Question Number 88603 by hovero clinton last updated on 11/Apr/20
Answered by TANMAY PANACEA. last updated on 11/Apr/20
![x=sina ∫_0 ^(π/2) ((lnsina×cosada)/((1+8sin^2 a)cosa)) ∫_0 ^(π/2) lnsina.(1/({1+4(1−cos2a)})) ∫_0 ^(π/2) lnsina.(1/({5−4.((1−tan^2 a)/(1+tan^2 a))}))da ∫_0 ^(π/2) lnsina.((sec^2 a)/((5+5tan^2 a−4+4tan^2 a)))da ∫_0 ^(π/2) lnsina.((sec^2 a)/(1+9tan^2 a))da now lnsina∫((sec^2 a da)/(1+9tan^2 a))−∫[cota∫((sec^2 a)/(1+9tan^2 a))da]da lnsina∫((d(tana))/(9((1/9)+tan^2 a)))−∫cota.∫((d(tana))/(9((1/9)+tan^2 a)))da lnsina×(1/9)×3×tan^(−1) (((tana)/(1/3)))−∫cota.(1/9).3tan^(−1) (((tana)/(1/3)))da ((lnsina.tan^(−1) (3tana))/3)−(1/3)∫cota.tan^(−1) (3tana) wait...](https://www.tinkutara.com/question/Q88608.png)
$${x}={sina} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{lnsina}×{cosada}}{\left(\mathrm{1}+\mathrm{8}{sin}^{\mathrm{2}} {a}\right){cosa}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsina}.\frac{\mathrm{1}}{\left\{\mathrm{1}+\mathrm{4}\left(\mathrm{1}−{cos}\mathrm{2}{a}\right)\right\}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsina}.\frac{\mathrm{1}}{\left\{\mathrm{5}−\mathrm{4}.\frac{\mathrm{1}−{tan}^{\mathrm{2}} {a}}{\mathrm{1}+{tan}^{\mathrm{2}} {a}}\right\}}{da} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsina}.\frac{{sec}^{\mathrm{2}} {a}}{\left(\mathrm{5}+\mathrm{5}{tan}^{\mathrm{2}} {a}−\mathrm{4}+\mathrm{4}{tan}^{\mathrm{2}} {a}\right)}{da} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsina}.\frac{{sec}^{\mathrm{2}} {a}}{\mathrm{1}+\mathrm{9}{tan}^{\mathrm{2}} {a}}{da} \\ $$$${now} \\ $$$${lnsina}\int\frac{{sec}^{\mathrm{2}} {a}\:{da}}{\mathrm{1}+\mathrm{9}{tan}^{\mathrm{2}} {a}}−\int\left[{cota}\int\frac{{sec}^{\mathrm{2}} {a}}{\mathrm{1}+\mathrm{9}{tan}^{\mathrm{2}} {a}}{da}\right]{da} \\ $$$${lnsina}\int\frac{{d}\left({tana}\right)}{\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{9}}+{tan}^{\mathrm{2}} {a}\right)}−\int{cota}.\int\frac{{d}\left({tana}\right)}{\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{9}}+{tan}^{\mathrm{2}} {a}\right)}{da} \\ $$$${lnsina}×\frac{\mathrm{1}}{\mathrm{9}}×\mathrm{3}×{tan}^{−\mathrm{1}} \left(\frac{{tana}}{\frac{\mathrm{1}}{\mathrm{3}}}\right)−\int{cota}.\frac{\mathrm{1}}{\mathrm{9}}.\mathrm{3}{tan}^{−\mathrm{1}} \left(\frac{{tana}}{\frac{\mathrm{1}}{\mathrm{3}}}\right){da} \\ $$$$\frac{{lnsina}.{tan}^{−\mathrm{1}} \left(\mathrm{3}{tana}\right)}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\int{cota}.{tan}^{−\mathrm{1}} \left(\mathrm{3}{tana}\right) \\ $$$$\boldsymbol{{wait}}… \\ $$