JEE Main 2021 (25 Feb, Shift 1) Maths Q67: Integrals

Question

The value of the integral
$\int \frac{\sin	heta . \sin 2	heta\ (\sin^6	heta + \sin^4	heta + \sin^2	heta)\sqrt{2\sin^4	heta + 3\sin^2	heta + 6}}{1 - \cos 2	heta}\, d	heta$ is :
(where c is a constant of integration)

Accepted Answer

Option 4: $\frac{1}{18}\left[11 - 18\cos^2	heta + 9\cos^4	heta - 2\cos^6	hetaight]^{\frac{3}{2}} + c$

Answer

$\frac{1}{18}\left[9 - 2\sin^6	heta - 3\sin^4	heta - 6\sin^2	hetaight]^{\frac{3}{2}} + c$

Answer

$\frac{1}{18}\left[9 - 2\cos^6	heta - 3\cos^4	heta - 6\cos^2	hetaight]^{\frac{3}{2}} + c$

Answer

$\frac{1}{18}\left[11 - 18\sin^2	heta + 9\sin^4	heta - 2\sin^6	hetaight]^{\frac{3}{2}} + c$

Answer

$\frac{1}{18}\left[11 - 18\cos^2	heta + 9\cos^4	heta - 2\cos^6	hetaight]^{\frac{3}{2}} + c$

JEE Main 25 February 2021, Shift 1, Maths Q67

Same idea in other shifts