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Q 32.

Expert-verified
Calculus
Found in: Page 429
Calculus

Calculus

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

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Short Answer

Solve the integral: x2cos x dx.

The required answer is x2sin x+2x cos x-2sin x+c.

See the step by step solution

Step by Step Solution

Step 1. Given information. 

We have given integral is x2cos x dx.

Step 2. Solve the integration by parts . 

We have, u=x2dv=cos x dx

u=x2du=2xdx

and

dv=cosx dxv=cos x dxv=sin x

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

x2cosx dx=x2sinx -2xsinx dx=x2sinx-2xsin x dx

Step 3. Integration by parts.

We have,

u=x, dv=sinxdxu=xdu=dx

and

dv=sinxdxv=sinxdxv=-cosx

So,

x2sinx-2xsinxdx=x2sin x-2((x(-cos x))-(-cos x)dx)=x2sin x-2((x(-cos x))+cos xdx)=x2sin x+2xcos x-2sin x+c

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