Correct option is (4) (1/18)[11- 18 sin 2 θ + 9 sin 4 θ - 2 cos 6 θ]3/2 + c
\(I=\int\frac{sin\theta. sin\,2\theta(sin^6\theta+sin^4\theta+sin^2\theta)\sqrt{2sin^4\theta+3sin^2\theta+6}}{1-cos\,2\theta}d\theta\)
⇒ \(I=\int\frac{sin\theta. 2sin\,\theta cos\,\theta sin^2\theta(sin^4\theta+sin^2\theta+1)(2sin^4\theta+3sin^2\theta+6)^{1/2}}{2sin^2\theta}d\theta\)
\(=\int{sin^2\theta. cos\,\theta(sin^4\theta+sin^2\theta+1){(2sin^4\theta+3sin^2\theta+6)}^{1/2}}d\theta\)
Let sin θ = t ⇒ cos θ dθ = dt
\(\therefore\) I = ∫t2 (t 4 + t2 +1)(2t4 + 3t2 + 6)1/2 dt
= ∫(t5 + t3 + t )t (2t4 + 3t2 + 6)1/2 dt
= ∫(t5 + t3 + t)(t2 )1/2 (2t4 + 3t2 + 6)1/2 dt
= ∫(t5 + t3+ t) (2t6 + 3t4 + 6t2)1/2dt
Let 2t6 + 3t4 + 6t2 = u2
⇒ 12 (t5 + t3 + t) dt = 2udu
\(\therefore\) I = ∫(u2)1/2.\(\frac{2udu}{12}\)
\(=\int\frac{u^2}6{du}=\frac{u^3}{18}+C\)
\(=\frac{(2t^6+3t^4+6t^2)}{18}+C\)
when t = sin θ
and t2 = 1 - cos2q will give option (4)