0-pi-2-dx-1-tan-x-

Question Number 155686 by john_santu last updated on 03/Oct/21

\int_{0}^{π / 2} \frac{d x}{1 + \tan x} = ?

Answered by peter frank last updated on 03/Oct/21

\int_{0}^{\frac{π}{2}} \frac{\cos x}{\cos x + \sin x} dx

by t - substitution

\cos x = \frac{1 - t^{2}}{1 + t^{2}}

\sin x = \frac{2 t}{1 + t^{2}}

t = \tan \frac{x}{2}

Answered by puissant last updated on 03/Oct/21

Q = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + t a n x}; x = \frac{π}{2} - u \to d x = - d u

\Rightarrow Q = \int_{\frac{π}{2}}^{0} \frac{(- d u)}{1 + t a n (\frac{π}{2} - u)} = \int_{0}^{\frac{π}{2}} \frac{d u}{1 + \frac{1}{t a n u}}

= \int_{0}^{\frac{π}{2}} \frac{t a n u}{1 + t a n u} d u = \int_{0}^{\frac{π}{2}} \frac{1 + t a n u - 1}{1 + t a n u} d u

\Rightarrow 2 Q = \int_{0}^{\frac{π}{2}} d u \Rightarrow Q = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4} . .

∴∵ Q = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + t a n x} = \frac{π}{4} \dots

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