Question Number 200418 by mnjuly1970 last updated on 18/Nov/23
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{calculus}\:\:\left(\:\:\mathrm{I}\:\:\right)\:\: \\ $$$$\:\:\mathrm{I}{f}\:,\:\:\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\:\pi} \:\frac{\:{x}\:}{\mathrm{1}\:\:+\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}\:\mathrm{d}{x}\:=\:{a}\:\zeta\:\left(\:\mathrm{2}\:\right)\:\: \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\:\:{a}\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:{where}\:\:,\:\:\:\zeta\:\left({s}\:\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\mathrm{1}}{{n}^{\:{s}} } \\ $$$$ \\ $$
Answered by witcher3 last updated on 18/Nov/23
![I=∫_0 ^π ((π−x)/(1+sin^2 (x)))dx=π∫_0 ^π (dx/(1+sin^2 (x)))−I 2I=π∫_0 ^π (dx/(cos^2 (x)+sin^2 (x)+sin^2 (x))) =π∫_0 ^(π/2) (dx/(cos^2 (x)+2sin^2 (x)))+π∫_(π/2) ^π (dx/(cos^2 (x)+2sin^2 (x)))dx t=π−x..in 2nd =2π{∫_0 ^(π/2) (dx/(cos^2 (x)+2sin^2 (x)))} =π∫_0 ^(π/2) (1/(cos^2 (x))).(dx/(1+((√2).tg(x))^2 )) =2π[(1/( (√2)))tan^(−1) ((√2)tan (x))]_0 ^(π/2) =(π^2 /( (√2))) I=(π^2 /(2(√2)))=(6/(2(√2)))ζ(2)∴a=(3/( (√2)))](Q200431.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi} \frac{\pi−\mathrm{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx}=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}−\mathrm{I} \\ $$$$\mathrm{2I}=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)} \\ $$$$=\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)}+\pi\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$\mathrm{t}=\pi−\mathrm{x}..\mathrm{in}\:\mathrm{2nd} \\ $$$$=\mathrm{2}\pi\left\{\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2sin}^{\mathrm{2}} \left(\mathrm{x}\right)}\right\} \\ $$$$=\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}.\frac{\mathrm{dx}}{\mathrm{1}+\left(\sqrt{\mathrm{2}}.\mathrm{tg}\left(\mathrm{x}\right)\right)^{\mathrm{2}} } \\ $$$$=\mathrm{2}\pi\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\mathrm{tan}\:\left(\mathrm{x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{2}}}\zeta\left(\mathrm{2}\right)\therefore\mathrm{a}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 18/Nov/23
$$\:\:{thanks}\:{alot}\:{sir}\:... \\ $$