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Let x=atanθ⇒dx=asec2θdθ,a2+x2=a∣secθ∣.
Since the integrand is an even function, then it can be simplified to
I:4aII==================2∫0aln(−a+a2+x2a+a2+x2)dx2∫0π/4ln(−a+a∣secθ∣a+a∣secθ∣)⋅asec2θdθ2a∫0π/4ln(−1+∣secθ∣1+∣secθ∣)sec2θdθ2a∫0π/4ln(−1+secθ1+secθ)sec2θdθ2a∫0π/4ln(1−cosθcosθ+1)sec2θdθ2a∫0π/4ln(2sin22x2cos22x)sec2θdθ2a∫0π/4ln(cot2θ)2sec2θdθ∫0π/4=ulncot2θ⋅dvsec2θdθ, Integrate by parts[ln(cot2θ)tanθ]θ→0θ→π/4−∫0π/4dθd(lncot2θ)⋅tanθdθ⎣⎢⎢⎡(tan4π)ln(cot8π)−=0, See (★)z→0limtanzlncot(2z)⎦⎥⎥⎤−∫0π/4cot2θ−csc22θ⋅21⋅tanθdθ[ln(cot8π)−0]+21∫0π/4cot2θcsc22θ⋅tanθdθln(cot8π)+21∫0π/42secθdθln⎝⎜⎜⎛=2+1, See (★★)cot8π⎠⎟⎟⎞+∫0π/4secθ+tanθsecθ(secθ+tanθ)dθln(2+1)+∫0π/4dθdln∣secθ+tanθ∣dθln(2+1)+[ln∣secθ+tanθ∣]0π/4ln(2+1)+[ln(sec4π+tan4π)−ln(sec0+tan0)]ln(2+1)+ln(1+2)=2ln(1+2)8aln(1+2)≈7.050988a.
Notes:
For (★): We can rewrite the limit as z→0limcotz1ln(cot2z), then apply L'Hôpital's rule to obtain a value of 0.
For (★★): To prove that cot8π=2+1 is equivalent to proving that tan8π=2−1. This can be shown by applying the double angle identitytan(2A)=1−tan2A2tanA, where A=8π followed by the quadratic formula.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Let x=atanθ⇒dx=asec2θdθ,a2+x2=a∣secθ∣.
Since the integrand is an even function, then it can be simplified to
I:4aII==================2∫0aln(−a+a2+x2a+a2+x2)dx2∫0π/4ln(−a+a∣secθ∣a+a∣secθ∣)⋅asec2θdθ2a∫0π/4ln(−1+∣secθ∣1+∣secθ∣)sec2θdθ2a∫0π/4ln(−1+secθ1+secθ)sec2θdθ2a∫0π/4ln(1−cosθcosθ+1)sec2θdθ2a∫0π/4ln(2sin22x2cos22x)sec2θdθ2a∫0π/4ln(cot2θ)2sec2θdθ∫0π/4=ulncot2θ⋅dvsec2θdθ, Integrate by parts[ln(cot2θ)tanθ]θ→0θ→π/4−∫0π/4dθd(lncot2θ)⋅tanθdθ⎣⎢⎢⎡(tan4π)ln(cot8π)−=0, See (★)z→0limtanzlncot(2z)⎦⎥⎥⎤−∫0π/4cot2θ−csc22θ⋅21⋅tanθdθ[ln(cot8π)−0]+21∫0π/4cot2θcsc22θ⋅tanθdθln(cot8π)+21∫0π/42secθdθln⎝⎜⎜⎛=2+1, See (★★)cot8π⎠⎟⎟⎞+∫0π/4secθ+tanθsecθ(secθ+tanθ)dθln(2+1)+∫0π/4dθdln∣secθ+tanθ∣dθln(2+1)+[ln∣secθ+tanθ∣]0π/4ln(2+1)+[ln(sec4π+tan4π)−ln(sec0+tan0)]ln(2+1)+ln(1+2)=2ln(1+2)8aln(1+2)≈7.050988a.
Notes:
For (★): We can rewrite the limit as z→0limcotz1ln(cot2z), then apply L'Hôpital's rule to obtain a value of 0.
For (★★): To prove that cot8π=2+1 is equivalent to proving that tan8π=2−1. This can be shown by applying the double angle identity tan(2A)=1−tan2A2tanA, where A=8π followed by the quadratic formula.
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Its one of the best Latex i've seen till now(more than 3-4 years) on brilliant
Thanks a lot. !
Why is it +1/2∫secθ? Shouldn't it be sec theta alone?
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Thanks for spotting my mistake. I've made the necessary changes.