Question Number 214301 by efronzo1 last updated on 04/Dec/24
$$\:\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{x}\right)+\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx}\:=? \\ $$
Commented by Frix last updated on 04/Dec/24
$$\mathrm{Should}\:\mathrm{be}\:\frac{\pi}{\mathrm{2}} \\ $$
Answered by mr W last updated on 04/Dec/24
![∫_0 ^(π/2) [sin^2 (sin x)+cos^2 (cos x)]dx =∫_0 ^(π/2) [sin^2 (sin x)+1−sin^2 (cos x)]dx =(π/2)+∫_0 ^(π/2) sin^2 (sin x) dx−∫_0 ^(π/2) sin^2 (cos x) dx =(π/2)+∫_0 ^(π/2) sin^2 (sin x) dx−∫_(π/2) ^0 sin^2 (cos ((π/2)−t)) d((π/2)−t) =(π/2)+∫_0 ^(π/2) sin^2 (sin x) dx+∫_(π/2) ^0 sin^2 (sin t) dt =(π/2)+∫_0 ^(π/2) sin^2 (sin x) dx−∫_0 ^(π/2) sin^2 (sin x) dx =(π/2)](https://www.tinkutara.com/question/Q214316.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)+\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{cos}\:{x}\right)\right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)+\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{cos}\:{x}\right)\right]{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{cos}\:{x}\right)\:{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:{dx}−\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−{t}\right)\right)\:{d}\left(\frac{\pi}{\mathrm{2}}−{t}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:{dx}+\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{t}\right)\:{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:{x}\right)\:{dx} \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$