The value of the integral ∫ sinθ · 2θ( ^ 6 θ + ^ 4 θ + ^ 2 θ) 2 ^ 4...

$⇒ I = ∫ \dfrac{sin\theta . 2sin\theta cos\theta \cdot \sin^{2}\theta\left(\sin^{4}\theta + \sin^{2}\theta + 1\right){\left(2\sin^{4}\theta + 3\sin^{2}\theta + 6\right)}^{1 / 2}}{2\sin^{2}\theta}d\theta$

$= ∫ \sin^{2}\theta \cdot cos\theta\left(\sin^{4}\theta + \sin^{2}\theta + 1\right){\left(2\sin^{4}\theta + 3\sin^{2}\theta + 6\right)}^{1 / 2}d\theta$

Let $sin\theta = t ⇒ cos\thetad\theta = dt$

$∴ I = ∫ {t}^{2}\left({t}^{4} + {t}^{2} + 1\right){\left(2{t}^{4} + 3{t}^{2} + 6\right)}^{1 / 2}dt$

$= ∫ \left({t}^{5} + {t}^{3} + t\right)t{\left(2{t}^{4} + 3{t}^{2} + 6\right)}^{1 / 2}dt$

$= ∫ \left({t}^{5} + {t}^{3} + t\right){\left({t}^{2}\right)}^{1 / 2}{\left(2{t}^{4} + 3{t}^{2} + 6\right)}^{1 / 2}dt$

$= ∫ \left({t}^{5} + {t}^{3} + t\right){\left(2{t}^{6} + 3{t}^{4} + 6{t}^{2}\right)}^{1 / 2}dt$

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

$⇒ 12\left({t}^{5} + {t}^{3} + t\right)dt = 2udu$

$∴ I = ∫ {\left({u}^{2}\right)}^{1 / 2} \cdot \dfrac{2udu}{12}$

$= ∫ \dfrac{{u}^{2}}{6}du = \dfrac{{u}^{3}}{18} + C$

$= \dfrac{{\left(2{t}^{6} + 3{t}^{4} + 6{t}^{2}\right)}^{3 / 2}}{18} + C$

when $t = sin\theta$

and ${t}^{2} = 1 - \cos^{2}\theta$ will give

$= \dfrac{1}{18}\left[11 - 18\cos^{2}\theta + 9\cos^{4}\theta - 2\cos^{6}\theta\right]\dfrac{3}{2} + c$