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Class-12-science»Maths

integrate 1/sinx(5-4 cos x)

$\int \frac{1}{\sin x (5 - 4 \cos x)} ⅆ x \int \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2} (5 - 4 (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}))} ⅆ x \int \frac{\sec^{4} \frac{x}{2}}{2 \tan \frac{x}{2} (5 (1 + \tan^{2} \frac{x}{2}) - 4 (1 - \tan^{2} \frac{x}{2}))} ⅆ x \int \frac{\sec^{2} \frac{x}{2} (1 + \tan^{2} \frac{x}{2})}{2 \tan \frac{x}{2} (1 + 9 \tan^{2} \frac{x}{2})} ⅆ x \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \int \frac{2 (1 + t^{2})}{2 t (1 + 9 t^{2})} ⅆ t = \int \frac{1 + t^{2}}{t (1 + 9 t^{2})} ⅆ t = \int \frac{1 + 9 t^{2} - 8 t^{2}}{t (1 + 9 t^{2})} ⅆ t = \int \frac{1}{t} ⅆ t - \int \frac{8 t}{1 + 9 t^{2}} ⅆ t = \ln t - \frac{4}{9} \ln (1 + 9 t^{2}) = \ln (\tan \frac{x}{2}) - \frac{4}{9} \ln (1 + 9 \tan^{2} \frac{x}{2}) + c$

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