Simplify the integrand
We need to evaluate
I = ∫ sin θ sin 2 θ ( sin 6 θ + sin 4 θ + sin 2 θ ) 2 sin 4 θ + 3 sin 2 θ + 6 1 − cos 2 θ d θ . I=\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. I = ∫ 1 − cos 2 θ sin θ sin 2 θ ( sin 6 θ + sin 4 θ + sin 2 θ ) 2 sin 4 θ + 3 sin 2 θ + 6 d θ . Use the identities
sin 2 θ = 2 sin θ cos θ , 1 − cos 2 θ = 2 sin 2 θ . \sin 2\theta=2\sin\theta\cos\theta,
\qquad 1-\cos 2\theta=2\sin^2\theta. sin 2 θ = 2 sin θ cos θ , 1 − cos 2 θ = 2 sin 2 θ . So,
sin θ sin 2 θ 1 − cos 2 θ = sin θ ⋅ 2 sin θ cos θ 2 sin 2 θ = cos θ . \frac{\sin\theta\sin 2\theta}{1-\cos 2\theta}
=\frac{\sin\theta\cdot 2\sin\theta\cos\theta}{2\sin^2\theta}=\cos\theta. 1 − cos 2 θ sin θ sin 2 θ = 2 sin 2 θ sin θ ⋅ 2 sin θ cos θ = cos θ . Hence,
I = ∫ cos θ ( sin 6 θ + sin 4 θ + sin 2 θ ) 2 sin 4 θ + 3 sin 2 θ + 6 d θ . I=\int \cos\theta\, (\sin^6\theta+\sin^4\theta+\sin^2\theta)\sqrt{2\sin^4\theta+3\sin^2\theta+6}\,d\theta. I = ∫ cos θ ( sin 6 θ + sin 4 θ + sin 2 θ ) 2 sin 4 θ + 3 sin 2 θ + 6 d θ . Also,
sin 6 θ + sin 4 θ + sin 2 θ = sin 2 θ ( sin 4 θ + sin 2 θ + 1 ) . \sin^6\theta+\sin^4\theta+\sin^2\theta
=\sin^2\theta(\sin^4\theta+\sin^2\theta+1). sin 6 θ + sin 4 θ + sin 2 θ = sin 2 θ ( sin 4 θ + sin 2 θ + 1 ) .
Substitute
Let
t = sin θ ⇒ d t = cos θ d θ . t=\sin\theta \quad\Rightarrow\quad dt=\cos\theta\,d\theta. t = sin θ ⇒ d t = cos θ d θ . Then
I = ∫ ( t 6 + t 4 + t 2 ) 2 t 4 + 3 t 2 + 6 d t . I=\int (t^6+t^4+t^2)\sqrt{2t^4+3t^2+6}\,dt. I = ∫ ( t 6 + t 4 + t 2 ) 2 t 4 + 3 t 2 + 6 d t . Now set
u = 2 t 4 + 3 t 2 + 6. u=2t^4+3t^2+6. u = 2 t 4 + 3 t 2 + 6. Then
d ν d t = 8 t 3 + 6 t = 2 t ( 4 t 2 + 3 ) . \frac{d\nu}{dt}=8t^3+6t=2t(4t^2+3). d t d ν = 8 t 3 + 6 t = 2 t ( 4 t 2 + 3 ) . This does not directly match, so instead observe a better substitution structure from the options.
Match with option forms
Option C contains
1 18 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3 / 2 + C . \frac{1}{18}\left(11-18\cos^2\theta+9\cos^4\theta-2\cos^6\theta\right)^{3/2}+C. 18 1 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3/2 + C . Let
F ( θ ) = 1 18 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3 / 2 . F(\theta)=\frac{1}{18}\left(11-18\cos^2\theta+9\cos^4\theta-2\cos^6\theta\right)^{3/2}. F ( θ ) = 18 1 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3/2 . Differentiate it.
