Question

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

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

Answer 1

The correct option is B $\frac{1}{18}{[11-18{\cos }^2\theta +9{\cos }^4\theta -2{\cos }^6\theta ]}^{\frac{3}{2}}+c$

$\int \frac{2{\sin }^2\theta \cos \theta ({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta )\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{2{\sin }^2\theta }d\theta$

Let $\sin \theta =t,\cos \theta d\theta =dt=\int (t^6+t^4+t^2)\sqrt{2t^4+3t^2+6}dt=\int (t^5+t^3+t)\sqrt{2t^6+3t^4+6t^2}dt\text{Let}2t^6+3t^4+6t^2=z12(t^5+t^3+t)dt=dz=\frac{1}{12}\int \sqrt{z}dz=\frac{1}{18}{z}^{3/2}+c=\frac{1}{18}\left[\right(2{\sin }^6\theta +3{\sin }^4\theta +6{\sin }^2\theta {)}^{\frac{3}{2}}+c]=\frac{1}{18}[(1-{\cos }^2\theta )\left\{2\right(1-{\cos }^2\theta {)}^2+3-3{\cos }^2\theta +6\}{]}^{\frac{3}{2}}+c=\frac{1}{18}[(1-{\cos }^2\theta )(2{\cos }^4\theta -7{\cos }^2\theta +11){]}^{\frac{3}{2}}+c=\frac{1}{18}[-2{\cos }^6\theta +9{\cos }^4\theta -18{\cos }^2\theta +11{]}^{\frac{3}{2}}+c$