The value of the integral {θ \; 2θ (^{6}θ+^{4}θ+²θ) {2^{4}θ+3²θ+6}}{1-2θ} dθ is (where, c is a constant of integration) [25 Feb 2021 Shift 1]

Correct Answer is : 1/18[11−18cos⁡2θ+9cos⁡4θ−2cos⁡6θ] 3/2+c\;{1}/{18}[11-18 ² θ+9 ^{4} θ-2 ^{6} θ]^{\;{3}/{2}}+c1/18[11−18cos2θ+9cos4θ−2cos6θ]3/2+c

1/18 [9-2^{6}θ-3^{4}θ-6²θ]³/2} + c

1/18 [11-18²θ+9^{4}θ-2^{6}θ]³/2} + c

\;{1}/{18}[11-18 ² θ+9 ^{4} θ-2 ^{6} θ]^{\;{3}/{2}}+c

Correct Answer is : 1/18[11−18cos⁡2θ+9cos⁡4θ−2cos⁡6θ] 3/2+c\;{1}/{18}[11-18 ² θ+9 ^{4} θ-2 ^{6} θ]^{\;{3}/{2}}+c1/18[11−18cos2θ+9cos4θ−2cos6θ]3/2+c

Test Index

Indefinite Integrals

Section: Mathematics

Question : 45 of 90

Marks: +1, -0

\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-\cos2\theta} d\theta

c

Solution:

Let

\int \left[ \frac{\sin\theta \cdot \sin 2\theta (\sin^{6}\theta+\sin^{4}\theta+\sin^{2}\theta) \sqrt{2\sin^{4}\theta+3\sin^{2}\theta+6}}{1-\cos2\theta} \right] d\theta

\because \;\; \sin 2A = 2 \sin A \cos A

and

1-\cos 2A = 2 \sin^{2} A

\sin\theta \cdot 2 \sin\theta (\sin^{6}\theta+\sin^{4}\theta+\sin^{2}\theta)

I = \int \; \frac{\sqrt{2\sin^{4}\theta+3\sin^{2}\theta+6}}{2\sin^{2}\theta} d\theta

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

= \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

2t^{6}+3t^{4}+6t^{2}=z

Let

\therefore \;\; dz = (12t^{5}+12t^{3}+12t) dt

\therefore \;\; dz = 12(t^{5}+t^{3}+t) dt

\frac{1}{12} \int \sqrt{z} dz = \frac{1}{12} \times \frac{z^{3/2}}{3/2} + c

Now,

= \frac{1}{18} z^{3/2} + c

= \frac{1}{18} [2t^{6}+3t^{4}+6t^{2}]^{3/2} + c

= \frac{1}{18} [2\sin^{6}\theta+3\sin^{4}\theta+6\sin^{2}\theta]^{3/2} + c

= \frac{1}{18} \left[ (1-\cos^{2}\theta) \{ 2(1-\cos^{2}\theta)^{3}+3-3\cos^{2}\theta+6 \} \right]^{3/2} + c

= \frac{1}{18} [(1-\cos^{2}\theta)(2\cos^{4}\theta-7\cos^{2}\theta+11)]^{3/2} + c

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

= \frac{1}{18} [11-18\cos^{2}\theta+9\cos^{4}\theta-2\cos^{6}\theta]^{3/2}

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

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