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Question Number 133911 by liberty last updated on 25/Feb/21

A=∫ ((cos x)/(sin 5x+sin x)) dx ?

$$\mathcal{A}=\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{5x}+\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Answered by Ar Brandon last updated on 25/Feb/21

A=∫((cosx)/(sin5x+sinx))dx=∫((cosx)/(2sin3xcos2x))dx =(1/2)∫((cosx dx)/((3sinx−4sin^3 x)(1−2sin^2 x))) =(1/2)∫(du/((3u−4u^3 )(1−2u^2 ))) =(1/6)∫(du/(u(1−((2u)/( (√3))))(1+((2u)/( (√3))))(1−(√2)u)(1+(√2)u))) =(1/6)∫{(1/u)−(((√3)/2)/((1−2u/(√3))))+(((√3)/2)/((1+2u/(√3))))+(((√2)/3)/(1−(√2)u))−(((√2)/3)/(1+(√2)u))}du =(1/6){ln∣sinx∣−(3/4)ln∣(((√3)−2sinx)/( (√3)))∣+(3/4)ln∣(((√3)+2sinx)/( (√3)))∣+((ln∣1−(√2)sinx∣)/3)−((ln∣1+(√2)sinx∣)/3)+C}

$$\mathcal{A}=\int\frac{\mathrm{cosx}}{\mathrm{sin5x}+\mathrm{sinx}}\mathrm{dx}=\int\frac{\mathrm{cosx}}{\mathrm{2sin3xcos2x}}\mathrm{dx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cosx}\:\mathrm{dx}}{\left(\mathrm{3sinx}−\mathrm{4sin}^{\mathrm{3}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\left(\mathrm{3u}−\mathrm{4u}^{\mathrm{3}} \right)\left(\mathrm{1}−\mathrm{2u}^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{du}}{\mathrm{u}\left(\mathrm{1}−\frac{\mathrm{2u}}{\:\sqrt{\mathrm{3}}}\right)\left(\mathrm{1}+\frac{\mathrm{2u}}{\:\sqrt{\mathrm{3}}}\right)\left(\mathrm{1}−\sqrt{\mathrm{2}}\mathrm{u}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{u}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\int\left\{\frac{\mathrm{1}}{\mathrm{u}}−\frac{\sqrt{\mathrm{3}}/\mathrm{2}}{\left(\mathrm{1}−\mathrm{2u}/\sqrt{\mathrm{3}}\right)}+\frac{\sqrt{\mathrm{3}}/\mathrm{2}}{\left(\mathrm{1}+\mathrm{2u}/\sqrt{\mathrm{3}}\right)}+\frac{\sqrt{\mathrm{2}}/\mathrm{3}}{\mathrm{1}−\sqrt{\mathrm{2}}\mathrm{u}}−\frac{\sqrt{\mathrm{2}}/\mathrm{3}}{\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{u}}\right\}\mathrm{du} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{6}}\left\{\mathrm{ln}\mid\mathrm{sinx}\mid−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{3}}−\mathrm{2sinx}}{\:\sqrt{\mathrm{3}}}\mid+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\mid\frac{\sqrt{\mathrm{3}}+\mathrm{2sinx}}{\:\sqrt{\mathrm{3}}}\mid+\frac{\mathrm{ln}\mid\mathrm{1}−\sqrt{\mathrm{2}}\mathrm{sinx}\mid}{\mathrm{3}}−\frac{\mathrm{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\mathrm{sinx}\mid}{\mathrm{3}}+\mathrm{C}\right\} \\ $$

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