Feynman?

Calculus Level 5

$\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x} dx=\frac{\ln^A (\sqrt{B}+C)}D$

The equation above holds true for positive integers $A$ , $B$ , $C$ , and $D$ . Find $\min\left(A+B+C+D\right)$ .

The answer is 7.

1 solution

Rohan Shinde
Sep 1, 2019

$\color{#20A900}{\text{Hint:}}$

First Integrate by parts and then use $\color{#69047E}{\mathbf{\text{Legendre Chi Function}}}$

$\color{#D61F06}{\text{Edit:}}$

Using Integration by Parts we get that $\displaystyle \int_0^{\pi/2}\frac{x\cos x}{1+\sin^2x}dx$ $=\displaystyle \frac{\pi^2}{8}-\int_0^{\pi/2} \arctan(\sin x)dx$ $=\displaystyle \frac{\pi^2}{8}-2\chi_2\left(\sqrt{2}-1\right)$

From here we have that $\displaystyle \chi_2\left(\sqrt{2}-1\right)=\frac{\pi^2}{16}-\frac{\ln^2\left(\sqrt{2}+1\right)}{4}$

And hence we get $\displaystyle \int_0^{\pi/2}\frac{x\cos x}{1+\sin^2x}dx=\frac{\ln^2\left(\sqrt{2}+1\right)}{2}$

Can I view the solution somewhere?

Deepima Pathania - 1 year, 7 months ago

Thank you very much for your answers, which deepened my understanding of the Legendre chi function. https://en.wikipedia.org/wiki/Legendre chi function http://mathworld.wolfram.com/LegendresChi-Function.html

Wang xingyu - 1 year, 7 months ago

Please do this sum in a simple manner without using legendre chi function

Venkataramana gandu - 1 year, 4 months ago

Feynman?

The answer is 7.

1 solution

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