Chapter 7, Problem 19RE

To determine

To determine the integration of the function $\left(x^3\cos x\right)$

Answer to Problem 19RE

  $x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c$

Explanation of Solution

Given information:

  $f\left(x\right)=x^3\cos x$

Calculation:

Integration by parts formula

  ${\displaystyle \int u}dv=uv-{\displaystyle \int v}du$

In the given question $I={\displaystyle \int x^3}\cos xdx$

  $u=x^3$ and $dv=\cos xdx$

Differentiating u, we get

  $u=x^3\Rightarrow du=3x^{2}dx\left(1\right)$

Integrating dv, we get

  $\begin{array}{l}dv=\cos xdx\Rightarrow {\displaystyle \int dv}={\displaystyle \int \cos x}dx\\ v=\sin x\left(2\right)\end{array}$

Now substituting values u and v into the formula,

  $\begin{array}{l}I={\displaystyle \int x^3}\cos xdx\\ I=uv-{\displaystyle \int v}du\\ I=x^3\sin x-{\displaystyle \int 3x^2}\sin xdx\end{array}$

Now ${\displaystyle \int 3x^2}\sin xdx$ can be evaluated using same integration by parts

Then $u=2x^2$ and $dv=\sin xdx$

Differntriating u, $du=6xdx$

Integrating $dv,v=-\cos x$

Thus ${\displaystyle \int 3x^2}\sin xdx=3x^2\left(-\cos x\right)={\displaystyle \int 6x}\left(-\cos x\right)dx$

Substituting value of ${\displaystyle \int 3x^2}\sin xdx$ in I, we get

  $\begin{array}{l}I=x^3\sin x-\left[3x^2\left(-\cos x\right)-{\displaystyle \int 6x}\left(-\cos x\right)dx\right]\\ I=x^3\sin x+3x^2\cos x-{\displaystyle \int 6x}\cos xdx\left(I\right)\end{array}$

Again the value of ${\displaystyle \int 6x}\cos xdx$ can be found by integration by part.

Here $u=6x\Rightarrow du=6dx$

Here $dv=\cos xdx\Rightarrow v=\sin x$

So ${\displaystyle \int 6x}\cos xdx=6x\sin x-{\displaystyle \int 6}\sin xdx$

  $\begin{array}{l}=6x\sin x+6\cos x\left\{{\displaystyle \int \sin x}=-\cos x\right\}\\ {\displaystyle \int 6x}\cos xdx=6x\sin x+6\cos x\end{array}$

Substituting the value of ${\displaystyle \int 6x}\cos xdx$ in (I)

  $\begin{array}{l}I=x^3\sin x+3x^2\cos x-\left[6x\sin x+6\cos x\right]\\ I=x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c\end{array}$

Where c is an integration constant.

  ${\displaystyle \int x^3}\cos xdx=x^3\sin x+3x^2\cos x-6x\sin x-6\cos x+c$