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Question-200606




Question Number 200606 by Calculusboy last updated on 20/Nov/23
Answered by Frix last updated on 20/Nov/23
∫((cos x sin^2  x)/(cos x +sin x))dx =^(t=x−(π/4))   =∫((1/4)+((cos t sin t)/2)−((sin^2  t)/2)−((tan t)/4))dt=  =(t/4)−((cos^2  t)/4)−((t+cos t sin t)/4)+((ln cos t)/4)=  =((ln ∣cos x +sin x∣)/4)−(((cos x +sin x)cos x)/4)+C
$$\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}\:\overset{{t}={x}−\frac{\pi}{\mathrm{4}}} {=} \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}}{\mathrm{2}}−\frac{\mathrm{sin}^{\mathrm{2}} \:{t}}{\mathrm{2}}−\frac{\mathrm{tan}\:{t}}{\mathrm{4}}\right){dt}= \\ $$$$=\frac{{t}}{\mathrm{4}}−\frac{\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{4}}−\frac{{t}+\mathrm{cos}\:{t}\:\mathrm{sin}\:{t}}{\mathrm{4}}+\frac{\mathrm{ln}\:\mathrm{cos}\:{t}}{\mathrm{4}}= \\ $$$$=\frac{\mathrm{ln}\:\mid\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\mid}{\mathrm{4}}−\frac{\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)\mathrm{cos}\:{x}}{\mathrm{4}}+{C} \\ $$
Commented by Calculusboy last updated on 21/Nov/23
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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