Question

Find the Integration of the following :-

$\frac{ \cos(x) \times dx }{ ({ \sin(x) })^{3} \times ({1 + ({ \sin(x) })^{6} })^{ \frac{2}{3} } }$​

Answer 1

To Find

$\displaystyle \int\: \dfrac{ \cos(x) \times dx }{ ({ \sin(x) })^{3} \times ({1 + ({ \sin(x) })^{6} })^{ \frac{2}{3} } } \\ \\ \\ \\ Put\:\sin(x)\:=\:t\\ \\ \\ => \:\cos(x)\:dx\:=\:dt \\ \\ \\ Therefore,\:the\:integral\:becomes: \\ \\ \\ \\ \displaystyle \int\: \dfrac{ dt }{ ({ t })^{3} \times ({1 + (t)^{6} })^{ \frac{2}{3} } }\:\:(\because\:\cos(x)\:dx\:=\:dt)$

$\\ \\ \\ Now\:,Observe\:the\:important\:changes\:made\:to\:the\:integral\\ \\ \\ \\ = \: \displaystyle \int \dfrac{\dfrac{dt}{t^3}}{(\dfrac{(t^6)(1\:+\:t^6)}{(t^6)})^{\frac{2}{3}}} \:\:(Obtained\:by\:multiplying\:and\:dividing\:t^6) \\ \\ \\ \\= \: \displaystyle \int \dfrac{\dfrac{dt}{t^7}}{(\dfrac{1}{t^6}\:+\:1)^{\frac{2}{3}}}\\ \\ \\ \\ Put\: \dfrac{1}{t^6}\:+\:1=\:m\\ \\ \\ \\ => \: \dfrac{-6}{t^7}\:dt\:=\:dm \\ \\ \\ \\ \displaystyle \int \dfrac{1}{-6} \: \dfrac{\dfrac{-6\:dt}{t^7}}{(\dfrac{1}{t^6}\:+\:1)^{\frac{2}{3}}}$

$\\ \\ \\ \\ \displaystyle \int \dfrac{-1}{6}\: \dfrac{dm}{m^{\frac{2}{3}}} \\ \\ \\ \\ \:\displaystyle \int \dfrac{-1}{6}\:m^{\frac{-2}{3}}\:dm \\ \\ \\ \\ =\: \dfrac{-3}{6}m^{\frac{1}{3}}\:+\:C\\ \\ \\ Put\: m\:=\:1\:+\: \dfrac{1}{t^6}=1\:+\: \dfrac{1}{{\sin(x)}^6},\:we\:get \\ \\ \\ Integration\:= \dfrac{-1}{2}(1\:+\:\: \dfrac{1}{{\sin(x)}^6}})^{\frac{1}{3}$