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91Ó°ÊÓ

Solve each of the integrals in Exercises 27–70. Some integrals require integration by parts, and some do not. (The last two exercises involve hyperbolic functions.)

∫x3cosxdx

Short Answer

Expert verified

The value isx3sinx+3x2cosx-6xsinx-6cosx+C.

Step by step solution

01

Step 1. Given information.

The given integral is∫x3cosxdx.

02

Step 2. Substitution.

Now,

u=x3,dv=cosxdx,du=3x2dx,v=∫cosxdx=sinx∫x3cosxdx=x3sinx-∫sin(x)·3x2dx=x3sinx-3∫x2sin(x)dxletu=x2,dv=sinxdx.x3sinx-3∫x2sin(x)dx==x3sinx-3x2·(-cosx)-∫2x(-cos(x))dx=x3sinx-3-x2cosx+2∫xcosxdx=x3sinx+3x2cosx-6∫xcosxdx

03

Step 3. Value of the integral.

Again,

u=x,dv=cosxdxv=∫cosxdx=sinxx3sinx+3x2cosx-6∫xcosxdx==x3sinx+3x2cosx-6xsinx-∫sinxdx=x3sinx+3x2cosx-6xsinx+6(-cosx)+C=x3sinx+3x2cosx-6xsinx-6cosx+C

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