Question

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

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

Answer 1

Answer: (D)

$I=\int \frac{\sin \theta .\sin 2\theta \left({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta \right)\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta$
$\Rightarrow I=\int \frac{\sin \theta .2\sin \theta \cos \theta \cdot {\sin }^2\theta \left({\sin }^4\theta +{\sin }^2\theta +1\right){\left(2{\sin }^4\theta +3{\sin }^2\theta +6\right)}^{1/2}}{2{\sin }^2\theta }d\theta$
$=\int {\sin }^2\theta \cdot \cos \theta \left({\sin }^4\theta +{\sin }^2\theta +1\right){\left(2{\sin }^4\theta +3{\sin }^2\theta +6\right)}^{1/2}d\theta$
Let $\sin \theta =t$
$\Rightarrow \cos \theta d\theta =dt$
$\therefore I=\int t^2\left(t^4+t^2+1\right){\left(2t^4+3t^2+6\right)}^{1/2}dt$
$=\int \left(t^5+t^3+t\right)t{\left(2t^4+3t^2+6\right)}^{1/2}dt$
$=\int \left(t^5+t^3+t\right){\left(t^2\right)}^{1/2}{\left(2t^4+3t^2+6\right)}^{1/2}dt$
$=\int \left(t^5+t^3+t\right){\left(2t^6+3t^4+6t^2\right)}^{1/2}dt$
Let $2t^6+3t^4+6t^2=u^2$
$\Rightarrow 12\left(t^5+t^3+t\right)dt=2udu$
$\therefore I=\int {\left(u^2\right)}^{1/2}\cdot \frac{2udu}{12}$
$=\int \frac{u^2}{6}du=\frac{u^3}{18}+C$
$=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C$
when $t=sin\theta$
and $t^2=1-co{s}^2\theta$