yungman
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I am having some sort of blindness today! I just don't see how I can go from the first step to the second:
2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du = \int_{-1}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du
Where "a" is just a constant.
As you can see, the numerator does not change. so it really boils down to
2\int_0^1\frac{1}{1-u^2} du = \int_{-1}^1\frac{1}{1+u} du
I tried using partial fraction where:
\frac 1 { (1+u)(1-u)}=\frac A {(1+u)}+\frac B {(1-u)}
where A=B=##\frac 1 2##.
\Rightarrow\; 2\int_0^1\frac{1}{1-u^2} du=\int_0^1\frac{1}{1+u} du+\int_0^1\frac{1}{1-u} du
I just don't see it. Please help.
2\int_0^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1-u^2} du = \int_{-1}^1\frac{[\cos^2 (au) +\cos^2a-2\cos (au) \cos a]}{1+u} du
Where "a" is just a constant.
As you can see, the numerator does not change. so it really boils down to
2\int_0^1\frac{1}{1-u^2} du = \int_{-1}^1\frac{1}{1+u} du
I tried using partial fraction where:
\frac 1 { (1+u)(1-u)}=\frac A {(1+u)}+\frac B {(1-u)}
where A=B=##\frac 1 2##.
\Rightarrow\; 2\int_0^1\frac{1}{1-u^2} du=\int_0^1\frac{1}{1+u} du+\int_0^1\frac{1}{1-u} du
I just don't see it. Please help.
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