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Question Number 210593 by efronzo1 last updated on 13/Aug/24

∫ ((sin x cos x)/( ((cos 2x))^(1/3) + (√(cos 2x)))) dx =? ∫ (dx/(sec x ((sin x))^(1/2) + cos x ((cosec^5 x))^(1/3) )) =?

$$\:\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{2x}}\:+\:\sqrt{\mathrm{cos}\:\mathrm{2x}}}\:\mathrm{dx}\:=? \\ $$$$\:\:\int\:\frac{\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}\:\sqrt[{\mathrm{2}}]{\mathrm{sin}\:\mathrm{x}}\:+\:\mathrm{cos}\:\mathrm{x}\:\sqrt[{\mathrm{3}}]{\mathrm{cosec}\:^{\mathrm{5}} \mathrm{x}}}\:=? \\ $$

Answered by Sutrisno last updated on 30/Aug/24

misal cos2x=y^6 −2sin2xdx=6y^5 dy dx=((6y^5 )/(−4sinxcosx))dy =∫((sinxcosx)/( (y^6 )^(1/3) +(√y^6 ))).((6y^5 )/(−4sinxcosx))dy =−(3/2)∫(y^5 /( y^2 +y^3 ))dy =−(3/2)∫(y^3 /( 1+y))dy =−(3/2)∫(((y−1)^3 )/( y))dy =−(3/2)∫((y^3 −3y^2 +3y−1)/( y))dy =−(3/2)∫y^2 −3y+3−(1/y)dy =−(3/2)((1/3)y^3 −(3/2)y^2 +3y−lny)+c =−(3/2)((1/3)(((cos2x))^(1/6) )^3 −(3/2)(((cos2x))^(1/6) )^2 +3((cos2x))^(1/6) −ln((cos2x))^(1/6) )+c =−(3/2)((1/3)(√(cos2x))−(3/2)((cos2x))^(1/3) +3((cos2x))^(1/6) −ln((cos2x))^(1/6) )+c

$${misal} \\ $$$${cos}\mathrm{2}{x}={y}^{\mathrm{6}} \\ $$$$−\mathrm{2}{sin}\mathrm{2}{xdx}=\mathrm{6}{y}^{\mathrm{5}} {dy} \\ $$$${dx}=\frac{\mathrm{6}{y}^{\mathrm{5}} }{−\mathrm{4}{sinxcosx}}{dy} \\ $$$$=\int\frac{{sinxcosx}}{\:\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} }+\sqrt{{y}^{\mathrm{6}} }}.\frac{\mathrm{6}{y}^{\mathrm{5}} }{−\mathrm{4}{sinxcosx}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{y}^{\mathrm{5}} }{\:{y}^{\mathrm{2}} +{y}^{\mathrm{3}} }{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{y}^{\mathrm{3}} }{\:\mathrm{1}+{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\left({y}−\mathrm{1}\right)^{\mathrm{3}} }{\:{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}−\mathrm{1}}{\:{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int{y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{3}−\frac{\mathrm{1}}{{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{3}{y}−{lny}\right)+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}\left(\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)^{\mathrm{2}} +\mathrm{3}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}−{ln}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\sqrt{{cos}\mathrm{2}{x}}−\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{{cos}\mathrm{2}{x}}+\mathrm{3}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}−{ln}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)+{c} \\ $$$$ \\ $$

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