Integration Questions

Question Number 160767 by cortano last updated on 06/Dec/21

$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=?\: \\$$

Answered by MJS_new last updated on 06/Dec/21

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{4sin}^{\mathrm{2}} \:{x}}{dx}= \\$$ $$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\$$ $$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}= \\$$ $$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{2arctan}\:\mathrm{2}{t}\:−\mathrm{arctan}\:{t}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{6}} \\$$