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Question

Solve sinθdθ(4+cos2θ)(2sin2θ)

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Solution

sinθdθ(4+cos2θ)(2sin2θ)
Lety x=cosθdx=sinθdθ
dx(4+x2)(1+x2) |2sin2θ=1+(1sin2θ)=1+cos2θ=1+x2
=13(11+x214+x2)dx
=13[tan1x12tan1(x2)]
=13[tan1(cosθ)12tan1(cosθ2)]

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