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Find:\(\int\frac{sin\,x(1-cos\,x)}{(1+cos\,x)(5-cos\,x)}dx\) |
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Answer» Let I = \(\int\frac{sin\,x(1-cos\,x)}{(1+cos\,x)(5-cos\,x)}dx\) \(=\int\frac{sin\,x(1-cos\,x)}{(2-(1-cos\,x))(4+(1-cos\,x))}dx\) Let 1 - cos x = t sin x dx = dt ∴ I \(=\int\frac{t\,dt}{(2-t)(4+t)}\) \(=\int\frac{t\,dt}{8-2t-t^2}\) = -1/2 ∫\(\frac{-2t\,dt}{8-2t-t^2}\) = -1/2 ∫\(\frac{(-2t-2+2)\,dt}{8-2t-t^2}\) = -1/2 [∫\(\frac{-2t-2}{8-2t-t^2}dt\) + ∫\(\frac{2}{8-2t-t^2}dt\)] = -1/2 log |8 - 2t - t2| - ∫\(\frac1{9-(t+1)^2}dt\) = -1/2 log |8 - 2t - t2| - \(\frac1{2\times3}\) log \(|\frac{3+(t-1)}{3-(t-1)}|+c\) = -1/2 log |8 - 2t - t2| - 1/6 log \(|\frac{2+t}{4-t}|+c\) = -1/2 log |8 - 2(1 - cos x) - (1 - cos x)2| - 1/6 log \(|\frac{2+1-cos\,x}{4-(1-cos\,x)}|+c\) = -1/2 log |5 - cos2x + 4 cos x| - 1/6 log \(|\frac{3-cos\,x}{3+cos\,x}|+c.\) |
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