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0 votes
14.5k views
in Mathematics by (34.9k points)
edited by

The value of the integral

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

(where c is a constant of integration)

(1) (1/18)[11 - 18 sin θ + 9 sin θ - 2 sin 6 θ]3/2 + c

(2) (1/18)[9 - 2 cos θ - 3 cos θ - 6 cos 2 θ]3/2 + c

(3) (1/18)[9 - 2 sin θ + 3 sin θ - sin 2 θ]3/2 + c

(4) (1/18)[11- 18 sin θ + 9 sin θ - 2 cos 6 θ]3/2 + c

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1 Answer

+1 vote
by (34.4k points)

Correct option is (4) (1/18)[11- 18 sin 2 θ + 9 sin 4 θ - 2 cos 6 θ]3/2 + c

\(I=\int\frac{sin\theta. sin\,2\theta(sin^6\theta+sin^4\theta+sin^2\theta)\sqrt{2sin^4\theta+3sin^2\theta+6}}{1-cos\,2\theta}d\theta\)

⇒ \(I=\int\frac{sin\theta. 2sin\,\theta cos\,\theta sin^2\theta(sin^4\theta+sin^2\theta+1)(2sin^4\theta+3sin^2\theta+6)^{1/2}}{2sin^2\theta}d\theta\)

\(=\int{sin^2\theta. cos\,\theta(sin^4\theta+sin^2\theta+1){(2sin^4\theta+3sin^2\theta+6)}^{1/2}}d\theta\)

Let sin θ = t ⇒ cos θ dθ = dt

\(\therefore\) I = ∫t2 (t 4 + t2 +1)(2t4 + 3t2 + 6)1/2 dt

=  ∫(t5 + t3 + t )t (2t4 + 3t2 + 6)1/2 dt

= ∫(t5 + t3 + t)(t2 )1/2 (2t4 + 3t2 + 6)1/2 dt

=  ∫(t+ t3+ t) (2t+ 3t+ 6t2)1/2dt

Let 2t6 + 3t4 + 6t2 = u2
⇒ 12 (t5 + t3 + t) dt = 2udu

\(\therefore\) I = ∫(u2)1/2.\(\frac{2udu}{12}\)

\(=\int\frac{u^2}6{du}=\frac{u^3}{18}+C\)

\(=\frac{(2t^6+3t^4+6t^2)}{18}+C\)

when t = sin θ
and t2 = 1 - cos2q will give option (4)

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