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Question-215963




Question Number 215963 by RoseAli last updated on 23/Jan/25
Answered by Frix last updated on 23/Jan/25
∫(cos x +sin x)^n (1−2sin^2  x)dx=  =∫(cos x +sin x)^n (sin^2  x −cos^2  x)dx=  =∫(cos x +sin x)^(n+1) (sin x −cos x)dx=  =(1/(n+2))(cos x +sin x)^(n+2) +C
$$\int\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)^{{n}} \left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}\right){dx}= \\ $$$$=\int\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)^{{n}} \left(\mathrm{sin}^{\mathrm{2}} \:{x}\:−\mathrm{cos}^{\mathrm{2}} \:{x}\right){dx}= \\ $$$$=\int\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)^{{n}+\mathrm{1}} \left(\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{{n}+\mathrm{2}}\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right)^{{n}+\mathrm{2}} +{C} \\ $$
Commented by mehdee7396 last updated on 23/Jan/25
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$$\:\underline{\underbrace{\lesseqgtr}} \\ $$

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