Integration of #int e^sin(x) dx#?

Question

Isabella Ackerman · Accepted Answer

#sum_(n=0)^(oo)intsin^n(x)/(n!)dx#

There's really no way to integrate this. The way to integrate is to think "this is the derivative of what?" Since your original equation is

#e^sin(x)#

You can't actually apply this, because it would mean:

#inte^sin(x)dx=-e^sin(x)/cos(x)#

This isn't the case, however, because this becomes a quotient rule, which leads to a much more complex function afterwards when integrated, of

#(e^sin(x)*sin(x))/cos^2(x)+e^sin(x)#

So we're going to need another approach. Because we can't actually integrate this, we need to turn this into something more integrable. We can rewrite #e^sin(x)# as a power series in order to make this integrable.

A Power Series is an infinite series written in the form:

#sum_(n=0)^(oo)a_nx^n#

To build one of these, you have to continuously take the derivative of your original function, and it's pretty complex. I recommend watching videos on it from Khan Academy. They do a great job explaining it.

Just know that

#e^x=sum_(n=0)^(oo)x^n/(n!)#

Since we can replace x with sin(x), we can deduce

#e^sinx=sum_(n=0)^(oo)sin^n(x)/(n!)#

And we can integrate both sides to get

#inte^sinxdx=sum_(n=0)^(oo)intsin^n(x)/(n!)dx#

This may not be what you were searching for, but this is as far as I know how to take you. Maybe you'll wanna take another step and use the Power Series for sin(x). Here it is, in case it may help:

#sin(x)=sum_(n=0)^(oo)((-1)^nx^(2n+1))/((1+2n)!)#

Best of luck!