The integral ∫ e ( 1 + i ) x d x .

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter I, Problem 50E

(a)

To determine

Expert Solution

Answer to Problem 50E

The value of the integral is 1i2e(1+i)x+c.

Explanation of Solution

Calculation:

From the exercise 49, it is known that if F(x)=e(1+i)x, then F(x)=(1+i)e(1+i)x.

Therefore, the given integral becomes,

e(1+i)xdx=F(x)1+idx=11+iF(x)dx=11+iF(x)+c=11+ie(1+i)x+c

Simplify further as,

e(1+i)xdx=11+i(1i1i)e(1+i)x+c=1i12i2e(1+i)x+c=1i1+1e(1+i)x+c=1i2e(1+i)x+c

Thus, the value of the integral is 1i2e(1+i)x+c.

(b)

To determine

Expert Solution

Answer to Problem 50E

The value of the integral is excosxdx=ex2(cosx+sinx) and exsinxdx=ex2(sinxcosx).

Explanation of Solution

Calculation:

Rewrite the given integral as,

e(1+i)xdx=exeixdx=ex(cosx+isinx)dx=excosxdx+iexsinxdx

Thus, e(1+i)xdx=excosxdx+iexsinxdx.                                                            (1)

From part (a), it is known that e(1+i)xdx=1i2e(1+i)x+c.

Simplify this as,

1i2e(1+i)x=e(1+i)x2ie(1+i)x2=12[ex(cosx+isinx)iex(cosx+isinx)]=ex2[cosx+isinxicosx+sinx]=ex2[cosx+sinx+i(sinxcosx)]

Thus, 1i2e(1+i)x=ex2(cosx+sinx)+iex2(sinxcosx).                                         (2)

Compare the equations (1) and (2), it is observed that excosxdx=ex2(cosx+sinx) and exsinxdx=ex2(sinxcosx).

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