• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q 32.

Expert-verified
Found in: Page 429

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Solve the integral: $\int {x}^{2}\mathrm{cos}xdx$.

The required answer is ${x}^{2}\mathrm{sin}x+2x\mathrm{cos}x-2\mathrm{sin}x+c$.

See the step by step solution

## Step 1. Given information.

We have given integral is $\int {x}^{2}\mathrm{cos}xdx$.

## Step 2. Solve the integration by parts .

We have, $u={x}^{2}dv=\mathrm{cos}xdx$

$u={x}^{2}\phantom{\rule{0ex}{0ex}}du=2xdx$

and

$dv=\mathrm{cos}xdx\phantom{\rule{0ex}{0ex}}v=\int \mathrm{cos}xdx\phantom{\rule{0ex}{0ex}}v=\mathrm{sin}x$

The formula of integration by parts is $\int udv=uv-\int vdu$.

$\int {x}^{2}\mathrm{cos}xdx\phantom{\rule{0ex}{0ex}}={x}^{2}\mathrm{sin}x-\int 2x\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}={x}^{2}\mathrm{sin}x-2\int x\mathrm{sin}xdx$

## Step 3. Integration by parts.

We have,

$u=x,dv=\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}u=x\phantom{\rule{0ex}{0ex}}du=dx$

and

$dv=\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}v=\int \mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}v=-\mathrm{cos}x$

So,

${x}^{2}\mathrm{sin}x-2\int x\mathrm{sin}xdx\phantom{\rule{0ex}{0ex}}={x}^{2}\mathrm{sin}x-2\left(\left(x\left(-\mathrm{cos}x\right)\right)-\int \left(-\mathrm{cos}x\right)dx\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\mathrm{sin}x-2\left(\left(x\left(-\mathrm{cos}x\right)\right)+\int \mathrm{cos}xdx\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\mathrm{sin}x+2x\mathrm{cos}x-2\mathrm{sin}x+c$

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.