JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 11)
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 :
$$\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta $$ is :
$${1 \over {18}}{\left[ {9 - 2{{\cos }^6}\theta - 3{{\cos }^4}\theta - 6{{\cos }^2}\theta } \right]^{{3 \over 2}}} + c$$
$${1 \over {18}}{\left[ {11 - 18{{\sin }^2}\theta + 9{{\sin }^4}\theta - 2{{\sin }^6}\theta } \right]^{{3 \over 2}}} + c$$
$${1 \over {18}}{\left[ {11 - 18{{\cos }^2}\theta + 9{{\cos }^4}\theta - 2{{\cos }^6}\theta } \right]^{{3 \over 2}}} + c$$
$${1 \over {18}}{\left[ {9 - 2{{\sin }^6}\theta - 3{{\sin }^4}\theta - 6{{\sin }^2}\theta } \right]^{{3 \over 2}}} + c$$
Explanation
$$\int {{{2{{\sin }^2}\theta \cos \theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {2{{\sin }^2}\theta }}d\theta } $$
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 $$2{t^6} + 3{t^4} + 6{t^2} = z$$
$$12({t^5} + {t^3} + t)dt = dz$$
$$ = {1 \over {12}}\int {\sqrt z dz = {1 \over {18}}{z^{3/2}} + c} $$
$$ = {1 \over {18}}[{(2{\sin ^6}\theta + 3{\sin ^4}\theta + 6{\sin ^2}\theta )^{3/2}} + C$$
$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{(1 - {\cos ^2}\theta )^2} + 3 - 3{\cos ^2}\theta + 6)]^{3/2}} + C$$
$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{\cos ^4}\theta - 7{\cos ^2}\theta + 11)]^{3/2}} + C$$
$$ = {1 \over {18}}{[ - 2{\cos ^6}\theta + 9{\cos ^4}\theta - 18{\cos ^2}\theta + 11]^{3/2}} + C$$
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 $$2{t^6} + 3{t^4} + 6{t^2} = z$$
$$12({t^5} + {t^3} + t)dt = dz$$
$$ = {1 \over {12}}\int {\sqrt z dz = {1 \over {18}}{z^{3/2}} + c} $$
$$ = {1 \over {18}}[{(2{\sin ^6}\theta + 3{\sin ^4}\theta + 6{\sin ^2}\theta )^{3/2}} + C$$
$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{(1 - {\cos ^2}\theta )^2} + 3 - 3{\cos ^2}\theta + 6)]^{3/2}} + C$$
$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{\cos ^4}\theta - 7{\cos ^2}\theta + 11)]^{3/2}} + C$$
$$ = {1 \over {18}}{[ - 2{\cos ^6}\theta + 9{\cos ^4}\theta - 18{\cos ^2}\theta + 11]^{3/2}} + C$$
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