How do you integrate $int x^3 sin x dx$ using integration by parts?

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Answer 1 · 2016-08-13T06:13:07

For the integration of a product of two functions (First function)*(Second function) = $f(x)$ ,

$int f(x)*dx = (First function)*int (Second function)*dx - int (d/dx(First function)*int(Second function)*dx$ .

The choice of first function and second function is arbitrary in case of most functions.

Here, we have $f(x) = x^3*Sin x$ and we will choose $x^3$ as the first function and $Sin x$ as the second.

Thus, $int f(x)*dx = int x^3*Sin x*dx$

$implies int f(x)*dx = x^3 int Sin x*dx - int (d(x^3)/dx*int Sin x*dx)*dx$

$= -x^3*Cos x + int 3x^2*Cos x*dx$

$= -x^3*Cos x + 3[x^2int Cos x*dx - int (d/dx(x^2) intCos x*dx)*dx]$

$= -x^3*Cos x + 3[x^2Sin x - int 2xSin x*dx]$

$= -x^3*Cos x + 3[x^2Sin x - 2(x int Sin x*dx - int (d/dx(x)int Sin x*dx)*dx]$

$= -x^3Cos x + 3[x^2Sin x - 2(-xCos x + int Cos x*dx)]$

$= -x^3Cos x + 3[x^2Sin x + 2xCos x - 2Sin x]$

Since it is an indefinite integral, we add an arbitrary constant to it.

$int x^3Sin x*dx = - x^3Cos x + 3x^2Sin x + 6xCos x - 6Sin x + C$ where $C$ is the integration constant.

How do you integrate int x^3 sin x dx int x^3 sin x dx using integration by parts?