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dx-sin-3-x-cos-5-x-




Question Number 160902 by blackmamba last updated on 08/Dec/21
  ∫ (dx/( (√(sin^3 x)) (√(cos^5 x)))) =?
$$\:\:\int\:\frac{{dx}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{3}} {x}}\:\sqrt{\mathrm{cos}\:^{\mathrm{5}} {x}}}\:=? \\ $$
Answered by chhaythean last updated on 09/Dec/21
=∫(dx/( (√(((sin^3 x)/(cos^3 x))×cos^8 x))))  =∫((sec^4 xdx)/(tan^(3/2) x))  let u=tanx⇒du=sec^2 xdx  =∫((1+u^2 )/u^(3/2) )du=−(2/( (√u)))+(2/3)u^(3/2) +c  =−(2/( (√(tanx))))+(2/3)tan^(3/2) x+c
$$=\int\frac{\mathrm{dx}}{\:\sqrt{\frac{\mathrm{sin}^{\mathrm{3}} \mathrm{x}}{\mathrm{cos}^{\mathrm{3}} \mathrm{x}}×\mathrm{cos}^{\mathrm{8}} \mathrm{x}}} \\ $$$$=\int\frac{\mathrm{sec}^{\mathrm{4}} \mathrm{xdx}}{\mathrm{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{x}} \\ $$$$\mathrm{let}\:\mathrm{u}=\mathrm{tanx}\Rightarrow\mathrm{du}=\mathrm{sec}^{\mathrm{2}} \mathrm{xdx} \\ $$$$=\int\frac{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }{\mathrm{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }\mathrm{du}=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{u}}}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{u}^{\frac{\mathrm{3}}{\mathrm{2}}} +\mathrm{c} \\ $$$$=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{tanx}}}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{tan}^{\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{x}+\mathrm{c} \\ $$
Commented by peter frank last updated on 09/Dec/21
great
$$\mathrm{great} \\ $$

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