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integrate-cosx-cos2x-1-cosx-dx-




Question Number 22151 by j.masanja06@gmail.com last updated on 12/Oct/17
integrate  ∫((cosx−cos2x)/(1+cosx))dx
$${integrate} \\ $$$$\int\frac{{cosx}−{cos}\mathrm{2}{x}}{\mathrm{1}+{cosx}}{dx} \\ $$
Answered by $@ty@m last updated on 12/Oct/17
=∫((cosx−(2cos^2 x−1))/(1+cosx))dx  =∫((1+cosx −2cos^2 x)/(1+cos x))dx  =∫(1−((2cos^2 x)/(1+cos x)))dx  =x−∫  ((2(2cos ^2 (x/2)−1)^2 )/(2cos^2 (x/2) ))dx  =x−∫(((2cos^2 (x/2)−1)/(cos (x/2))))^2 dx  =x−∫(2cos (x/2)−sec (x/2))^2 dx  =x−∫(4cos^2 (x/2)+sec^2 (x/2)−4)dx  =x−2∫(1+cos x)dx−∫sec^2 (x/2)dx+4∫dx  =x−2x−2sin x−2tan (x/2)+4x+C  =3x−2sin x−2tan (x/2)+C Ans.
$$=\int\frac{{cosx}−\left(\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\right)}{\mathrm{1}+{cosx}}{dx} \\ $$$$=\int\frac{\mathrm{1}+\mathrm{cos}{x}\:−\mathrm{2cos}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{cos}\:{x}}{dx} \\ $$$$=\int\left(\mathrm{1}−\frac{\mathrm{2cos}\:^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{cos}\:{x}}\right){dx} \\ $$$$={x}−\int\:\:\frac{\mathrm{2}\left(\mathrm{2cos}\:\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:}{dx} \\ $$$$={x}−\int\left(\frac{\mathrm{2cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{1}}{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}\right)^{\mathrm{2}} {dx} \\ $$$$={x}−\int\left(\mathrm{2cos}\:\frac{{x}}{\mathrm{2}}−\mathrm{sec}\:\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} {dx} \\ $$$$={x}−\int\left(\mathrm{4cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{sec}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{4}\right){dx} \\ $$$$={x}−\mathrm{2}\int\left(\mathrm{1}+\mathrm{cos}\:{x}\right){dx}−\int\mathrm{sec}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}+\mathrm{4}\int{dx} \\ $$$$={x}−\mathrm{2}{x}−\mathrm{2sin}\:{x}−\mathrm{2tan}\:\frac{{x}}{\mathrm{2}}+\mathrm{4}{x}+{C} \\ $$$$=\mathrm{3}{x}−\mathrm{2sin}\:{x}−\mathrm{2tan}\:\frac{{x}}{\mathrm{2}}+{C}\:{Ans}. \\ $$

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