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Question Number 53358 by gunawan last updated on 20/Jan/19

∫  (1/(√(sin^3 x cos x))) dx =

$$\int\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{sin}^{\mathrm{3}} {x}\:\mathrm{cos}\:{x}}}\:{dx}\:= \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

∫(dx/(√(sin^3 xcosx)))  ∫(dx/(√(((sin^3 x)/(cos^3 x))×cos^4 x)))  ∫((sec^2 xdx)/(√(tan^3 x)))  ∫((d(tanx))/((tanx)^(3/2) ))  =(((tanx)^(((−3)/2)+1) )/(((−3)/2)+1))+c  =(((tanx)^((−1)/2) )/((−1)/2))=((−2)/(√(tanx)))+c

$$\int\frac{{dx}}{\sqrt{{sin}^{\mathrm{3}} {xcosx}}} \\ $$$$\int\frac{{dx}}{\sqrt{\frac{{sin}^{\mathrm{3}} {x}}{{cos}^{\mathrm{3}} {x}}×{cos}^{\mathrm{4}} {x}}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} {xdx}}{\sqrt{{tan}^{\mathrm{3}} {x}}} \\ $$$$\int\frac{{d}\left({tanx}\right)}{\left({tanx}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=\frac{\left({tanx}\right)^{\frac{−\mathrm{3}}{\mathrm{2}}+\mathrm{1}} }{\frac{−\mathrm{3}}{\mathrm{2}}+\mathrm{1}}+{c} \\ $$$$=\frac{\left({tanx}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\frac{−\mathrm{1}}{\mathrm{2}}}=\frac{−\mathrm{2}}{\sqrt{{tanx}}}+{c} \\ $$

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