Relation and Functions Questions

Question Number 156793 by alcohol last updated on 15/Oct/21

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{sin}\mathrm{2}{x}}{\mathrm{2}−{sin}^{\mathrm{2}} \mathrm{2}{x}}{dx} \\$$

Answered by FongXD last updated on 15/Oct/21

$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin2x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{2x}}\mathrm{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\mathrm{cos2x}\right)'}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{2x}}\mathrm{dx} \\$$ $$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{arctan}\left(\mathrm{cos2x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{4}} \\$$ $$\mathrm{therefore}.\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin2x}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{2x}}\mathrm{dx}=\frac{\pi}{\mathrm{4}} \\$$