Question

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

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

Answer 1

The correct option is C

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

Explanation for the correct option:

Step 1. Reduction of the integration by substitution

Given:$\int \frac{\sin \theta \sin 2\theta ({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta )\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta$

$$\begin{gathered} =\int \frac{\sin \theta 2\sin \theta \cos \theta ({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta )\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{2{\sin }^2\theta }d\theta \left[\because \sin 2\theta =2\sin \theta \cos \theta \&1-\cos 2\theta =2{\sin }^2\theta \right] \\ =\int \frac{\sin \theta 2\sin \theta \cos \theta ({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta )\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{2{\sin }^2\theta }d\theta \end{gathered}$$

Let $\sin \theta =t$

$\therefore \cos \theta d\theta =dt$

$$\begin{gathered} =\int (t^6+t^4+t^2)\sqrt{2t^4+3t^2+6}dt \\ =\int (t^5+t^3+t)\sqrt{2t^6+3t^4+6t^2}dt \end{gathered}$$

Step 2. Further reduction of the integration by substitution

$\int (t^5+t^3+t)\sqrt{2t^6+3t^4+6t^2}dt$

Let $2t^6+3t^4+6t^2=z$

$12(t^5+t^3+t)dt=dz$

$=\frac{1}{12}\int \sqrt{z}dz$

$=\frac{1}{18}{z}^{\frac{3}{2}}+c$ [Where $c$ is a constant of integration]

Step 3. Substitute the value

$\therefore \frac{1}{18}{z}^{\frac{3}{2}}+c$

$$\begin{gathered} =\frac{1}{18}{(2t^6+3t^4+6t^2)}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[2{\sin }^6\theta +3{\sin }^4\theta +6{\sin }^2\theta \right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[{\sin }^2\theta \left\{2{\left({\sin }^2\theta \right)}^2+3{\sin }^2\theta +6\right\}\right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[\left(1-{\cos }^2\theta \right)\left\{2{\left(1-{\cos }^2\theta \right)}^2+3\left(1-{\cos }^2\theta \right)+6\right\}\right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[\left(1-{\cos }^2\theta \right)\left\{2\left(1+{\cos }^4\theta -2{\cos }^2\theta \right)+3\left(1-{\cos }^2\theta \right)+6\right\}\right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[\left(1-{\cos }^2\theta \right)\left\{2+2{\cos }^4\theta -4{\cos }^2\theta +3-3{\cos }^2\theta +6\right\}\right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[\left(1-{\cos }^2\theta \right)\left(2{\cos }^4\theta -7{\cos }^2\theta +11\right)\right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[2{\cos }^4\theta -7{\cos }^2\theta +11-2{\cos }^6\theta +7{\cos }^4\theta -11{\cos }^2\theta \right]}^{\frac{3}{2}}+c \\ =\frac{1}{18}{\left[11–18{\cos }^2\theta +9{\cos }^4\theta –2{\cos }^6\theta \right]}^{\frac{3}{2}}+c \end{gathered}$$

Hence, option (C) is the correct answer.