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Question 226 of 255

Q.Evaluate: ∫dθsin⁡θ+sin⁡2θ\displaystyle\int \dfrac{d\theta}{\sin\theta + \sin 2\theta}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2017Subjective· 4mImportance★★★★★
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Factor the denominator, then use the Weierstrass substitution t=tan⁡(θ/2)t=\tan(\theta/2) and partial fractions.

sin⁡θ+sin⁡2θ=sin⁡θ+2sin⁡θcos⁡θ=sin⁡θ (1+2cos⁡θ)\sin\theta+\sin2\theta = \sin\theta+2\sin\theta\cos\theta = \sin\theta\,(1+2\cos\theta)

I=∫dθsin⁡θ(1+2cos⁡θ)I = \int\dfrac{d\theta}{\sin\theta(1+2\cos\theta)}

Substitute t=tan⁡θ2t=\tan\dfrac\theta2, so sin⁡θ=2t1+t2\sin\theta=\dfrac{2t}{1+t^2}, cos⁡θ=1−t21+t2\cos\theta = \dfrac{1-t^2}{1+t^2}, dθ=2 dt1+t2d\theta = \dfrac{2\,dt}{1+t^2}.

sin⁡θ(1+2cos⁡θ)=2t1+t2(1+2(1−t2)1+t2)=2t1+t2⋅3−t21+t2=2t(3−t2)(1+t2)2\sin\theta(1+2\cos\theta) = \dfrac{2t}{1+t^2}\left(1+\dfrac{2(1-t^2)}{1+t^2}\right) = \dfrac{2t}{1+t^2}\cdot\dfrac{3-t^2}{1+t^2} = \dfrac{2t(3-t^2)}{(1+t^2)^2}

I=∫2 dt/(1+t2)2t(3−t2)/(1+t2)2=∫(1+t2) dtt(3−t2)I = \int \dfrac{2\,dt/(1+t^2)}{2t(3-t^2)/(1+t^2)^2} = \int \dfrac{(1+t^2)\,dt}{t(3-t^2)}

Write 1+t2=−(3−t2)+41+t^2 = -(3-t^2)+4:

1+t2t(3−t2)=−1t+4t(3−t2)\dfrac{1+t^2}{t(3-t^2)} = -\dfrac1t + \dfrac{4}{t(3-t^2)}

Partial fractions for 1t(3−t2)=1t(3−t)(3+t)\dfrac{1}{t(3-t^2)} = \dfrac{1}{t(\sqrt3-t)(\sqrt3+t)}:

1t(3−t2)=13t+16(3−t)+16(3+t)\dfrac{1}{t(3-t^2)} = \dfrac{1}{3t} + \dfrac{1}{6(\sqrt3-t)} + \dfrac{1}{6(\sqrt3+t)}

So 4t(3−t2)=43t+23(3−t)+23(3+t)\dfrac{4}{t(3-t^2)} = \dfrac{4}{3t} + \dfrac{2}{3(\sqrt3-t)} + \dfrac{2}{3(\sqrt3+t)}

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