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Question Number 209687 by uuuuu last updated on 18/Jul/24
Answered by lepuissantcedricjunior last updated on 18/Jul/24
k=∫((sin2x)/(sin^4 x+cos^2 x))dx =∫((2sinxcosx)/(cos^4 x(1+tan^4 x)))dx =∫((2sinx)/(cos^3 x))×(1/(1+tan^4 x))dx =>k=arctan(tan^2 x)+k
$$\boldsymbol{{k}}=\int\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{si}}\overset{\mathrm{4}} {\boldsymbol{{n}x}}+\boldsymbol{{co}}\overset{\mathrm{2}} {\boldsymbol{{s}x}}}\boldsymbol{{dx}} \\ $$$$\:\:=\int\frac{\mathrm{2}\boldsymbol{{sinxcosx}}}{\boldsymbol{{co}}\overset{\mathrm{4}} {\boldsymbol{{s}x}}\left(\mathrm{1}+\boldsymbol{{ta}}\overset{\mathrm{4}} {\boldsymbol{{n}x}}\right)}\boldsymbol{{dx}} \\ $$$$\:\:=\int\frac{\mathrm{2}\boldsymbol{{sinx}}}{\boldsymbol{{co}}\overset{\mathrm{3}} {\boldsymbol{{s}x}}}×\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{ta}}\overset{\mathrm{4}} {\boldsymbol{{n}x}}}\boldsymbol{{dx}} \\ $$$$=>\boldsymbol{{k}}=\boldsymbol{{arctan}}\left(\boldsymbol{{ta}}\overset{\mathrm{2}} {\boldsymbol{{n}x}}\right)+\boldsymbol{{k}} \\ $$
Answered by mr W last updated on 19/Jul/24
=∫((sin 2x)/((sin^2 x+cos^2 x)^2 −2 sin^2 x cos^2 x))dx =∫((sin 2x)/(1−((sin^2 2x)/2)))dx =∫((2 sin 2x)/(2−sin^2 2x))dx =−∫((d(cos 2x))/(1+cos^2 2x))=−∫(du/(1+u^2 )) =−tan^(−1) u+C =−tan^(−1) (cos 2x)+C =tan^(−1) (tan^2 x)+C
$$=\int\frac{\mathrm{sin}\:\mathrm{2}{x}}{\left(\mathrm{sin}^{\mathrm{2}} \:{x}+\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} −\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}}{dx} \\ $$$$=\int\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}}{\mathrm{2}}}{dx} \\ $$$$=\int\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}}{dx} \\ $$$$=−\int\frac{{d}\left(\mathrm{cos}\:\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}}=−\int\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=−\mathrm{tan}^{−\mathrm{1}} {u}+{C} \\ $$$$=−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{2}{x}\right)+{C} \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}^{\mathrm{2}} \:{x}\right)+{C} \\ $$
Answered by mathmax last updated on 19/Jul/24
I=∫ ((sin(2x))/(sin^4 x+cos^4 x))dx ⇒I=∫((sin(2x))/((sin^2 x+cos^2 x)^2 −2sin^2 x cos^2 x))dx =∫ ((sin(2x))/(1−(1/2)sin^2 (2x)))dx =∫ ((2sin(2x))/(2−sin^2 (2x)))dx (2x=t) =∫ ((sint)/(2−sin^2 t))dt =∫((sint)/(((√2)−sint)((√2)+sint)))dt =(1/(2(√2)))∫((1/( (√2)−sint))+(1/( (√2)+sint)))dt =(1/(2(√2))){ I_1 −I_2 } I_1 =_(tan((t/2))=y) ∫ ((2dy)/((1+y^2 )((√2)−((2y)/(1+y^2 ))))) =∫ ((2dy)/( (√2)+(√2)y^2 −2y))=(√2)∫(dy/(y^2 −(√2)y+1)) Δ=2−1=1 ⇒y_1 =(((√2)+1)/2) et y_2 =(((√2)−1)/2) ⇒I_1 =(√2)∫((1/(y−y_1 ))−(1/(y−y_2 )))dy =(√2)ln∣((y−y_1 )/(y−y_2 ))∣+c_1 =(√2)ln∣((tanx−(((√2)+1)/2))/(tanx−(((√2)−1)/2)))∣+c_1 =(√2)ln∣((2tanx−(√2)−1)/(2tanx−(√2)+1))∣+c_1 I_2 =∫ (dt/( (√2)+sint)) =_(tan((t/2))=y) ∫ ((2dy)/((1+y^2 )((√2)+((2y)/(1+y^2 ))))) =∫ ((2dy)/( (√2)+(√2)y^2 +2y))=(√2)∫(dy/(y^2 +(√2)y+1)) y_1 =((−(√2)+1)/2) and y_2 =((−(√2)−1)/2) I_2 =(√2)∫((1/(y−y_1 ))−(1/(y−y_2 )))dy =(√2)ln∣((y−y_1 )/(y−y_2 ))∣ +c_2 =(√2)ln∣((tanx+(((√2)−1)/2))/(tanx+(((√2)+1)/2)))∣+c_2 =(√2)ln∣((2tanx+(√2)−1)/(2tanx+(√2)+1))∣ +c_2 ⇒ I=(1/(2(√2))){I_1 −I_2 }=(1/2){ln∣((2tanx−(√2)−1)/(2tanx−(√2)+1))∣ −ln∣((2tanx+(√2)−1)/(2tanx+(√2)+1))∣ +c
