- #1
pierce15
- 315
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1. The problem, the whole problem, and nothing but the problem
[tex] \int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx [/tex]
Integration by parts
trig substitution
My first idea was to break up the integral by letting [itex]u=x[/itex] and [itex]dv=sin(x)/(1+cos^2 x)[/itex]. I will omit the work, but it got me here:
[tex] \int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx = (x \cdot tan^{-1} (cos(x)) \big|_0^\pi - \int_0^\pi tan^{-1}(cos(x)) \, dx [/tex]
I am fairly sure that I arrived at the second integral correctly. Is it integrateable, if that's a word?
[tex] \int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx [/tex]
Homework Equations
Integration by parts
trig substitution
The Attempt at a Solution
My first idea was to break up the integral by letting [itex]u=x[/itex] and [itex]dv=sin(x)/(1+cos^2 x)[/itex]. I will omit the work, but it got me here:
[tex] \int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx = (x \cdot tan^{-1} (cos(x)) \big|_0^\pi - \int_0^\pi tan^{-1}(cos(x)) \, dx [/tex]
I am fairly sure that I arrived at the second integral correctly. Is it integrateable, if that's a word?
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