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Question Number 59575 by maxmathsup by imad last updated on 12/May/19
find ∫ ((sin(2x))/(1+cos^2 x))dx
$${find}\:\int\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by Smail last updated on 12/May/19
∫((sin(2x))/(1+cos^2 x))dx=∫((sin(2x))/(1+((1+cos2x)/2)))dx =∫((2sin(2x))/(3+cos2x))dx=−∫((d(3+cos2x))/(3+cos2x)) =−ln∣3+cos2x∣+C Or ∫((sin2x)/(1+cos^2 x))dx=∫((2sinxcosx)/(1+cos^2 x))dx =−∫((d(1+cos^2 x))/(1+cos^2 x))=−ln∣1+cos^2 x∣+C
$$\int\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}=\int\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}{dx} \\ $$$$=\int\frac{\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}{\mathrm{3}+{cos}\mathrm{2}{x}}{dx}=−\int\frac{{d}\left(\mathrm{3}+{cos}\mathrm{2}{x}\right)}{\mathrm{3}+{cos}\mathrm{2}{x}} \\ $$$$=−{ln}\mid\mathrm{3}+{cos}\mathrm{2}{x}\mid+{C} \\ $$$${Or} \\ $$$$\int\frac{{sin}\mathrm{2}{x}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx}=\int\frac{\mathrm{2}{sinxcosx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=−\int\frac{{d}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}=−{ln}\mid\mathrm{1}+{cos}^{\mathrm{2}} {x}\mid+{C} \\ $$
Commented by malwaan last updated on 12/May/19
3+cos2x=3+(2cos^2 x−1) =2+2cos^2 x=2(1+cos^2 x)≠1+cos^2 x What does that mean?
$$\mathrm{3}+{cos}\mathrm{2}{x}=\mathrm{3}+\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right) \\ $$$$=\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} {x}=\mathrm{2}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)\neq\mathrm{1}+{cos}^{\mathrm{2}} {x} \\ $$$${What}\:{does}\:{that}\:{mean}? \\ $$
Commented by Smail last updated on 12/May/19
ln(3+cos2x)+C=ln(2(1+cos^2 x))+C =ln(1+cos^2 x)+ln2+C=ln(1+cos^2 x)+K with K=ln2+C
$${ln}\left(\mathrm{3}+{cos}\mathrm{2}{x}\right)+{C}={ln}\left(\mathrm{2}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)\right)+{C} \\ $$$$={ln}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)+{ln}\mathrm{2}+{C}={ln}\left(\mathrm{1}+{cos}^{\mathrm{2}} {x}\right)+{K} \\ $$$${with}\:{K}={ln}\mathrm{2}+{C} \\ $$
Commented by malwaan last updated on 13/May/19
thank you sir
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}} \\ $$

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