Chapter C, Problem 1P

To determine

To find:

The integral ${\displaystyle \int x\cos xdx}$.

Answer to Problem 1P

Solution:

The integral is ${\displaystyle \int x\cos x}dx=x\sin x+\cos x+c$.

Explanation of Solution

The given integral is,

${\displaystyle \int x\cos xdx}$

Approach:

Integration by parts: the basic formula for integration by parts can be written in the form.

${\displaystyle \int udv}=uv-{\displaystyle \int vdu}$ …$\left(1\right)$

Calculation:

Consider the given integral.

${\displaystyle \int x\cos xdx}$

Assume the value of $u$ as,

$\begin{array}{rll}u& =& x\\ \frac{du}{dx}& =& 1\end{array}$

Assume the value of $dv$ as,

$\begin{array}{rll}dv& =& \cos xdx\\ {\displaystyle \int dv}& =& {\displaystyle \int \cos xdx}\\ v& =& \sin x\end{array}$

Substitute $x$ for $u$, $\cos xdx$ for $dv$ and $\sin x$ for $v$ in equation $\left(1\right)$.

$\begin{array}{rll}{\displaystyle \int x\cos x}dx& =& x\sin x-{\displaystyle \int \sin xdx}\\ & =& x\sin x+\cos x+c\end{array}$

Therefore, the integral is ${\displaystyle \int x\cos x}dx=x\sin x+\cos x+c$.

Conclusion:

Hence, the integral is ${\displaystyle \int x\cos x}dx=x\sin x+\cos x+c$.