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Q 32.

Expert-verifiedFound in: Page 429

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 ${x}^{2}\mathrm{sin}x+2x\mathrm{cos}x-2\mathrm{sin}x+c$.

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

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 $\int udv=uv-\int vdu$.

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

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,

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

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