pi-4-pi-4-sec-x-e-x-1-dx-

Question Number 128610 by john_santu last updated on 08/Jan/21

\int_{- π / 4}^{π / 4} \frac{\sec x}{e^{x} + 1} dx

Commented by liberty last updated on 09/Jan/21

I = - \int_{- π / 4}^{π / 4} \frac{\sec (- x)}{e^{- x} + 1} d (- x)

I = \int_{- π / 4}^{π / 4} \frac{\sec x}{e^{- x} + 1} dx = \int_{- π / 4}^{π / 4} (\frac{\sec x}{e^{- x} (1 + e^{x})}) dx

I = \int_{- π / 4}^{π / 4} \frac{e^{x} \sec x}{e^{x} + 1} dx

adding together two equation

2 I = \int_{- π / 4}^{π / 4} \frac{\sec x + e^{x} \sec x}{e^{x} + 1} dx = \int_{- π / 4}^{π / 4} \sec x dx

2 I = 2 \int_{0}^{π / 4} \sec x dx

{I = \ln ∣ \sec x + \tan x ∣]}_{0}^{π / 4} = \ln (1 + \sqrt{2})

Answered by mathmax by abdo last updated on 09/Jan/21

I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{cosx (1 + e^{x})} dx =_{x = - t} \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{cost (1 + e^{- t})} dt \Rightarrow

2 I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{cosx} (\frac{1}{1 + e^{x}} + \frac{1}{1 + e^{- x}}) dx = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{cosx} (\frac{1 + e^{- x} + 1 + e^{x}}{1 + e^{- x} + e^{x} + 1}) dx

= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{dx}{cosx} = 2 \int_{0}^{\frac{π}{4}} \frac{dx}{cosx} \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{dx}{cosx} =_{\tan (\frac{x}{2}) = t}

\int_{0}^{\sqrt{2} - 1} \frac{2 dt}{(1 + t^{2}) . \frac{1 - t^{2}}{1 + t^{2}}} = \int_{0}^{\sqrt{2} - 1} (\frac{1}{1 - t} + \frac{1}{1 + t}) dt = {[\ln ∣ \frac{1 + t}{1 - t} ∣]}_{0}^{\sqrt{2} - 1}

= \ln ∣ \frac{\sqrt{2}}{2 - \sqrt{2}} ∣