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Question Number 111616 by mathdave last updated on 04/Sep/20
solve the integral problem ∫((cosx)/(2−sin2x))dx
$${solve}\:{the}\:{integral}\:{problem} \\ $$$$\int\frac{\mathrm{cos}{x}}{\mathrm{2}−\mathrm{sin2}{x}}{dx} \\ $$
Answered by Her_Majesty last updated on 05/Sep/20
∫((cos x)/(2−sin 2x))dx= (use t=tan (x/2)) =−∫((t^2 −1)/(t^4 +2t^3 +2t^2 −2t+1))dt= =((√3)/6)∫(((t−1)/(t^2 +(1−(√3))t+2−(√3)))−((t−1)/(t^2 +(1+(√3))t+2+(√3))))dt= =((√3)/(12))ln ((t^2 +(1−(√3))t+2−(√3))/(t^2 +(1+(√3))t+2+(√3))) −(1/2)(arctan ((1−(√3))t−1) +arctan ((1+(√3))t−1)) now put t=tan (x/2)
$$\int\frac{\mathrm{cos}\:{x}}{\mathrm{2}−\mathrm{sin}\:\mathrm{2}{x}}{dx}= \\ $$$$\left(\mathrm{use}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right) \\ $$$$=−\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}{dt}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\int\left(\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){t}+\mathrm{2}−\sqrt{\mathrm{3}}}−\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){t}+\mathrm{2}+\sqrt{\mathrm{3}}}\right){dt}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\:\frac{{t}^{\mathrm{2}} +\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){t}+\mathrm{2}−\sqrt{\mathrm{3}}}{{t}^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){t}+\mathrm{2}+\sqrt{\mathrm{3}}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arctan}\:\left(\left(\mathrm{1}−\sqrt{\mathrm{3}}\right){t}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){t}−\mathrm{1}\right)\right) \\ $$$$\mathrm{now}\:\mathrm{put}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$
Answered by ajfour last updated on 05/Sep/20
let x=(π/4)−θ I=−∫((cos ((π/4)−θ)dθ)/(2−cos 2θ)) =−(1/(2(√2)))∫((cos θ+sin θ)/(sin^2 θ))dθ =(1/(2(√2))){cosec θ−ln ∣cosec θ−cot θ∣}+c I =(1/(2(√2)sin ((π/4)−x)))−(1/(2(√2)))ln∣tan ((π/8)−(x/2))∣+c
$${let}\:{x}=\frac{\pi}{\mathrm{4}}−\theta \\ $$$${I}=−\int\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}−\theta\right){d}\theta}{\mathrm{2}−\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta}{d}\theta \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\mathrm{cosec}\:\theta−\mathrm{ln}\:\mid\mathrm{cosec}\:\theta−\mathrm{cot}\:\theta\mid\right\}+{c} \\ $$$$\:\:{I}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−{x}\right)}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}−\frac{{x}}{\mathrm{2}}\right)\mid+{c}\: \\ $$
Commented by Her_Majesty last updated on 05/Sep/20
great, I haven′t thought of this
$${great},\:{I}\:{haven}'{t}\:{thought}\:{of}\:{this} \\ $$

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