QJEEM21Ma32c9: JEE Main 2021 Mathematics - Class 12 JEEM21Ma32c9 Solution: The value of the integral is :

Question

Verified Solution • 2025–26 Session moderate

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 :

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

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

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