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Question-67617




Question Number 67617 by aliesam last updated on 29/Aug/19
Commented by kaivan.ahmadi last updated on 29/Aug/19
∫(((x^2 −2(x^6 )^(1/4) +x))^(1/4) /( (x^3 )^(1/4) ))dx=∫((1/( (x)^(1/4) ))−2(x^3 )^(1/4) +(1/( (x^2 )^(1/4) )))dx=  ∫(x^((−1)/4) −2x^(3/4) +x^((−1)/2) )dx=(4/3)x^(3/4) −(8/7)x^(7/4) +2x^(1/2) +C=  (4/3)(x^3 )^(1/4) −(8/7)x(x^3 )^(1/4) +2(√x)+C
$$\int\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{2}} −\mathrm{2}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} }+{x}}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }}{dx}=\int\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}}−\mathrm{2}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{2}} }}\right){dx}= \\ $$$$\int\left({x}^{\frac{−\mathrm{1}}{\mathrm{4}}} −\mathrm{2}{x}^{\frac{\mathrm{3}}{\mathrm{4}}} +{x}^{\frac{−\mathrm{1}}{\mathrm{2}}} \right){dx}=\frac{\mathrm{4}}{\mathrm{3}}{x}^{\frac{\mathrm{3}}{\mathrm{4}}} −\frac{\mathrm{8}}{\mathrm{7}}{x}^{\frac{\mathrm{7}}{\mathrm{4}}} +\mathrm{2}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +{C}= \\ $$$$\frac{\mathrm{4}}{\mathrm{3}}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }−\frac{\mathrm{8}}{\mathrm{7}}{x}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }+\mathrm{2}\sqrt{{x}}+{C} \\ $$
Commented by mathmax by abdo last updated on 29/Aug/19
let A =∫  ((cos(6x))/(cos(3x)−sin(3x)))dx  changement 3x =t give  A =(1/3)∫   ((cos(2t))/(cost−sint))dx  =(1/3)∫  ((cos(2t)(cost +sint))/(cos^2 t−sin^2 t))dt  =(1/3) ∫   ((cos(2t)(cost +sint))/(cos(2t)))dt =(1/3)∫ (cost +sint)dt  =(1/3)(sint−cost)+C.
$${let}\:{A}\:=\int\:\:\frac{{cos}\left(\mathrm{6}{x}\right)}{{cos}\left(\mathrm{3}{x}\right)−{sin}\left(\mathrm{3}{x}\right)}{dx}\:\:{changement}\:\mathrm{3}{x}\:={t}\:{give} \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{{cost}−{sint}}{dx}\:\:=\frac{\mathrm{1}}{\mathrm{3}}\int\:\:\frac{{cos}\left(\mathrm{2}{t}\right)\left({cost}\:+{sint}\right)}{{cos}^{\mathrm{2}} {t}−{sin}^{\mathrm{2}} {t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\int\:\:\:\frac{{cos}\left(\mathrm{2}{t}\right)\left({cost}\:+{sint}\right)}{{cos}\left(\mathrm{2}{t}\right)}{dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\int\:\left({cost}\:+{sint}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({sint}−{cost}\right)+{C}. \\ $$

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