calculate ∫_0 ^(π/2) ((cos^2 x)/(cosx +sinx))dx


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Question Number 65691 by mathmax by abdo last updated on 02/Aug/19

$c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x}{c o s x + s i n x} d x$
Commented by mathmax by abdo last updated on 02/Aug/19

$t a n (\frac{u}{2}) m e t h o d l e t I = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x}{c o s x + s i n x} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{c o s x}{1 + t a n x} d x$ $c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e I = \int_{0}^{1} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{1 + \frac{2 t}{1 - t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= 2 \int_{0}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{2}} \frac{1 - t^{2}}{1 - t^{2} + 2 t} d t = 2 \int_{0}^{1} \frac{{(1 - t^{2})}^{2}}{(1 + t^{2}) (- t^{2} + 2 t + 1)} d t$ $= - 2 \int_{0}^{1} \frac{{(1 - t^{2})}^{2}}{(1 + t^{2}) (t^{2} - 2 t - 1)} d t l e t d e c o m p o s e$ $F (t) = \frac{t^{4} - 2 t^{2} + 1}{(t^{2} + 1) (t^{2} - 2 t - 1)} \Rightarrow F (t) = \frac{t^{4} - 2 t^{2} + 1}{t^{4} - 2 t^{3} - t^{2} + t^{2} - 2 t - 1}$ $= \frac{t^{4} - 2 t^{2} + 1}{t^{4} - 2 t^{3} - 2 t - 1} \Rightarrow F (t) - 1 = \frac{t^{4} - 2 t^{2} + 1 - t^{4} + 2 t^{3} + 2 t + 1}{t^{4} - 2 t^{3} - 2 t - 1}$ $= \frac{2 t^{3} - 2 t^{2} + 2 t + 2}{(t^{2} + 1) (t^{2} - 2 t - 1)} = G (t) = \frac{a t + b}{t^{2} + 1} + \frac{c}{t - t_{1}} + \frac{d}{t - t_{2}}$ $Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} a n d t_{2} = 1 - \sqrt{2}$ $c = {l i m}_{t \to t_{1}} (t - t_{1}) G (t) = \frac{2 t_{1}^{3} - 2 t_{1}^{2} + 2 t_{1} + 2}{(t_{1}^{2} + 1) (t_{1} - t_{2})} (t_{1} - t_{2} = 2 \sqrt{2})$ $d = {l i m}_{t \to t_{2}} (t - t_{2}) G (t_{2}) = \frac{2 t_{2}^{3} - 2 t_{2}^{2} + 2 t_{2} + 2}{(t_{2}^{2} + 1) (t_{2} - t_{1})}$ ${l i m}_{t \to + \infty} t G (t) = 2 = a + c + d \Rightarrow a = 2 - c - d$ $G (0) = - 2 = b - \frac{c}{t_{1}} - \frac{d}{t_{2}} \Rightarrow b = \frac{c}{t_{1}} + \frac{d}{t_{2}} - 2 r e s t t o s i m p l i f y t h e$ $c a l c u l u s \Rightarrow \int F (t) d t = t + \int G (t) d t$ $= t + c l n ∣ t - t_{1} ∣ + d ∣ t - t_{2} ∣ + \frac{a}{2} l n (t^{2} + 1) + b a r c t a n t + λ$ $. . . . b e c o n t i n u e d . . . .$
	Answered by Tanmay chaudhury last updated on 02/Aug/19

	$I = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} x}{c o s x + s i n x} d x$ $I = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{2} (\frac{π}{2} - x)}{c o s (\frac{π}{2} - x) + s i n (\frac{π}{2} - x)} d x$ $= \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} x}{s i n x + c o s x} d x$ $2 I = \int_{0}^{\frac{π}{2}} \frac{d x}{s i n x + c o s x}$ $2 I = \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{2} (\frac{1}{\sqrt{2}} s i n x + \frac{1}{\sqrt{2}} c o s x)}$ $I = \frac{1}{2 \sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{d x}{s i n (\frac{π}{4} + x)}$ $I = \frac{1}{2 \sqrt{2}} \int_{0}^{\frac{π}{4}} c o s e c (\frac{π}{4} + x) d x$ $I = \frac{1}{2 \sqrt{2}} ∣ l n t a n (\frac{\frac{π}{4} + x}{2}) ∣_{0}^{\frac{π}{4}}$ $= \frac{1}{2 \sqrt{2}} [l n t a n (\frac{π}{4}) - l n t a n (\frac{π}{8})]$ $= \frac{- 1}{2 \sqrt{2}} l n t a n (\frac{π}{8})$
	Commented by mathmax by abdo last updated on 02/Aug/19

	$t h a n k y o u s i r t a n m a y$
	Commented by Tanmay chaudhury last updated on 02/Aug/19

	$m o s t e e l c o m e s i r . . .$
	Commented by peter frank last updated on 02/Aug/19

	$t h a n k y o u$
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