limit 0 to pie/2 integral dx/2cosx+4sinx

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\int_{0}^{π / 2} \frac{1}{2 cosx + 4 sinx} d x = \int_{0}^{π / 2} \frac{1}{\frac{2 (1 - \tan^{2} \frac{x}{2})}{1 + \tan^{2} \frac{x}{2}} + \frac{4 (2 \tan \frac{x}{2})}{1 + \tan^{2} \frac{x}{2}}} dx = \int_{0}^{π / 2} \frac{1 + \tan^{2} \frac{x}{2}}{2 - 2 \tan^{2} \frac{x}{2} + 8 \tan \frac{x}{2}} dx = \int_{0}^{π / 2} \frac{\sec^{2} \frac{x}{2}}{2 - 2 \tan^{2} \frac{x}{2} + 8 \tan \frac{x}{2}} dx Let \tan \frac{x}{2} = t . Then \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt Also, x = 0 \Rightarrow t = \tan 0 = 0 and x = \frac{π}{2} \Rightarrow t = \tan \frac{π}{4} = 1 Therefore, I = \int_{0}^{1} \frac{\sec^{2} \frac{x}{2}}{2 - 2 t^{2} + 8 t} . \frac{2 dt}{\sec^{2} \frac{x}{2}} = \int_{0}^{1} \frac{1}{1 - t^{2} + 4 t} dt = \int_{0}^{1} \frac{- 1}{t^{2} - 4 t - 1} dt = \int_{0}^{1} \frac{- dt}{t^{2} - 4 t + 4 - 4 - 1} = \int_{0}^{1} \frac{- dt}{[{(t - 2)}^{2} - 5]} = \int_{0}^{1} \frac{dt}{{(\sqrt{5})}^{2} - {(t - 2)}^{2}} dt = \frac{1}{2 \sqrt{5}} {[\log |\frac{\sqrt{5} + t - 2}{\sqrt{5} - t + 2}|]}^{1}_{0} = \frac{1}{2 \sqrt{5}} [\log (\frac{\sqrt{5} - 1}{\sqrt{5} + 1}) - \log (\frac{\sqrt{5} - 2}{\sqrt{5} + 2})] = \frac{1}{2 \sqrt{5}} [\log \{\frac{(\sqrt{5} - 1) (\sqrt{5} + 2)}{(\sqrt{5} + 1) (\sqrt{5} - 2)}\}] = \frac{1}{2 \sqrt{5}} \log (\frac{3 + \sqrt{5}}{3 - \sqrt{5}}) = \frac{1}{2 \sqrt{5}} \log (\frac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}}) = \frac{1}{2 \sqrt{5}} \log {(\frac{3 + \sqrt{5}}{2})}^{2} = \frac{1}{\sqrt{5}} \log (\frac{3 + \sqrt{5}}{2})

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