Integration Questions

Question Number 123896 by john_santu last updated on 29/Nov/20

$$\:\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\:\sqrt{{x}}}\:{dx} \\$$

Answered by mindispower last updated on 29/Nov/20

$${sin}^{\mathrm{2}} \left({x}\right)=\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}} \\$$$$\Leftrightarrow\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}\sqrt{{x}}}{dx} \\$$$$=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}−\int_{\mathrm{0}} ^{\pi} \frac{{cos}\left(\mathrm{2}{x}\right)}{\:\mathrm{2}\sqrt{{x}}}{dx}={A}−{B} \\$$$${x}=\frac{\pi{t}^{\mathrm{2}} }{\mathrm{4}} \\$$$$=\left[\sqrt{{x}}\right]_{\mathrm{0}} ^{\pi} −\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{cos}\left(\frac{\pi}{\mathrm{2}}{t}^{\mathrm{2}} \right)}{\:\sqrt{\pi}{t}}.\frac{\pi}{\mathrm{2}}{tdt} \\$$$$=\sqrt{\pi}−\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{cos}\left(\frac{\pi{t}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{2}}\sqrt{\pi}{dt} \\$$$$=\sqrt{\pi}−\frac{\sqrt{\pi}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} {cos}\left(\pi\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt} \\$$$${Recall}\:\int_{\mathrm{0}} ^{{u}} {cos}\left(\pi\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right){dt}={C}\left({u}\right)\:{Fernesl}\:{integral} \\$$$${we}\:{get} \\$$$$\int_{\mathrm{0}} ^{\pi} \frac{{sin}^{\mathrm{2}} \left({t}\right)}{\mathrm{2}\sqrt{{t}}}{dt}=\sqrt{\pi}−\frac{\sqrt{\pi}}{\mathrm{2}}{C}\left(\mathrm{2}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\left(\mathrm{2}−{C}\left(\mathrm{2}\right)\right) \\$$