Question Number 140998 by mohammad17 last updated on 14/May/21
Answered by Ar Brandon last updated on 14/May/21
$$\mathcal{I}=\int\frac{\mathrm{sin4x}}{\mathrm{cos}^{\mathrm{4}} \mathrm{x}+\mathrm{4}}\mathrm{dx}=\int\frac{\mathrm{2sin2xcos2x}}{\left(\frac{\mathrm{1}+\mathrm{cos2x}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}}\mathrm{dx},\:\mathrm{c}=\mathrm{cos2x} \\ $$$$\:\:\:=−\int\frac{\mathrm{c}}{\frac{\left(\mathrm{c}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{4}}\mathrm{dc}=−\mathrm{4}\int\frac{\mathrm{cdc}}{\mathrm{c}^{\mathrm{2}} +\mathrm{2c}+\mathrm{17}}\mathrm{dc} \\ $$
Commented by mohammad17 last updated on 14/May/21
$${chose}\left(\:{a}\:\right){or}\left(\:{b}\right)\:{or}\:\left({c}\right)\:{or}\left(\:{d}\right) \\ $$$$ \\ $$$${i}\:{think}\:{d}\: \\ $$