Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 18307 by mondodotto@gmail.com last updated on 18/Jul/17

Answered by ajfour last updated on 18/Jul/17

Q.2  I=∫_0 ^(  π) ((ysin ydy)/(1+cos^2 y))=∫_0 ^(  π) (((π−y)sin ydy)/(1+cos^2 y))    =π∫_0 ^(  π) ((sin ydy)/(1+cos^2 y))−I  ⇒   2I=−πtan^(−1) (cos y)∣_0 ^π        or   I=−(π/2)[tan^(−1) (−1)−tan^(−1) 1]                =−(π/2)(−(π/4)−(π/4))              I=(π^2 /4) .

$$\mathrm{Q}.\mathrm{2} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\:\pi} \frac{\mathrm{ysin}\:\mathrm{ydy}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}=\int_{\mathrm{0}} ^{\:\:\pi} \frac{\left(\pi−\mathrm{y}\right)\mathrm{sin}\:\mathrm{ydy}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}} \\ $$$$\:\:=\pi\int_{\mathrm{0}} ^{\:\:\pi} \frac{\mathrm{sin}\:\mathrm{ydy}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}−\mathrm{I} \\ $$$$\Rightarrow\:\:\:\mathrm{2I}=−\pi\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{y}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$\:\:\:\:\:\mathrm{or}\:\:\:\mathrm{I}=−\frac{\pi}{\mathrm{2}}\left[\mathrm{tan}^{−\mathrm{1}} \left(−\mathrm{1}\right)−\mathrm{tan}^{−\mathrm{1}} \mathrm{1}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\pi}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com