The value of the integral [ . 2 ( ^6 + ^4 + ^2 ) 2 ^4 + 3 ^2 + 6 1 - 2 ,d ] 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 :

Accepted Answer

$$\int {{{2{{\sin }^2}\theta \cos \theta ({{\sin }^6}\theta  + {{\sin }^4}\theta  + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta  + 3{{\sin }^2}\theta  + 6} } \over {2{{\sin }^2}\theta }}d\theta } $$

Let sin$$\theta$$ = t, cos$$\theta$$ d$$\theta$$ = dt

$$ = \int {({t^6} + {t^4} + {t^2})\sqrt {2{t^4} + 3{t^2} + 6} \,dt}  $$

$$= \int {({t^5} + {t^3} + t)\sqrt {2{t^6} + 3{t^4} + 6{t^2}} dt} $$

Let $2{t^6} + 3{t^4} + 6{t^2} = z$

$12({t^5} + {t^3} + t)dt = dz$

$$ = {1 \over {12}}\int {\sqrt z dz = {1 \over {18}}{z^{3/2}} + c} $$

$$ = {1 \over {18}}[{(2{\sin ^6}\theta  + 3{\sin ^4}\theta  + 6{\sin ^2}\theta )^{3/2}} + C$$

$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{(1 - {\cos ^2}\theta )^2} + 3 - 3{\cos ^2}\theta  + 6)]^{3/2}} + C$$

$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{\cos ^4}\theta  - 7{\cos ^2}\theta  + 11)]^{3/2}} + C$$

$$ = {1 \over {18}}{[ - 2{\cos ^6}\theta  + 9{\cos ^4}\theta  - 18{\cos ^2}\theta  + 11]^{3/2}} + C$$

Solutions · 1