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Question-97892




Question Number 97892 by pranesh last updated on 10/Jun/20
Answered by john santu last updated on 10/Jun/20
sin^2 x = (1/2)−(1/2)cos 2x  sin^4 x = (1/4)−(1/2)cos 2x+(1/8)+(1/8)cos 4x  = (3/8)−(1/2)cos 2x+(1/8)cos 4x  ∫ sin^4 x dx = (3/8)x−(1/4)sin 2x+(1/(32))sin 4x + c
$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x} \\ $$$$\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4x} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4x} \\ $$$$\int\:\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\:\mathrm{dx}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2x}+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\:\mathrm{4x}\:+\:\mathrm{c}\: \\ $$

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