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

Aug 13, 2016

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

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

This is called integration by parts.

Explanation:

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

Here, we have $f \left(x\right) = {x}^{3} \cdot S \in x$ and we will choose ${x}^{3}$ as the first function and $S \in x$ as the second.

Thus, $\int f \left(x\right) \cdot \mathrm{dx} = \int {x}^{3} \cdot S \in x \cdot \mathrm{dx}$

$\implies \int f \left(x\right) \cdot \mathrm{dx} = {x}^{3} \int S \in x \cdot \mathrm{dx} - \int \left(d \frac{{x}^{3}}{\mathrm{dx}} \cdot \int S \in x \cdot \mathrm{dx}\right) \cdot \mathrm{dx}$

$= - {x}^{3} \cdot C o s x + \int 3 {x}^{2} \cdot C o s x \cdot \mathrm{dx}$

$= - {x}^{3} \cdot C o s x + 3 \left[{x}^{2} \int C o s x \cdot \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} \left({x}^{2}\right) \int C o s x \cdot \mathrm{dx}\right) \cdot \mathrm{dx}\right]$

$= - {x}^{3} \cdot C o s x + 3 \left[{x}^{2} S \in x - \int 2 x S \in x \cdot \mathrm{dx}\right]$

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

$= - {x}^{3} C o s x + 3 \left[{x}^{2} S \in x - 2 \left(- x C o s x + \int C o s x \cdot \mathrm{dx}\right)\right]$

$= - {x}^{3} C o s x + 3 \left[{x}^{2} S \in x + 2 x C o s x - 2 S \in x\right]$

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

$\int {x}^{3} S \in x \cdot \mathrm{dx} = - {x}^{3} C o s x + 3 {x}^{2} S \in x + 6 x C o s x - 6 S \in x + C$ where $C$ is the integration constant.