Differentiation Questions

Question Number 167164 by mnjuly1970 last updated on 08/Mar/22

$$\\$$$$\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\:\:\:\:{sin}^{\:\mathrm{2}} \left({x}\right)}{\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)^{\:\mathrm{6}} }\:{dx}=? \\$$

Answered by MJS_new last updated on 08/Mar/22

$$\int\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\left(\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{6}} }{dx}=\int\frac{\mathrm{tan}^{\mathrm{2}} \:{x}\:\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{tan}\:{x}\right)^{\mathrm{6}} }{dx} \\$$$$\:\:\:\:\:\left[{t}=\mathrm{1}+\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\$$$$=\int\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}\right)}{{t}^{\mathrm{6}} }{dt}= \\$$$$=\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{4}}{{t}^{\mathrm{3}} }+\frac{\mathrm{7}}{{t}^{\mathrm{4}} }−\frac{\mathrm{6}}{{t}^{\mathrm{5}} }+\frac{\mathrm{2}}{{t}^{\mathrm{6}} }\right){dt}= \\$$$$=−\frac{\mathrm{1}}{{t}}+\frac{\mathrm{2}}{{t}^{\mathrm{2}} }−\frac{\mathrm{7}}{\mathrm{3}{t}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{2}{t}^{\mathrm{4}} }−\frac{\mathrm{2}}{\mathrm{5}{t}^{\mathrm{5}} }= \\$$$$... \\$$$$=−\frac{\mathrm{30tan}^{\mathrm{4}} \:{x}\:+\mathrm{60tan}^{\mathrm{3}} \:{x}\:+\mathrm{70tan}^{\mathrm{2}} \:{x}\:+\mathrm{35tan}\:{x}\:+\mathrm{7}}{\mathrm{30}\left(\mathrm{1}+\mathrm{tan}\:{x}\right)^{\mathrm{5}} }+{C} \\$$