calculus-I-If-I-0-pi-x-1-sin-2-x-dx-a-2-a-where-s-n-1-1-n-s-

Question Number 200418 by mnjuly1970 last updated on 18/Nov/23

calculus (I)

I f, I = \int_{0}^{π} \frac{x}{1 + \sin^{2} (x)} d x = a ζ (2)

\Rightarrow a = ?

w h e r e, ζ (s) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{s}}

Answered by witcher3 last updated on 18/Nov/23

I = \int_{0}^{π} \frac{π - x}{1 + \sin^{2} (x)} dx = π \int_{0}^{π} \frac{dx}{1 + \sin^{2} (x)} - I

2 I = π \int_{0}^{π} \frac{dx}{\cos^{2} (x) + \sin^{2} (x) + \sin^{2} (x)}

= π \int_{0}^{\frac{π}{2}} \frac{dx}{\cos^{2} (x) + {2 \sin}^{2} (x)} + π \int_{\frac{π}{2}}^{π} \frac{dx}{\cos^{2} (x) + {2 \sin}^{2} (x)} dx

t = π - x . . in 2 nd

= 2 π {\int_{0}^{\frac{π}{2}} \frac{dx}{\cos^{2} (x) + {2 \sin}^{2} (x)}}

= π \int_{0}^{\frac{π}{2}} \frac{1}{\cos^{2} (x)} . \frac{dx}{1 + {(\sqrt{2} . tg (x))}^{2}}

= 2 π {[\frac{1}{\sqrt{2}} \tan^{- 1} (\sqrt{2} \tan (x))]}_{0}^{\frac{π}{2}}

= \frac{π^{2}}{\sqrt{2}}

I = \frac{π^{2}}{2 \sqrt{2}} = \frac{6}{2 \sqrt{2}} ζ (2) ∴ a = \frac{3}{\sqrt{2}}

Commented by mnjuly1970 last updated on 18/Nov/23

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