Q. 13 - Calculus

Question

If $u\left(x\right)=\sin 3x$ and $v\left(x\right)=x$, what are du and dv? Write down $\int udv$ and $\int vdu$ in this situation. Which of these integrals would be easier to find? What does this exercise have to do with integration by parts?

Step-by-Step Solution

Answer

• $du=3\cos 3xdx$, $dv=dx$
• $\int udv=\int \sin 3xdx$, $\int vdu=\int 3x\cos 3xdx$
• $\int udv$ is easier to find.
• $\int vdu$ can be solved by using integration by parts.

Step 1. Given information

The given functions are $u\left(x\right)=\sin 3x$ and $v\left(x\right)=x$.

Step 2. Evaluate the derivatives to obtain du and dv.

Differentiate the given functions separately with respect to x.

$\begin{array}{rcl}\frac{d}{dx}u\left(x\right)& =& \cos 3x·3\\ du& =& 3\cos 3xdx\end{array}$

$\begin{array}{rcl}\frac{d}{dx}v\left(x\right)& =& 1\\ dv& =& dx\\ & & \end{array}$

Step 3. Obtain the expression for ∫ u d v and ∫ v d u .

Substitute the given and obtained values to obtain $\int udv$ and $\int vdu$.

$\int udv=\int \sin 3xdx$

$\int vdu=\int 3x\cos 3xdx$

Step 4. Explanation

The integral $\int udv=\int \sin 3xdx$ is easier to find because it involves only sine function, whereas the other integral, that is, $\int vdu=\int 3x\cos 3xdx$ involves multiplication of two functions. Thus, integration by parts would be used in solving $\int vdu$.