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Question: The value of the integral $\int \frac{\sin \theta.\sin 2\theta (\sin^6 \theta + \sin^4 \theta + \sin...

The value of the integral ∫sin⁡θ.sin⁡2θ(sin⁡6θ+sin⁡4θ+sin⁡2θ)2sin⁡4θ+3sin⁡2θ+61−cos⁡2θdθ\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

118[11−18sin⁡2θ+9sin⁡4θ−2sin⁡6θ]3/2+c\frac{1}{18}[11-18\sin^2 \theta + 9\sin^4 \theta - 2\sin^6 \theta]^{3/2} + c

B

118[9−2cos⁡6θ−3cos⁡4θ−6cos⁡2θ]3/2+c\frac{1}{18}[9-2\cos^6 \theta - 3\cos^4 \theta - 6\cos^2 \theta]^{3/2} + c

C

118[9−2sin⁡6θ−3sin⁡4θ−6sin⁡2θ]3/2+c\frac{1}{18}[9-2\sin^6 \theta - 3\sin^4 \theta - 6\sin^2 \theta]^{3/2} + c

D

118[11−18cos⁡2θ+9cos⁡4θ−2cos⁡6θ]3/2+c\frac{1}{18}[11-18\cos^2 \theta + 9\cos^4 \theta - 2\cos^6 \theta]^{3/2} + c

Answer

(4)

Explanation

Solution

The given integral is I=∫sin⁡θ.sin⁡2θ(sin⁡6θ+sin⁡4θ+sin⁡2θ)2sin⁡4θ+3sin⁡2θ+61−cos⁡2θdθI = \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.

Simplify the integrand using sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta and 1−cos⁡2θ=2sin⁡2θ1 - \cos 2\theta = 2\sin^2 \theta:

I=∫sin⁡θ⋅(2sin⁡θcos⁡θ)(sin⁡6θ+sin⁡4θ+sin⁡2θ)2sin⁡4θ+3sin⁡2θ+62sin⁡2θdθI = \int \frac{\sin \theta \cdot (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

I=∫cos⁡θ(sin⁡6θ+sin⁡4θ+sin⁡2θ)2sin⁡4θ+3sin⁡2θ+6dθI = \int \cos \theta (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta)\sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6} d\theta.

To find the integral, we can differentiate each option and check if it matches the integrand. Let's check option (4).

Option (4) is 118[11−18cos⁡2θ+9cos⁡4θ−2cos⁡6θ]3/2+c\frac{1}{18}[11-18\cos^2 \theta + 9\cos^4 \theta - 2\cos^6 \theta]^{3/2} + c.

Let x=sin⁡2θx = \sin^2 \theta. Then cos⁡2θ=1−sin⁡2θ=1−x\cos^2 \theta = 1-\sin^2 \theta = 1-x.

Substitute cos⁡2θ=1−x\cos^2 \theta = 1-x into the expression inside the bracket:

11−18(1−x)+9(1−x)2−2(1−x)311-18(1-x) + 9(1-x)^2 - 2(1-x)^3

=11−18+18x+9(1−2x+x2)−2(1−3x+3x2−x3)= 11-18+18x + 9(1-2x+x^2) - 2(1-3x+3x^2-x^3)

=−7+18x+9−18x+9x2−2+6x−6x2+2x3= -7+18x + 9-18x+9x^2 - 2+6x-6x^2+2x^3

=2x3+(9x2−6x2)+(18x−18x+6x)+(−7+9−2)= 2x^3 + (9x^2-6x^2) + (18x-18x+6x) + (-7+9-2)

=2x3+3x2+6x= 2x^3 + 3x^2 + 6x.

So, the term inside the bracket in option (4) is 2sin⁡6θ+3sin⁡4θ+6sin⁡2θ2\sin^6 \theta + 3\sin^4 \theta + 6\sin^2 \theta.

Let Y=2sin⁡6θ+3sin⁡4θ+6sin⁡2θY = 2\sin^6 \theta + 3\sin^4 \theta + 6\sin^2 \theta.

The derivative of option (4) is ddθ(118Y3/2)\frac{d}{d\theta} \left( \frac{1}{18} Y^{3/2} \right).

