Question Number 59575 by maxmathsup by imad last updated on 12/May/19
$${find}\:\int\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by Smail last updated on 12/May/19
$$\int\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}=\int\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}{dx} \\ $$$$=\int\frac{\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}{\mathrm{3}+{cos}\mathrm{2}{x}}{dx}=−\int\frac{{d}\left(\mathrm{3}+{cos}\mathrm{2}{x}\right)}{\mathrm{3}+{cos}\mathrm{2}{x}} \\ $$$$=−{ln}\mid\mathrm{3}+{cos}\mathrm{2}{x}\mid+{C} \\ $$$${Or} \\ $$$$\int\frac{{sin}\mathrm{2}{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}=\int\frac{\mathrm{2}{sinxcosx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=−\int\frac{{d}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}=−{ln}\mid\mathrm{1}+{cos}^{\mathrm{2}} {x}\mid+{C} \\ $$
Commented by malwaan last updated on 12/May/19
$$\mathrm{3}+{cos}\mathrm{2}{x}=\mathrm{3}+\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right) \\ $$$$=\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} {x}=\mathrm{2}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)\neq\mathrm{1}+{cos}^{\mathrm{2}} {x} \\ $$$${What}\:{does}\:{that}\:{mean}? \\ $$
Commented by Smail last updated on 12/May/19
$${ln}\left(\mathrm{3}+{cos}\mathrm{2}{x}\right)+{C}={ln}\left(\mathrm{2}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)\right)+{C} \\ $$$$={ln}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)+{ln}\mathrm{2}+{C}={ln}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)+{K} \\ $$$${with}\:{K}={ln}\mathrm{2}+{C} \\ $$
Commented by malwaan last updated on 13/May/19
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}} \\ $$