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Question Number 103512 by bemath last updated on 15/Jul/20

∫ ((x dx)/((cot x+tan x)^2 )) = (a) (x/(16))−((x sin 4x)/(32))−((cos 4x)/(128))+c (b) (x/(16))+((x sin 4x)/(32))−((cos 4x)/(128))+c (c) (x/(16))+((xsin 4x)/(64))+((cos 4x)/(128))+c (d)(x/(16))+((xcos 4x)/(32))+((sin 4x)/(128))+c

$$\int\:\frac{{x}\:{dx}}{\left(\mathrm{cot}\:{x}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }\:= \\ $$$$\left({a}\right)\:\frac{{x}}{\mathrm{16}}−\frac{{x}\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{32}}−\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$$$\left({b}\right)\:\frac{{x}}{\mathrm{16}}+\frac{{x}\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{32}}−\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$$$\left({c}\right)\:\frac{{x}}{\mathrm{16}}+\frac{{x}\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{64}}+\frac{\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$$$\left({d}\right)\frac{{x}}{\mathrm{16}}+\frac{{x}\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{32}}+\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{128}}+{c} \\ $$

Commented by bobhans last updated on 15/Jul/20

nothing answer too

$${nothing}\:{answer}\:{too} \\ $$

Commented by bemath last updated on 15/Jul/20

yes. i think the question wrong

$${yes}.\:{i}\:{think}\:{the}\:{question}\:{wrong} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Jul/20

∫(x/((((cosx)/(sinx))+((sinx)/(cosx)))^2 ))dx=(1/4)∫xsin^2 2x=∫(x/8)(1−cos4x) ∫(x/8)−(1/8)∫xcos4x (x^2 /(16))−(1/(32))xsin4x+(1/(32))∫sin4x (x^2 /(16))−(1/(32))xsin4x−(1/(128))cos4x+Constant

$$\int\frac{{x}}{\left(\frac{{cosx}}{{sinx}}+\frac{{sinx}}{{cosx}}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{4}}\int{xsin}^{\mathrm{2}} \mathrm{2}{x}=\int\frac{{x}}{\mathrm{8}}\left(\mathrm{1}−{cos}\mathrm{4}{x}\right) \\ $$$$\int\frac{{x}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\int{xcos}\mathrm{4}{x} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{32}}{xsin}\mathrm{4}{x}+\frac{\mathrm{1}}{\mathrm{32}}\int{sin}\mathrm{4}{x} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{32}}{xsin}\mathrm{4}{x}−\frac{\mathrm{1}}{\mathrm{128}}{cos}\mathrm{4}{x}+\mathrm{Constant} \\ $$

Answered by bramlex last updated on 15/Jul/20

cot x+tan x = ((cos x)/(sin x))+((sin x)/(cos x)) = (2/(sin 2x)) ; I = ∫ ((x dx)/(((4/(sin^2 2x))))) = (1/4)∫ x((1/2)−(1/2)cos 4x) dx = (1/8)∫ (x−xcos 4x) dx = (1/8)((x^2 /2)−((xsin 4x)/4)−((cos 4x)/(16)))+c = (x^2 /(16))−((xsin 4x)/(32))−((cos 4x)/(128))+c ■

$$\mathrm{cot}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\:=\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2x}}\:;\:\mathrm{I}\:=\:\int\:\frac{\mathrm{x}\:\mathrm{dx}}{\left(\frac{\mathrm{4}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\int\:\mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{4x}\right)\:\mathrm{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\int\:\left(\mathrm{x}−\mathrm{xcos}\:\mathrm{4x}\right)\:\mathrm{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{xsin}\:\mathrm{4x}}{\mathrm{4}}−\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{16}}\right)+\mathrm{c} \\ $$$$=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{16}}−\frac{\mathrm{xsin}\:\mathrm{4x}}{\mathrm{32}}−\frac{\mathrm{cos}\:\mathrm{4x}}{\mathrm{128}}+\mathrm{c}\:\blacksquare \\ $$

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