Using the chain rule, ddθ(118Y3/2)=118⋅32Y1/2⋅dYdθ=112Y1/2dYdθ\frac{d}{d\theta} \left( \frac{1}{18} Y^{3/2} \right) = \frac{1}{18} \cdot \frac{3}{2} Y^{1/2} \cdot \frac{dY}{d\theta} = \frac{1}{12} Y^{1/2} \frac{dY}{d\theta}.

Now, calculate dYdθ\frac{dY}{d\theta}:

dYdθ=ddθ(2sin⁡6θ+3sin⁡4θ+6sin⁡2θ)\frac{dY}{d\theta} = \frac{d}{d\theta} (2\sin^6 \theta + 3\sin^4 \theta + 6\sin^2 \theta)

=2(6sin⁡5θcos⁡θ)+3(4sin⁡3θcos⁡θ)+6(2sin⁡θcos⁡θ)= 2(6\sin^5 \theta \cos \theta) + 3(4\sin^3 \theta \cos \theta) + 6(2\sin \theta \cos \theta)

=12sin⁡5θcos⁡θ+12sin⁡3θcos⁡θ+12sin⁡θcos⁡θ= 12\sin^5 \theta \cos \theta + 12\sin^3 \theta \cos \theta + 12\sin \theta \cos \theta

Factor out 12sin⁡θcos⁡θ12\sin \theta \cos \theta:

=12sin⁡θcos⁡θ(sin⁡4θ+sin⁡2θ+1)= 12\sin \theta \cos \theta (\sin^4 \theta + \sin^2 \theta + 1).

Substitute this back into the derivative of option (4):

112(2sin⁡6θ+3sin⁡4θ+6sin⁡2θ)1/2⋅12sin⁡θcos⁡θ(sin⁡4θ+sin⁡2θ+1)\frac{1}{12} (2\sin^6 \theta + 3\sin^4 \theta + 6\sin^2 \theta)^{1/2} \cdot 12\sin \theta \cos \theta (\sin^4 \theta + \sin^2 \theta + 1)

=(2sin⁡6θ+3sin⁡4θ+6sin⁡2θ)1/2⋅sin⁡θcos⁡θ(sin⁡4θ+sin⁡2θ+1)= (2\sin^6 \theta + 3\sin^4 \theta + 6\sin^2 \theta)^{1/2} \cdot \sin \theta \cos \theta (\sin^4 \theta + \sin^2 \theta + 1).

Factor out sin⁡2θ\sin^2 \theta from inside the square root:

=(sin⁡2θ(2sin⁡4θ+3sin⁡2θ+6))1/2⋅sin⁡θcos⁡θ(sin⁡4θ+sin⁡2θ+1)= (\sin^2 \theta (2\sin^4 \theta + 3\sin^2 \theta + 6))^{1/2} \cdot \sin \theta \cos \theta (\sin^4 \theta + \sin^2 \theta + 1)

=∣sin⁡θ∣2sin⁡4θ+3sin⁡2θ+6⋅sin⁡θcos⁡θ(sin⁡4θ+sin⁡2θ+1)= |\sin \theta| \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6} \cdot \sin \theta \cos \theta (\sin^4 \theta + \sin^2 \theta + 1).

Assuming sin⁡θ>0\sin \theta > 0 (which is typically the case for such integrals unless specified otherwise), ∣sin⁡θ∣=sin⁡θ|\sin \theta| = \sin \theta:

=sin⁡2θcos⁡θ(sin⁡4θ+sin⁡2θ+1)2sin⁡4θ+3sin⁡2θ+6= \sin^2 \theta \cos \theta (\sin^4 \theta + \sin^2 \theta + 1) \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6}.

Rearrange the terms:

=cos⁡θ(sin⁡2θ(sin⁡4θ+sin⁡2θ+1))2sin⁡4θ+3sin⁡2θ+6= \cos \theta (\sin^2 \theta (\sin^4 \theta + \sin^2 \theta + 1)) \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6}

=cos⁡θ(sin⁡6θ+sin⁡4θ+sin⁡2θ)2sin⁡4θ+3sin⁡2θ+6= \cos \theta (\sin^6 \theta + \sin^4 \theta + \sin^2 \theta) \sqrt{2\sin^4 \theta + 3\sin^2 \theta + 6}.

This matches the simplified integrand.