(1 - (sin x)^2)*cos x]dx  = sin x - (sin x)^3/3 +
C

Int x*(sinx)^3dx = x*[-cos x - (cos
x)^3/3] +sinx + (1/3)[sin x - (sin x)^3/3] + C