Integrating d/dx y = (e^x)(sin x)

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how do you anti-differentiate the equation in the title thread?

Let's do it one step at a time. if y = f(x) * g(x) then dy/dx = f'(x) * g(x) + f(x) * g'(x).

Making the obvious substitutions we get:

dy/dx = exp(x) * cos x + sin x * exp (x) = exp(x) * ( cos x + sin x)

since d exp (x)/dx = exp(x) and d sin x /dx = cos x.

I do hope this is not a homework problem.

ruveyn

Not sure if this is what you're looking for...

[youtube]http://www.youtube.com/watch?v=sSE6_fK3mu0[/youtube]

Aw shucks! Anti-differentiate.

Try integration by parts.

ruveyn

You need to do integration by parts twice, and after the second time, you should wind up with a term that looks like your original integral. Move that term to the left hand side of the equation and divide by the coefficient (should be 2) to get your answer.

so with integration by parts:

S (u' v) dx = u v - S (u v') dx

where S is the integrand, u = exp(x), v = sin(x) and a ' (prime) means that substitution's derivative.

therefore:

S (exp(x) sin(x)) dx = (exp(x) sin(x)) - S (exp(x) (-cos(x))) dx = exp(x)sin(x) + S (exp(x)cos(x)) dx

= exp(x)sin(x) + exp(x)cos(x) - S (exp(x)sin(x)) dx

this is where i got unstuck, as I would then integrate again and again...

following what mcg says:

2 S (exp(x) sin(x)) dx = exp(x)sin(x) + exp(x)cos(x) = exp(x) ( sin(x) + cos(x) )

S (exp(x) sin(x)) dx = exp(x) ( sin(x) + cos(x) ) / 2


we haven't done integration by parts in class, i'm self studying and didn't understand what to do with repeating terms.

Is there a Calculus Thread? If not i'll start one and post further questions there.

I came up with a solution that does not use integration by parts. I used Euler's formula

e^ix = cos x + i * sin x

e^-ix = cos x - i * sin x

so e^i x - e^-ix = 2*i*sin x
hence ( e^i x - e^-ix )/2*i = sin x

therefore e^x * ( e^i x - e^-ix )/2*i = e^x * sin x

Now integrate all those exponentials without using integration by parts and do some algebra and you have it. (I leave the end calculations for you as an exercise).

The moral of the story: think outside the box.

ruveyn

This Link has some sample/explained examples of various calculus techniques. It works things out in detail without skipping steps. When I was taking calculus, it was probably more helpful than the book.

I like ruveyn's method better than the by parts method. In class, they pretty much told you that you'd never know if or when you could do it by parts, so just guess and hope you guess right. It seems kind of silly to do that much tedious calculation when a computer could do the same thing, but much faster.