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Question Number 1350 by 112358 last updated on 24/Jul/15

Evaluate the following integral: I=∫_(π/4) ^( π/2) (cos2x+sin2x)ln(cosx+sinx) dx

$${Evaluate}\:{the}\:{following}\:{integral}: \\ $$$${I}=\int_{\pi/\mathrm{4}} ^{\:\pi/\mathrm{2}} \left({cos}\mathrm{2}{x}+{sin}\mathrm{2}{x}\right){ln}\left({cosx}+{sinx}\right)\:{dx} \\ $$

Commented by prakash jain last updated on 25/Jul/15

∫sin 2xln (cos x+sin x)dx =ln (cos x+sin x)(−((cos 2x)/2))+∫((cos 2x)/2) ∙((cos x−sin x)/(cos x+sin x))dx ∫((cos 2x)/2) ∙((cos x−sin x)/(cos x+sin x))dx=(1/2)∫(1−sin 2x)dx =(x/2)+((cos 2x)/4) From above ∫sin 2xln (cos x+sin x)dx = =(x/2)+((cos 2x)/4)−((cos 2x)/2)ln (cos x+sin x)

$$\int\mathrm{sin}\:\mathrm{2}{x}\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right){dx} \\ $$$$=\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)\left(−\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\right)+\int\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\:\centerdot\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{dx} \\ $$$$\int\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\:\centerdot\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}{x}\right){dx} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{4}} \\ $$$$\mathrm{From}\:\mathrm{above} \\ $$$$\int\mathrm{sin}\:\mathrm{2}{x}\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right){dx}\:= \\ $$$$\:\:\:\:\:\:\:\:=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{4}}−\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right) \\ $$

Commented by prakash jain last updated on 25/Jul/15

∫cos 2xln (cos x+sin x)dx =ln (cos x+sin x)((sin 2x)/2)−∫ ((sin 2x)/2)∙((cos x−sin x)/(cos x+sin x))dx ∫ ((sin 2x)/2)∙((cos x−sin x)/(cos x+sin x))dx= (1/2)[sin xcos x−ln (cos x+sin x)

$$\int\mathrm{cos}\:\mathrm{2}{x}\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right){dx} \\ $$$$=\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}−\int\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\centerdot\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{dx} \\ $$$$\int\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}\centerdot\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{dx}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\:{x}\mathrm{cos}\:{x}−\mathrm{ln}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)\right. \\ $$

Commented by 112358 last updated on 25/Jul/15

Thanks.

$${Thanks}.\: \\ $$

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