The choice of first function and second function is arbitrary in case of most functions.
Here, we have f(x) = x^3*Sin xf(x)=x3⋅sinx and we will choose x^3x3 as the first function and Sin xsinx as the second.
Thus, int f(x)*dx = int x^3*Sin x*dx∫f(x)⋅dx=∫x3⋅sinx⋅dx
implies int f(x)*dx = x^3 int Sin x*dx - int (d(x^3)/dx*int Sin x*dx)*dx⇒∫f(x)⋅dx=x3∫sinx⋅dx−∫(dx3dx⋅∫sinx⋅dx)⋅dx
= -x^3*Cos x + int 3x^2*Cos x*dx=−x3⋅cosx+∫3x2⋅cosx⋅dx
= -x^3*Cos x + 3[x^2int Cos x*dx - int (d/dx(x^2) intCos x*dx)*dx]=−x3⋅cosx+3[x2∫cosx⋅dx−∫(ddx(x2)∫cosx⋅dx)⋅dx]
= -x^3*Cos x + 3[x^2Sin x - int 2xSin x*dx]=−x3⋅cosx+3[x2sinx−∫2xsinx⋅dx]
= -x^3*Cos x + 3[x^2Sin x - 2(x int Sin x*dx - int (d/dx(x)int Sin x*dx)*dx]=−x3⋅cosx+3[x2sinx−2(x∫sinx⋅dx−∫(ddx(x)∫sinx⋅dx)⋅dx]
= -x^3Cos x + 3[x^2Sin x - 2(-xCos x + int Cos x*dx)]=−x3cosx+3[x2sinx−2(−xcosx+∫cosx⋅dx)]
= -x^3Cos x + 3[x^2Sin x + 2xCos x - 2Sin x]=−x3cosx+3[x2sinx+2xcosx−2sinx]
Since it is an indefinite integral, we add an arbitrary constant to it.
int x^3Sin x*dx = - x^3Cos x + 3x^2Sin x + 6xCos x - 6Sin x + C∫x3sinx⋅dx=−x3cosx+3x2sinx+6xcosx−6sinx+C where CC is the integration constant.
