The value of the integral $\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta $ is :

Question

The value of the integral
$\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta $ is :

${1 \over {18}}{\left[ {9 - 2{{\cos }^6}\theta - 3{{\cos }^4}\theta - 6{{\cos }^2}\theta } \right]^{{3 \over 2}}} + c$

${1 \over {18}}{\left[ {11 - 18{{\sin }^2}\theta + 9{{\sin }^4}\theta - 2{{\sin }^6}\theta } \right]^{{3 \over 2}}} + c$

${1 \over {18}}{\left[ {11 - 18{{\cos }^2}\theta + 9{{\cos }^4}\theta - 2{{\cos }^6}\theta } \right]^{{3 \over 2}}} + c$

${1 \over {18}}{\left[ {9 - 2{{\sin }^6}\theta - 3{{\sin }^4}\theta - 6{{\sin }^2}\theta } \right]^{{3 \over 2}}} + c$

Solution

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