find ∫_0 ^π (dx/(cosx +sinx))


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Question Number 35059 by math khazana by abdo last updated on 14/May/18

$f i n d \int_{0}^{π} \frac{d x}{c o s x + s i n x}$
Commented by math khazana by abdo last updated on 15/May/18
$let put I = ∫_0 ^π (dx/(cosx +sinx)) changement tan((x/2))=t give x=2arctant and I = ∫_0 ^(+∞) (1/(((1−t^2 )/(1+t^2 )) +((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) I = ∫_0 ^(+∞) ((2dt)/(1−t^2 +2t)) = −2 ∫_0 ^∞ (dt/(t^2 −2t −1)) =−2 ∫_0 ^∞ (dt/((t−1)^2 −2)) =−2 ∫_0 ^∞ (dt/((t−3)(t+1))) =(1/2) ∫_0 ^∞ { (1/(t+1)) −(1/(t−3))}dt =(1/2) [ln∣((t+1)/(t−3))∣]_0 ^(+∞) =(1/2)( −ln((1/3))) I =(1/2)ln(3).$
$l e t p u t I = \int_{0}^{π} \frac{d x}{c o s x + s i n x} c h a n g e m e n t$ $t a n (\frac{x}{2}) = t g i v e x = 2 a r c t a n t a n d$ $I = \int_{0}^{+ \infty} \frac{1}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $I = \int_{0}^{+ \infty} \frac{2 d t}{1 - t^{2} + 2 t} = - 2 \int_{0}^{\infty} \frac{d t}{t^{2} - 2 t - 1}$ $= - 2 \int_{0}^{\infty} \frac{d t}{{(t - 1)}^{2} - 2}$ $= - 2 \int_{0}^{\infty} \frac{d t}{(t - 3) (t + 1)}$ $= \frac{1}{2} \int_{0}^{\infty} {\frac{1}{t + 1} - \frac{1}{t - 3}} d t$ $= \frac{1}{2} {[l n ∣ \frac{t + 1}{t - 3} ∣]}_{0}^{+ \infty} = \frac{1}{2} (- l n (\frac{1}{3}))$ $I = \frac{1}{2} l n (3) .$
Commented by prof Abdo imad last updated on 16/May/18
$error in the final lines I =−2 ∫_0 ^∞ (dt/((t−1)^2 −2)) =−2 ∫_0 ^∞ (dt/((t−1−_ (√2))(t−1 +(√2)))) =((−2)/(2(√2)))∫_0 ^∞ { (1/(t−1−(√2))) −(1/(t−1+(√2)))}dt =−(1/(√2))[ln∣((t−1−(√2))/(t−1+(√2)))]]_0 ^(+∞) =((−1)/(√2))( −ln(((1+(√2))/((√2) −1)))) I = (1/(√2))ln(((1+(√2))/(−1+(√2))))$
$e r r o r i n t h e f i n a l l i n e s$ $I = - 2 \int_{0}^{\infty} \frac{d t}{{(t - 1)}^{2} - 2}$ $= - 2 \int_{0}^{\infty} \frac{d t}{(t - 1 -_{} \sqrt{2}) (t - 1 + \sqrt{2})}$ $= \frac{- 2}{2 \sqrt{2}} \int_{0}^{\infty} {\frac{1}{t - 1 - \sqrt{2}} - \frac{1}{t - 1 + \sqrt{2}}} d t$ ${= - \frac{1}{\sqrt{2}} [l n ∣ \frac{t - 1 - \sqrt{2}}{t - 1 + \sqrt{2}}]]}_{0}^{+ \infty}$ $= \frac{- 1}{\sqrt{2}} (- l n (\frac{1 + \sqrt{2}}{\sqrt{2} - 1}))$ $I = \frac{1}{\sqrt{2}} l n (\frac{1 + \sqrt{2}}{- 1 + \sqrt{2}})$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18

	$I = \int_{0}^{Π} \frac{d x}{c o s (Π - x) + s i n (Π - x)}$ $2 I = \int_{0}^{Π} \frac{1}{c o s x + s i n x} + \frac{1}{- c o s x + s i n x} d x$ $= \int_{0}^{Π} \frac{2 s i n x}{{s i n}^{2} x - {c o s}^{2} x} d x$ $I = \frac{1}{2} \int_{0}^{Π} \frac{2 s i n x}{1 - 2 {c o s}^{2} x} d x$ $= \int_{0}^{Π} \frac{(- s i n x)}{2 {c o s}^{2} x - 1} d x$ $= \frac{1}{2} \int_{0}^{Π} \frac{d (c o s x)}{{c o s}^{2} x - {(\frac{1}{\sqrt{2}})}^{2}}$ $= \frac{1}{2} \times \frac{\sqrt{2}}{2} ∣ l n \frac{c o s x - \frac{1}{\sqrt{2}}}{c o s x + \frac{1}{\sqrt{2}}} ∣_{0}^{Π}$
	Commented by NECx last updated on 15/May/18

	$p l e a s e h o w d i d y o u c h a n g e$ $\int_{0}^{π} \frac{1}{c o s x + s i n x} d x t o$ $\int_{0}^{π} \frac{1}{c o s (π - x) + s i n (π - x)} d x ? ? ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 15/May/18

	$\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
	Commented by NECx last updated on 15/May/18

	$o h . . . T h a n k s$
	Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18

	$\int \frac{d x}{c o s x + s i n x}$ $t = t a n \frac{x}{2} d t = \frac{1}{2} {s e c}^{2} \frac{x}{2}$ $\int \frac{d x}{\frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}} + \frac{2 t a n \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}$ $\int \frac{{s e c}^{2} \frac{x}{2}}{1 - {t a n}^{2} \frac{x}{2} + 2 t a n \frac{x}{2}}$ $\int \frac{2 d t}{1 - t^{2} + 2 t}$ $- 2 \int \frac{d t}{t^{2} - 2 t + 1 - 2}$ $- 2 \int \frac{d t}{{(t - 1)}^{2} - {(\sqrt{2})}^{2}}$ $n o w u s e f o r m u l a a n d p u t l i m i t$ $o r m e t h o d$ $\int_{0}^{Π} \frac{d x}{c o s x + s i n x}$ $= \int_{0}^{Π} \frac{\frac{1}{\sqrt{2}} d x}{\frac{1}{\sqrt{2}} c o s x + \frac{1}{\sqrt{2}} s i n x}$ $= \frac{1}{\sqrt{2}} \int_{0}^{Π} \frac{d x}{s i n (Π / 4 + x)}$ $n o w u s e f o r m u l a a n d p u t l i m i t$
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