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Question

Find : sinθ dθ(4+cos2θ)(2sin2θ)

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Solution

Let I=sinθ dθ(4+cos2θ)(2(1cos2θ))
= sinθdθ(4+cos2θ)(1+cos2θ)
= sinθdθ(4+cos2θ) (1+cos2θ)
Put cosθ
sinθ dθ=dt sinθ dθ=dt
I=dt(4+t2)(1+t2)
For Partial Fraction, we have Let
1(4+t2)(1+t2)=At+B4+t2+Ct+D1+t2
1=(At+B)(1+t2)+(Ct+D)(4+t2)
1=(B+4D)+(A+4C)t+(B+D)t2+(A+C)t3

Put t = 0, we have B + 4D = 1 ...(i)
Equating the coeff. of t on both sides, we have A + 4 C = 0...(ii)
Equating the coeff. of t2andt3 respectively,
we obtain B+D = 0...(iii) and A + C = 0 ...(iv)
Solving (i), (ii),(iii) and (iv), we obtain
A = C = 0, B =13 and D=13
1(4+t2)(1+t2)=134+t2+131+t2
I=[1314+t2dt+1311+t2dt]
=[16tan1t2+13tan1t]+C
=16tan1(cosθ2)13tan1(cosθ) + C

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