$${I}=\int\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}^{\mathrm{4}} {x}+{cos}^{\mathrm{4}} {x}}{dx}\:\Rightarrow{I}=\int\frac{{sin}\left(\mathrm{2}{x}\right)}{\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {x}\:{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{dx} \\ $$$$=\int\:\frac{\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}−{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{dx}\:\:\:\:\:\left(\mathrm{2}{x}={t}\right) \\ $$$$=\int\:\:\frac{{sint}}{\mathrm{2}−{sin}^{\mathrm{2}} {t}}{dt}\:=\int\frac{{sint}}{\left(\sqrt{\mathrm{2}}−{sint}\right)\left(\sqrt{\mathrm{2}}+{sint}\right)}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}−{sint}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+{sint}}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:{I}_{\mathrm{1}} −{I}_{\mathrm{2}} \right\} \\ $$$${I}_{\mathrm{1}} =_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={y}} \:\:\int\:\:\:\:\frac{\mathrm{2}{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\sqrt{\mathrm{2}}−\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\:\:\frac{\mathrm{2}{dy}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{y}^{\mathrm{2}} −\mathrm{2}{y}}=\sqrt{\mathrm{2}}\int\frac{{dy}}{{y}^{\mathrm{2}} −\sqrt{\mathrm{2}}{y}+\mathrm{1}} \\ $$$$\Delta=\mathrm{2}−\mathrm{1}=\mathrm{1}\:\Rightarrow{y}_{\mathrm{1}} =\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}} \\ $$$${et}\:{y}_{\mathrm{2}} =\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}\:\Rightarrow{I}_{\mathrm{1}} =\sqrt{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{y}−{y}_{\mathrm{1}} }−\frac{\mathrm{1}}{{y}−{y}_{\mathrm{2}} }\right){dy} \\ $$$$=\sqrt{\mathrm{2}}{ln}\mid\frac{{y}−{y}_{\mathrm{1}} }{{y}−{y}_{\mathrm{2}} }\mid+{c}_{\mathrm{1}} =\sqrt{\mathrm{2}}{ln}\mid\frac{{tanx}−\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}}}{{tanx}−\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}}\mid+{c}_{\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}{ln}\mid\frac{\mathrm{2}{tanx}−\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}{tanx}−\sqrt{\mathrm{2}}+\mathrm{1}}\mid+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\:\:\frac{{dt}}{\:\sqrt{\mathrm{2}}+{sint}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={y}} \int\:\:\:\:\frac{\mathrm{2}{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\sqrt{\mathrm{2}}+\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\frac{\mathrm{2}{dy}}{\:\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{2}{y}}=\sqrt{\mathrm{2}}\int\frac{{dy}}{{y}^{\mathrm{2}} +\sqrt{\mathrm{2}}{y}+\mathrm{1}} \\ $$$${y}_{\mathrm{1}} =\frac{−\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}}\:{and}\:{y}_{\mathrm{2}} =\frac{−\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\sqrt{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{y}−{y}_{\mathrm{1}} }−\frac{\mathrm{1}}{{y}−{y}_{\mathrm{2}} }\right){dy} \\ $$$$=\sqrt{\mathrm{2}}{ln}\mid\frac{{y}−{y}_{\mathrm{1}} }{{y}−{y}_{\mathrm{2}} }\mid\:+{c}_{\mathrm{2}} =\sqrt{\mathrm{2}}{ln}\mid\frac{{tanx}+\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}}}{{tanx}+\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}}}\mid+{c}_{\mathrm{2}} \\ $$$$=\sqrt{\mathrm{2}}{ln}\mid\frac{\mathrm{2}{tanx}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}{tanx}+\sqrt{\mathrm{2}}+\mathrm{1}}\mid\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{{I}_{\mathrm{1}} −{I}_{\mathrm{2}} \right\}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\mid\frac{\mathrm{2}{tanx}−\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}{tanx}−\sqrt{\mathrm{2}}+\mathrm{1}}\mid\right. \\ $$$$−{ln}\mid\frac{\mathrm{2}{tanx}+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{2}{tanx}+\sqrt{\mathrm{2}}+\mathrm{1}}\mid\:+{c} \\ $$$$ \\ $$$$ \\ $$

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