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N-3-2cos-x-2-3cos-x-dx-

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Question Number 129746 by bramlexs22 last updated on 18/Jan/21
N = ∫ ((3+2cos x)/(2+3cos x)) dx
$$\:\mathrm{N}\:=\:\int\:\frac{\mathrm{3}+\mathrm{2cos}\:\mathrm{x}}{\mathrm{2}+\mathrm{3cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$
Answered by liberty last updated on 18/Jan/21
N= ∫ (((2/3)(2+3cos x)+(8/3))/(2+3cos x)) dx N= (2/3)x + (8/3)∫ (dx/(2+3(2cos^2 (x/2)−1))) N= (2/3)x+(8/3)∫ (dx/(6cos^2 (x/2)−1))
$$\:\mathrm{N}=\:\int\:\frac{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}+\mathrm{3cos}\:\mathrm{x}\right)+\frac{\mathrm{8}}{\mathrm{3}}}{\mathrm{2}+\mathrm{3cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$$$\:\mathrm{N}=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}\:+\:\frac{\mathrm{8}}{\mathrm{3}}\int\:\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{3}\left(\mathrm{2cos}\:^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$\:\mathrm{N}=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}+\frac{\mathrm{8}}{\mathrm{3}}\int\:\frac{\mathrm{dx}}{\mathrm{6cos}\:^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}}\: \\ $$
Answered by Dwaipayan Shikari last updated on 18/Jan/21
2∫((3+((2−2t^2 )/(1+t^2 )))/(2+((3(1−t^2 ))/(1+t^2 )))).(1/(1+t^2 ))dt t=tan(x/2) =2∫((5+t^2 )/(5−t^2 )).(1/(1+t^2 ))dt=(1/3)∫(5+t^2 )((1/(5−t^2 ))+(1/(1+t^2 )))dt =−(1/3)∫1+((10)/(t^2 −5)) +(1/3)∫1+(4/(1+t^2 ))dt =((√5)/3)log(((t+(√5))/(t−(√5))))+((2x)/3) =((√5)/3)log(((tan(x/2)+(√5))/(tan(x/2)−(√5))))+((2x)/3)+C
$$\mathrm{2}\int\frac{\mathrm{3}+\frac{\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}+\frac{\mathrm{3}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}} \\ $$$$=\mathrm{2}\int\frac{\mathrm{5}+{t}^{\mathrm{2}} }{\mathrm{5}−{t}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{\mathrm{3}}\int\left(\mathrm{5}+{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{\mathrm{5}−{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\int\mathrm{1}+\frac{\mathrm{10}}{{t}^{\mathrm{2}} −\mathrm{5}}\:+\frac{\mathrm{1}}{\mathrm{3}}\int\mathrm{1}+\frac{\mathrm{4}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}{log}\left(\frac{{t}+\sqrt{\mathrm{5}}}{{t}−\sqrt{\mathrm{5}}}\right)+\frac{\mathrm{2}{x}}{\mathrm{3}}\:=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}}{log}\left(\frac{{tan}\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{5}}}{{tan}\frac{{x}}{\mathrm{2}}−\sqrt{\mathrm{5}}}\right)+\frac{\mathrm{2}{x}}{\mathrm{3}}+{C}\: \\ $$

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