Solveeit Logo
Login

Question

Question: \[\int_{}^{}{\frac{2\sin 2\theta - \cos\theta}{6 - \cos^{2}\theta - 4\sin\theta}d\theta =}\]...

∫2sin⁡2θ−cos⁡θ6−cos⁡2θ−4sin⁡θdθ=\int_{}^{}{\frac{2\sin 2\theta - \cos\theta}{6 - \cos^{2}\theta - 4\sin\theta}d\theta =}

A

2log⁡∣sin⁡2θ−4sin⁡θ+5∣+7tan⁡−1(sin⁡θ−2)+c2\log|\sin^{2}\theta - 4\sin\theta + 5| + 7\tan^{- 1}(\sin\theta - 2) + c

B

2log⁡∣sin⁡2θ−4sin⁡θ+5∣−7tan⁡−1(sin⁡θ−2)+c2\log|\sin^{2}\theta - 4\sin\theta + 5| - 7\tan^{- 1}(\sin\theta - 2) + c

C

−2log⁡∣sin⁡2θ−4sin⁡θ+5∣+7tan⁡−1(sin⁡θ−2)+c- 2\log|\sin^{2}\theta - 4\sin\theta + 5| + 7\tan^{- 1}(\sin\theta - 2) + c

D

−2log⁡∣sin⁡2θ−4sin⁡θ+5∣−7tan⁡−1(sin⁡θ−2)+c- 2\log|\sin^{2}\theta - 4\sin\theta + 5| - 7\tan^{- 1}(\sin\theta - 2) + c

Answer

2log⁡∣sin⁡2θ−4sin⁡θ+5∣+7tan⁡−1(sin⁡θ−2)+c2\log|\sin^{2}\theta - 4\sin\theta + 5| + 7\tan^{- 1}(\sin\theta - 2) + c

Explanation

Solution

I=∫2(2sin⁡θcos⁡θ)−cos⁡θ6−(1−sin⁡2θ)−4sin⁡θdθ⇒I=∫(4sin⁡θ−1)cos⁡θsin⁡2θ−4sin⁡θ+5dθ\mathbf{I =}\int_{}^{}{\frac{\mathbf{2(2}\mathbf{\sin}\mathbf{\theta}\mathbf{\cos}\mathbf{\theta}\mathbf{) -}\mathbf{\cos}\mathbf{\theta}}{\mathbf{6 - (1 -}\mathbf{\sin}^{\mathbf{2}}\mathbf{\theta}\mathbf{) - 4}\mathbf{\sin}\mathbf{\theta}}\mathbf{d\theta}}\mathbf{\Rightarrow}\mathbf{I =}\int_{}^{}{\frac{\mathbf{(4}\mathbf{\sin}\mathbf{\theta}\mathbf{-}\mathbf{1)}\mathbf{\cos}\mathbf{\theta}}{\mathbf{\sin}^{\mathbf{2}}\mathbf{\theta}\mathbf{-}\mathbf{4}\mathbf{\sin}\mathbf{\theta}\mathbf{+ 5}}\mathbf{d\theta}}

Putsin⁡θ=t⇒cos⁡θdθ=dt\sin\theta = t \Rightarrow \cos\theta d\theta = dt, ∴I=∫4t−1t2−4t+5dt\therefore I = \int_{}^{}{\frac{4t - 1}{t^{2} - 4t + 5}dt}

Let 4t−1=Mddt(t2−4t+5)+N\mathbf{4t - 1 = M}\frac{\mathbf{d}}{\mathbf{dt}}\mathbf{(}\mathbf{t}^{\mathbf{2}}\mathbf{- 4t + 5) + N} ⇒4t−1=M(2t−4)+N\mathbf{\Rightarrow}\mathbf{4t}\mathbf{-}\mathbf{1 = M(2t}\mathbf{-}\mathbf{4) + N}

Comparing the coefficient of t and constant terms on both side, then M=2,N=7M = 2,N = 7 ∴I=∫2(2t−4)+7t2−4t+5dt\therefore I = \int_{}^{}{\frac{2(2t - 4) + 7}{t^{2} - 4t + 5}dt}

⇒I=2∫2t−4t2−4t+5dt+∫7dtt2−4t+5⇒I=2log⁡∣t2−4t+5∣+7∫dt(t−2)2+1⇒I=2log⁡∣t2−4t+5∣+7tan⁡−1(t−2)+c=2log⁡∣sin⁡2θ−4sin⁡θ+5∣+7tan⁡−1(sin⁡θ−2)+c\Rightarrow I = 2\int_{}^{}{\frac{2t - 4}{t^{2} - 4t + 5}dt + \int_{}^{}\frac{7dt}{t^{2} - 4t + 5}} \Rightarrow I = 2\log|t^{2} - 4t + 5| + 7\int_{}^{}\frac{dt}{(t - 2)^{2} + 1} \Rightarrow I = 2\log|t^{2} - 4t + 5| + 7\tan^{- 1}(t - 2) + c = 2\log|\sin^{2}\theta - 4\sin\theta + 5| + 7\tan^{- 1}(\sin\theta - 2) + c