   Chapter 8.2, Problem 76E

Chapter
Section
Textbook Problem

Proof In Exercises 71-76, use integration by parts to prove the formula. (For Exercises 71-74, assume that n is a positive integer.) ∫ e a x cos   b x   d x = e a x ( a   cos   b x + b   sin   b x ) a 2 + b 2 + C

To determine

To prove: The value of the integral eaxcosbxdx=eax(acosbx+bsinbx)a2+b2+C.

Explanation

Given:

The given value of integral is eaxcosbxdx=eax(acosbx+bsinbx)a2+b2+C.

Proof:

Consider the following given formula,

eaxcosbxdx=eax(acosbx+bsinbx)a2+b2+C

Consider the left hands side of the equation

We have.

eaxcosbxdx

Use the integration by part rule.

udv=uvvdu

Let

u=cosbx

On Differentiating with respect to x.

du=bsin(bx)dx

Let

dv=eaxdx.

On Integrating the equation.

We get,

v=eaxa

Now, Substitute u=cosbx, u=bsin(bx) and v=eaxa in equation (1).

We get,

I=[cosbx(eaxa)bsin(bx)(eaxa)dx]=[cosbx(eaxa)+basin(bx)(eax)dx]

Let u1=sinbx

On Differentiating with respect to x.

du1=bcos(bx)dx

Let

dv=eaxdx

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