Let
g ( θ ) = 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ . g(\theta)=11-18\cos^2\theta+9\cos^4\theta-2\cos^6\theta. g ( θ ) = 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ . Then
F ′ ( θ ) = 1 18 ⋅ 3 2 g ( θ ) 1 / 2 g ′ ( θ ) = 1 12 g ( θ ) 1 / 2 g ′ ( θ ) . F'(\theta)=\frac{1}{18}\cdot \frac{3}{2} g(\theta)^{1/2} g'(\theta)=\frac{1}{12}g(\theta)^{1/2}g'(\theta). F ′ ( θ ) = 18 1 ⋅ 2 3 g ( θ ) 1/2 g ′ ( θ ) = 12 1 g ( θ ) 1/2 g ′ ( θ ) . Now compute g ′ ( θ ) g'(\theta) g ′ ( θ )
d d θ ( cos 2 θ ) = − 2 sin θ cos θ , \frac{d}{d\theta}(\cos^2\theta)=-2\sin\theta\cos\theta, d θ d ( cos 2 θ ) = − 2 sin θ cos θ , d d θ ( cos 4 θ ) = − 4 sin θ cos 3 θ , \frac{d}{d\theta}(\cos^4\theta)=-4\sin\theta\cos^3\theta, d θ d ( cos 4 θ ) = − 4 sin θ cos 3 θ , d d θ ( cos 6 θ ) = − 6 sin θ cos 5 θ . \frac{d}{d\theta}(\cos^6\theta)=-6\sin\theta\cos^5\theta. d θ d ( cos 6 θ ) = − 6 sin θ cos 5 θ . So
g ′ ( θ ) = 36 sin θ cos θ − 36 sin θ cos 3 θ + 12 sin θ cos 5 θ . g'(\theta)=36\sin\theta\cos\theta-36\sin\theta\cos^3\theta+12\sin\theta\cos^5\theta. g ′ ( θ ) = 36 sin θ cos θ − 36 sin θ cos 3 θ + 12 sin θ cos 5 θ . Factor:
g ′ ( θ ) = 12 sin θ cos θ ( 3 − 3 cos 2 θ + cos 4 θ ) . g'(\theta)=12\sin\theta\cos\theta(3-3\cos^2\theta+\cos^4\theta). g ′ ( θ ) = 12 sin θ cos θ ( 3 − 3 cos 2 θ + cos 4 θ ) . Thus
F ′ ( θ ) = sin θ cos θ ( 3 − 3 cos 2 θ + cos 4 θ ) g ( θ ) . F'(\theta)=\sin\theta\cos\theta(3-3\cos^2\theta+\cos^4\theta)\sqrt{g(\theta)}. F ′ ( θ ) = sin θ cos θ ( 3 − 3 cos 2 θ + cos 4 θ ) g ( θ ) . Now simplify both factors in terms of sin θ \sin\theta sin θ
Since cos 2 θ = 1 − sin 2 θ \cos^2\theta=1-\sin^2\theta cos 2 θ = 1 − sin 2 θ
3 − 3 cos 2 θ + cos 4 θ = 3 − 3 ( 1 − sin 2 θ ) + ( 1 − sin 2 θ ) 2 = sin 4 θ + sin 2 θ + 1. 3-3\cos^2\theta+\cos^4\theta
=3-3(1-\sin^2\theta)+(1-\sin^2\theta)^2
=\sin^4\theta+\sin^2\theta+1. 3 − 3 cos 2 θ + cos 4 θ = 3 − 3 ( 1 − sin 2 θ ) + ( 1 − sin 2 θ ) 2 = sin 4 θ + sin 2 θ + 1. Also,
g ( θ ) = 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ . g(\theta)=11-18\cos^2\theta+9\cos^4\theta-2\cos^6\theta. g ( θ ) = 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ . Put x = cos 2 θ = 1 − sin 2 θ x=\cos^2\theta=1-\sin^2\theta x = cos 2 θ = 1 − sin 2 θ
g = 11 − 18 x + 9 x 2 − 2 x 3 . g=11-18x+9x^2-2x^3. g = 11 − 18 x + 9 x 2 − 2 x 3 . Substituting x = 1 − s 2 x=1-s^2 x = 1 − s 2 s = sin θ s=\sin\theta s = sin θ
g = 2 s 4 + 3 s 2 + 6. g=2s^4+3s^2+6. g = 2 s 4 + 3 s 2 + 6. So
g ( θ ) = 2 sin 4 θ + 3 sin 2 θ + 6 . \sqrt{g(\theta)}=\sqrt{2\sin^4\theta+3\sin^2\theta+6}. g ( θ ) = 2 sin 4 θ + 3 sin 2 θ + 6 . Therefore,
F ′ ( θ ) = sin θ cos θ ( sin 4 θ + sin 2 θ + 1 ) 2 sin 4 θ + 3 sin 2 θ + 6 . F'(\theta)=\sin\theta\cos\theta(\sin^4\theta+\sin^2\theta+1)\sqrt{2\sin^4\theta+3\sin^2\theta+6}. F ′ ( θ ) = sin θ cos θ ( sin 4 θ + sin 2 θ + 1 ) 2 sin 4 θ + 3 sin 2 θ + 6 . And since
sin θ cos θ ( sin 4 θ + sin 2 θ + 1 ) = cos θ ( sin 6 θ + sin 4 θ + sin 2 θ ) , \sin\theta\cos\theta(\sin^4\theta+\sin^2\theta+1)
=\cos\theta(\sin^6\theta+\sin^4\theta+\sin^2\theta), sin θ cos θ ( sin 4 θ + sin 2 θ + 1 ) = cos θ ( sin 6 θ + sin 4 θ + sin 2 θ ) , we get
F ′ ( θ ) = cos θ ( sin 6 θ + sin 4 θ + sin 2 θ ) 2 sin 4 θ + 3 sin 2 θ + 6 . F'(\theta)=\cos\theta(\sin^6\theta+\sin^4\theta+\sin^2\theta)\sqrt{2\sin^4\theta+3\sin^2\theta+6}. F ′ ( θ ) = cos θ ( sin 6 θ + sin 4 θ + sin 2 θ ) 2 sin 4 θ + 3 sin 2 θ + 6 . This is exactly the simplified integrand found in Step 1.
Hence,
I = 1 18 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3 / 2 + C . I=\frac{1}{18}\left(11-18\cos^2\theta+9\cos^4\theta-2\cos^6\theta\right)^{3/2}+C. I = 18 1 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3/2 + C .
Check options
This matches Option C .
Final Answer:
1 18 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3 / 2 + C \boxed{\frac{1}{18}\left(11-18\cos^2\theta+9\cos^4\theta-2\cos^6\theta\right)^{3/2}+C} 18 1 ( 11 − 18 cos 2 θ + 9 cos 4 θ − 2 cos 6 θ ) 3/2 + C