Dear student,let I=∫2sin2θ-cosθ6-cos2θ-4sinθdθ =∫2sinθcosθ-cosθ5+1-cos2θ-4sinθdθ =∫cosθ2sinθ-1sin2θ-4sinθ+5dθlet sinθ=t cosθdθ=dtI=∫2t-1t2-4t+5dt I=2log|sin2θ-4sinθ+5|+7tan-1(sinθ-2)+CRegards

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D e a r s t u d e n t, l e t I = \int \frac{2 \sin 2 θ - \cos θ}{6 - \cos^{2} θ - 4 \sin θ} d θ = \int \frac{2 \sin θ \cos θ - \cos θ}{5 + 1 - \cos^{2} θ - 4 \sin θ} d θ = \int \frac{\cos θ (2 \sin θ - 1)}{\sin^{2} θ - 4 \sin θ + 5} d θ l e t \sin θ = t \cos θ d θ = d t I = \int \frac{2 t - 1}{t^{2} - 4 t + 5} d t

I = 2 \log | \sin^{2} θ - 4 \sin θ + 5 | + 7 \tan^{- 1} (\sin θ - 2) + C R e g a r d s

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