Solve Integral & Complex Exponential Problems: Help Needed

[Geronimo85]

I have two homework problems that have been driving me nuts:

1.) evaluate the indefinite integral:

integral(dx(e^ax)cos^2(2bx))

where a and b are real positive constants. I just don't know where to start on it.

2.) Find all values of i^(2/3)

So far I have:

i^(2/3)
= e^(2/3*ln(i))
= e^(2/3*i*(Pi/2 + 2*n*Pi))
= e^(i*Pi/3)*e^(i*n*4Pi/3)

I know from the back of my book my three solutions should end up being (1+i*sqrt(3))/2, (1-i*sqrt(3))/2, -1. But I can't seem to get there. I'd really appreciate any help. Sorry if my shorthand is confusing.

 

[Max Eilerson]

For 1) use the exponential notation for cos x, then integrate. You can do it by integration by parts twice, then use 'the trick' but that would be pretty messy.

 

[kyle8921]

For the first problem, you can use the substitution u = e^ax, du = ae^ax dx to transform the integral into:

integral(cos^2(2bx)) du

Then, you can use the identity cos^2(x) = (1 + cos(2x))/2 to rewrite the integral as:

1/2 integral(1 + cos(4bx)) du

Integrating each term separately, you get:

u/2 + 1/8 sin(4bx) + C

Substituting back u = e^ax, the final solution is:

e^ax/2 + 1/8 sin(4bx) + C

For the second problem, your approach is correct. However, you need to be careful with the exponent of e. It should be:

e^(i*2Pi/3)*e^(i*n*4Pi/3)

Then, you can use Euler's formula e^(ix) = cos(x) + i*sin(x) to rewrite the expression as:

e^(i*2Pi/3) = cos(2Pi/3) + i*sin(2Pi/3) = -1/2 + i*sqrt(3)/2

e^(i*4Pi/3) = cos(4Pi/3) + i*sin(4Pi/3) = -1/2 - i*sqrt(3)/2

So, the three solutions are:

e^(i*2Pi/3) = -1/2 + i*sqrt(3)/2

e^(i*4Pi/3) = -1/2 - i*sqrt(3)/2

e^(i*0) = 1 + 0*i = 1

which can also be written as:

(1+i*sqrt(3))/2, (1-i*sqrt(3))/2, -1