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Question-141532




Question Number 141532 by sarkor last updated on 20/May/21
Answered by qaz last updated on 20/May/21
∫((3sin x−2cos x)/(1+cos x))dx  =∫((6sin (x/2)cos (x/2)−4cos^2 (x/2)+2)/(2cos^2 (x/2)))dx  =∫(3tan (x/2)−2+sec^2 (x/2))dx  =6ln∣cos (x/2)∣−2x+2tan (x/2)+C
$$\int\frac{\mathrm{3sin}\:{x}−\mathrm{2cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}{dx} \\ $$$$=\int\frac{\mathrm{6sin}\:\frac{{x}}{\mathrm{2}}\mathrm{cos}\:\frac{{x}}{\mathrm{2}}−\mathrm{4cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}}{\mathrm{2cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$=\int\left(\mathrm{3tan}\:\frac{{x}}{\mathrm{2}}−\mathrm{2}+\mathrm{sec}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\mathrm{6}{ln}\mid\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mid−\mathrm{2}{x}+\mathrm{2tan}\:\frac{{x}}{\mathrm{2}}+{C} \\ $$

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