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Question Number 38057 by ajfour last updated on 21/Jun/18

∫((cos 5x+cos 4x)/(1−2cos 3x))dx  = ?

$$\int\frac{\mathrm{cos}\:\mathrm{5}{x}+\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{1}−\mathrm{2cos}\:\mathrm{3}{x}}{dx}\:\:=\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jun/18

∫((sin3x(cos5x+coz4x))/(sin3x−sin6x))dx  ∫((sin3x(cos5x+cos4x))/(−2cos((9x)/2)sin((3x)/2)))dx  ∫((2sin((3x)/2)cos((3x)/2)×2cos((9x)/2)cos(x/2))/(−2cos((9x)/2)sin((3x)/2)))dx  =−1×∫((2cos((3x)/2)cos(x/2))/)dx  =−1×∫(cos2x+cosx)dx  =−1×{((sin2x)/2)+((sinx)/)}+c

$$\int\frac{{sin}\mathrm{3}{x}\left({cos}\mathrm{5}{x}+{coz}\mathrm{4}{x}\right)}{{sin}\mathrm{3}{x}−{sin}\mathrm{6}{x}}{dx} \\ $$$$\int\frac{{sin}\mathrm{3}{x}\left({cos}\mathrm{5}{x}+{cos}\mathrm{4}{x}\right)}{−\mathrm{2}{cos}\frac{\mathrm{9}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{dx} \\ $$$$\int\frac{\mathrm{2}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}×\mathrm{2}{cos}\frac{\mathrm{9}{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{−\mathrm{2}{cos}\frac{\mathrm{9}{x}}{\mathrm{2}}{sin}\frac{\mathrm{3}{x}}{\mathrm{2}}}{dx} \\ $$$$=−\mathrm{1}×\int\frac{\mathrm{2}{cos}\frac{\mathrm{3}{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{}{dx} \\ $$$$=−\mathrm{1}×\int\left({cos}\mathrm{2}{x}+{cosx}\right){dx} \\ $$$$=−\mathrm{1}×\left\{\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+\frac{{sinx}}{}\right\}+{c} \\ $$

Commented by ajfour last updated on 21/Jun/18

Very quick and right Sir.

$${Very}\:{quick}\:{and}\:{right}\:{Sir}. \\ $$

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