Question Number 73429 by Henri Boucatchou last updated on 12/Nov/19
![Solve : ∫(([cos^(−1) x(√(1−x^2 ))]^(−1) )/(log_e [2+((sin(2x(√(1−x^2 ))))/π)]))dx Evaluate ∫_(−π/2) ^( π/2) sin^2 xcos^2 x(cosx+sinx)dx](https://www.tinkutara.com/question/Q73429.png)
$$\:\:\:{Solve}\::\:\int\frac{\left[{cos}^{−\mathrm{1}} {x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right]^{−\mathrm{1}} }{{log}_{{e}} \left[\mathrm{2}+\frac{{sin}\left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\pi}\right]}{dx} \\ $$$$\:\:{Evaluate}\:\:\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\left({cosx}+{sinx}\right){dx} \\ $$
Commented by MJS last updated on 12/Nov/19
$$\int\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right){dx}= \\ $$$$=\int\mathrm{sin}^{\mathrm{3}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:{dx}+\int\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{3}} \:{x}\:{dx}= \\ $$$$=−\frac{\int\mathrm{sin}\:\mathrm{5}{x}\:{dx}}{\mathrm{16}}+\frac{\int\mathrm{sin}\:\mathrm{3}{x}\:{dx}}{\mathrm{16}}+\frac{\int\mathrm{sin}\:{x}\:{dx}}{\mathrm{8}}− \\ $$$$\:\:\:\:\:−\frac{\int\mathrm{cos}\:\mathrm{5}{x}\:{dx}}{\mathrm{16}}−\frac{\int\mathrm{cos}\:\mathrm{3}{x}\:{dx}}{\mathrm{16}}+\frac{\int\mathrm{cos}\:{x}\:{dx}}{\mathrm{8}}= \\ $$$$=\frac{\mathrm{cos}\:\mathrm{5}{x}}{\mathrm{80}}−\frac{\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{48}}−\frac{\mathrm{cos}\:{x}}{\mathrm{8}}−\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{80}}−\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{48}}+\frac{\mathrm{sin}\:{x}}{\mathrm{8}}+{C} \\ $$$$\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right){dx}=\frac{\mathrm{4}}{\mathrm{15}} \\ $$
Commented by MJS last updated on 12/Nov/19
$$\mathrm{please}\:\mathrm{check}\:\mathrm{the}\:\mathrm{first}\:\mathrm{one},\:\mathrm{is}\:\mathrm{it}\:\mathrm{sin}\:\mathrm{or}\:\mathrm{sin}^{−\mathrm{1}} ? \\ $$
Commented by MJS last updated on 12/Nov/19
$$…\mathrm{anyway}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it} \\ $$
Answered by MJS last updated on 12/Nov/19
![∫sin^2 x cos^2 x (cos x +sin x)dx= =∫sin^3 x cos^2 x dx+∫sin^2 x cos^3 x dx= [u=cos x → dx=−(du/(sin x)); v=sin x → dx=(dv/(cos x))] =∫u^4 −u^2 du+∫v^2 −v^4 dv= =(u^5 /5)−(u^3 /3)+(v^3 /3)−(v^5 /5)= =((cos^5 x)/5)−((cos^3 x)/3)+((sin^3 x)/3)−((sin^5 x)/5)+C ∫_(−(π/2)) ^(π/2) sin^2 x cos^2 x (cos x +sin x)dx=(4/(15))](https://www.tinkutara.com/question/Q73432.png)
$$\int\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right){dx}= \\ $$$$=\int\mathrm{sin}^{\mathrm{3}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:{dx}+\int\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{3}} \:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{cos}\:{x}\:\rightarrow\:{dx}=−\frac{{du}}{\mathrm{sin}\:{x}};\:{v}=\mathrm{sin}\:{x}\:\rightarrow\:{dx}=\frac{{dv}}{\mathrm{cos}\:{x}}\right] \\ $$$$=\int{u}^{\mathrm{4}} −{u}^{\mathrm{2}} {du}+\int{v}^{\mathrm{2}} −{v}^{\mathrm{4}} {dv}= \\ $$$$=\frac{{u}^{\mathrm{5}} }{\mathrm{5}}−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}+\frac{{v}^{\mathrm{3}} }{\mathrm{3}}−\frac{{v}^{\mathrm{5}} }{\mathrm{5}}= \\ $$$$=\frac{\mathrm{cos}^{\mathrm{5}} \:{x}}{\mathrm{5}}−\frac{\mathrm{cos}^{\mathrm{3}} \:{x}}{\mathrm{3}}+\frac{\mathrm{sin}^{\mathrm{3}} \:{x}}{\mathrm{3}}−\frac{\mathrm{sin}^{\mathrm{5}} \:{x}}{\mathrm{5}}+{C} \\ $$$$\underset{−\frac{\pi}{\mathrm{2}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right){dx}=\frac{\mathrm{4}}{\mathrm{15}} \\ $$