- #1

The Head

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- Homework Statement
- Evaluate ∫((xsin(x) - cos(x))/x^2. Hint: Use Integration by Parts for sin(x)/x

Problem also attached if that's clearer

- Relevant Equations
- IBP: ∫u dv = uv- ∫v du

I'm pretty confused here because after getting stuck on this problem, I tossed it into an integral calculator and it said the answer was 2 Si(x) + cos(x)/x + C. In intro calc we definitely haven't learned the Si(x) function or even gotten to any of the Taylor polynomial stuff yet.

I tried IBP for sin(x)/x and got -cos(x)/x - ∫cos(x)/x^2 dx. At first this looked promising, since that is my second term. So I had in total now: -cos(x)/x - 2∫cos(x)/x^2 dx.

To do the integral portion, I integrated again (by parts) to find -2∫cos(x)/x^2 dx= 2*cos(x)/x +2∫sin(x)/x dx.

Putting it all together, the original integral = -cos(x)/x + 2*cos(x)/x +2∫sin(x)/x dx= cos(x)/x +2∫sin(x)/x dx. I seem to be getting into a loop, where the integral of cos(x)/x^2 is related to the integral of the sin(x)/x, but can't seem to simplify the terms.

Help on how to proceed?

I tried IBP for sin(x)/x and got -cos(x)/x - ∫cos(x)/x^2 dx. At first this looked promising, since that is my second term. So I had in total now: -cos(x)/x - 2∫cos(x)/x^2 dx.

To do the integral portion, I integrated again (by parts) to find -2∫cos(x)/x^2 dx= 2*cos(x)/x +2∫sin(x)/x dx.

Putting it all together, the original integral = -cos(x)/x + 2*cos(x)/x +2∫sin(x)/x dx= cos(x)/x +2∫sin(x)/x dx. I seem to be getting into a loop, where the integral of cos(x)/x^2 is related to the integral of the sin(x)/x, but can't seem to simplify the terms.

Help on how to proceed?