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Q.

The value of the integral  sinθ.sin2θ(sin6θ+sin4θ+sin2θ)2sin4θ+3sin2θ+61cos2θdθ is : (where C is a constant of integration)

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a

118[1118sin2θ+9sin4θ2sin6θ]32+C

b

118[92cos6θ3cos4θ6cos2θ]32+C

c

118[92sin6θ3sin4θ6sin2θ]32+C

d

118[1118cos2θ+9cos4θ2cos6θ]32+C

answer is D.

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Detailed Solution

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I=sinθ.sin2θ(sin6θ+sin4θ+sin2θ)2sin4θ+3sin2θ+61cos2θdθ I=sinθ.2sinθcosθ.sin2θ(sin4θ+sin2θ+1)(2sin4θ+3sin2θ+6)122sin2θdθ Let  sinθ=tcosθdθ=dt I=t2(t4+t2+1)(2t4+3t2+6)12dt =(t5+t3+t)t(2t4+3t2+6)12dt =(t5+t3+t)(t2)12(2t4+3t2+6)12dt =(t5+t3+t)(2t6+3t4+6t2)12dt Let,  2t6+3t4+6t2=u2 12(t5+t3+t)dt=2udu  I=(u2)12.2udu12 =u26du=u318+C =(2t6+3t4+6t2)3218+C When   t = sin θ

And  t2=1cos2θ will give option (4) 

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