antiderivative for sinx/(1+x^2) ????

[borderking]

Hi, there...so I've been learning all about integration by partial fractions, substitution, and trig substitutions. I ran across this problem on a hw assignment though, and I can't figure out which category it falls under.
It asks for the indefinite integral of sin(x)/(1+x^2). That denominator looks like a trig substitution using tangent, but then there's that sin(x) on the top that's throwing me. Is this possible?
Thanks so much...

 

[galactus]

Looks like you're correct. This one isn't solvable by elementary means.

I ran it through Maple and it gave me:

\(\displaystyle \L\\\frac{1}{2}i[S\text{i}(x+i)cosh(1)-iC\text{i}(x+i)sinh(1)]-\frac{1}{2}i[S\text{i}(x-i)cosh(1)+iC\text{i}(x-i)sinh(1)]\)

 

[morson]

Or, you could look at the integral as this:

\(\displaystyle \L\ \int \sum_{k=0}^{\infty\\ \\) \(\displaystyle (-1)^k\frac{Tan^{2k + 1}m}{(2k + 1)!}\ dm\)

where \(\displaystyle x = Tan(m)\). This would be very difficult to represent in closed form. Edit: It took 9 times to get that thing up there right. :twisted:

 

[galactus]

Yes, Morson is on to something. Maybe you could use the Taylor series for sinx/(1+x^2).

 

[borderking]

Well, let me add something...I didn't think this was relevant, but the actual problem was a definite integral, and the integrand was [cosx/sqrt(1+sinx)] + sinx/(1+x^2).
The first fraction seemed simple, allowing u=(1+sinx). It was the second fraction that had me baffled. But the fact that there was another fraction in front or the fact that it was actually a definite integral doesn't matter, does it? I mean, sinx=(1+x^2) isn't an even or odd function, so I don't see how the limits could help. I don't remember them offhand...Does that change anything?

 

[galactus]

From what I can see, it doesn't make much difference. Then you have:

\(\displaystyle \L\\\int\frac{cos(x)}{\sqrt{1+sin(x)}}dx+\int\frac{sin(x)}{1+x^{2}}dx=2\sqrt{1+sin(x)}+\int\frac{sin(x)}{1+x^{2}}dx\)

Who knows, someone else may have some illuminating method.

 

[pka]

But the fact that there was another fraction in front or the fact that it was actually a definite integral doesn't matter, does it?

That certainly does make a real difference.
For example:\(\displaystyle \L
\int\limits_{ - a}^a {\left[ {\frac{{\cos (x)}}{{\sqrt {1 + \sin (x)} }} + \frac{{\sin (x)}}{{1 + x^2 }}} \right]dx} = \int\limits_{ - a}^a {\left[ {\frac{{\cos (x)}}{{\sqrt {1 + \sin (x)} }}} \right]dx}.\)
Because \(\displaystyle \L {\frac{{\sin (x)}}{{1 + x^2 }}}\) is an odd function.

 

[galactus]

DUH, I should've seen that. I thought there was probably something going on with it, but was to myopic to see it.

 

[borderking]

Holy cow, you're right! I didn't think of that. The limits were -pi/2 to pi/2. You're right, indeed, I guess the integral of that rightmost fraction is just 0? I didn't realize at first it was an odd function! Hmm, that teaches me not to leave information out when I ask a question...
Thanks