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Question Number 157257 by john_santu last updated on 21/Oct/21

∫ ((sin^6 x+cos^5 x)/(sin^2 x cos^2 x)) dx

$$\int\:\frac{\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{cos}\:^{\mathrm{5}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx} \\ $$

Answered by qaz last updated on 21/Oct/21

∫((sin^6 x+cos^5 x)/(sin^2 xcos^2 x))dx  =∫((sin^4 x)/(cos^2 x))dx+∫((cos^3 x)/(sin^2 x))dx  =∫((1−2cos^2 x+cos^4 x)/(cos^2 x))dx+∫((1−sin^2 x)/(sin^2 x))d(sin x)  =tan x−2x+(1/2)∫(1+cos 2x)dx−(1/(sin x))−sin x  =tan x−(3/2)x+(1/4)sin 2x−(1/(sin x))−sin x+C

$$\int\frac{\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{5}} \mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$=\int\frac{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\mathrm{dx}+\int\frac{\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$=\int\frac{\mathrm{1}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\mathrm{dx}+\int\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\mathrm{d}\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$=\mathrm{tan}\:\mathrm{x}−\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2x}\right)\mathrm{dx}−\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}−\mathrm{sin}\:\mathrm{x} \\ $$$$=\mathrm{tan}\:\mathrm{x}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2x}−\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}−\mathrm{sin}\:\mathrm{x}+\mathrm{C} \\ $$

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