Question Number 134987 by 0731619177 last updated on 09/Mar/21
Answered by Olaf last updated on 09/Mar/21
![I_2 (k) = ∫_0 ^(π/2) ((xsinxcosx)/( (√(1−k^2 sin^2 x)))) dx I_2 (k) = −(1/k^2 )∫_0 ^(π/2) x[((−2k^2 sinxcosx)/( 2(√(1−k^2 sin^2 x))))] dx I_2 (k) = −(1/k^2 )[x(√(1−k^2 sin^2 x))]_0 ^(π/2) +(1/k^2 )∫_0 ^(π/2) (√(1−k^2 sin^2 x)) dx I_2 (k) = −((π(√(1−k^2 )))/(2k^2 ))+((E(k))/k^2 ) Let k′ = (√(1−k^2 )) I_2 (k) = −((πk′)/(2k^2 ))+((E(k))/k^2 )](https://www.tinkutara.com/question/Q134995.png)
$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\mathrm{sin}{x}\mathrm{cos}{x}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}}\:{dx} \\ $$$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left[\frac{−\mathrm{2}{k}^{\mathrm{2}} \mathrm{sin}{x}\mathrm{cos}{x}}{\:\mathrm{2}\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}}\right]\:{dx} \\ $$$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left[{x}\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$+\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}\:{dx} \\ $$$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:−\frac{\pi\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }}{\mathrm{2}{k}^{\mathrm{2}} }+\frac{\mathrm{E}\left({k}\right)}{{k}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{k}'\:=\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} } \\ $$$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:−\frac{\pi{k}'}{\mathrm{2}{k}^{\mathrm{2}} }+\frac{\mathrm{E}\left({k}\right)}{{k}^{\mathrm{2}} } \\ $$
Commented by 0731619177 last updated on 09/Mar/21
$${tanks}\:{sir} \\ $$$$ \\ $$