Question

$\displaystyle \bold \red{ \int \frac{( \sin \theta. \sin2 \theta( { \sin}^{6} \theta + { \sin}^{4} \theta + { \sin}^{2} \theta) \sqrt{2 \: { \sin}^{4} \theta + 3 \: { \sin}^{2} \theta + 6} }{1 - \cos 2 \theta} \: d \theta}$​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\dfrac{\sin(\theta)\cdot\sin(2\,\theta)\left(\sin^6(\theta)+\sin^4(\theta)+\sin^2(\theta)\right)\sqrt{2\,\sin^4(\theta)+3\,\sin^2(\theta)+6}}{1-\cos(2\,\theta)}\,d\theta$

$\displaystyle=\int\dfrac{\sin(\theta)\cdot\sin(2\,\theta)\cdot\sin^2(\theta)\left(\sin^4(\theta)+\sin^2(\theta)+1\right)\sqrt{2\,\sin^4(\theta)+3\,\sin^2(\theta)+6}}{2\,\sin^2(\theta)}\,d\theta$

$\displaystyle=\dfrac{1}{2}\int\sin(\theta)\cdot\sin(2\,\theta)\left(\sin^4(\theta)+\sin^2(\theta)+1\right)\sqrt{2\,\sin^4(\theta)+3\,\sin^2(\theta)+6}\,d\theta$

$\bf{Put\,\,\,\sin^2(\theta)=t}\\\bf{\implies\,2\sin(\theta)\cos(\theta)\,d\theta=dt}\\\bf{\implies\,\sin(2\,\theta)d\theta=dt}$

So,

$\displaystyle=\dfrac{1}{2}\int\,\sqrt{t}\left(t^2+t+1\right)\sqrt{2\,t^2+3\,t+6}\,\,dt$

$\displaystyle=\dfrac{1}{2}\int\,\left(t^2+t+1\right)\sqrt{2\,t^3+3\,t^2+6t}\,\,dt$

$\displaystyle=\dfrac{1}{2\cdot6}\int\,\left(6t^2+6t+6\right)\sqrt{2\,t^3+3\,t^2+6t}\,\,dt$

$\bf{Put\,\,\,2t^3+3t^2+6t=u}\\\bf{\implies\,\left(6t^2+6t+6\right)dt=du}$

$\displaystyle=\dfrac{1}{12}\int\,\sqrt{u}\,\,du$

$=\dfrac{1}{12}\cdot\dfrac{2}{3}\,u^{\frac{3}{2} }+C$

$=\dfrac{1}{18}\,\left(2\,t^3+3\,t^2+6t\right)^{\frac{3}{2} }+C$

$=\dfrac{1}{18}\,\left\{2\,\sin^6(\theta)+3\,\sin^4(\theta)+6\,\sin^2(\theta)\right\}^{\frac{3}{2} }+C$

$=\dfrac{1}{18}\cdot\left\{2\,\sin^6(\theta)+3\,\sin^4(\theta)+6\,\sin^2(\theta)\right\}\cdot\sqrt{2\,\sin^6(\theta)+3\,\sin^4(\theta)+6\,\sin^2(\theta)}+C$