∫_0 ^(π/2) ((xdx)/(sin x+cos x)) = ?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 80921 by jagoll last updated on 08/Feb/20

$\underset{0}{\int^{\frac{π}{2}}} \frac{x d x}{\sin x + \cos x} = ?$
Commented by john santu last updated on 08/Feb/20

$c o r r e c t i o n$ $\underset{0}{\int^{\frac{π}{2}}} \frac{x d x}{{(\sin x + \cos x)}^{n}} = \frac{π}{4}$
Commented by mathmax by abdo last updated on 10/Feb/20
$A=∫_0 ^(π/2) ((xdx)/(sinx +cosx)) changement x=(π/2)−t give A =−∫_0 ^(π/2) ((((π/2)−t)(−dt))/(cost +sint)) =(π/2)∫_0 ^(π/2) (dt/(cost +sint)) −A ⇒ 2A=(π/2) ∫_0 ^(π/2) (dt/(cost +sint)) ⇒A=(π/4) ∫_0 ^(π/2) (dt/(cost +sint)) changement tan((t/2))=u give ∫_0 ^(π/2) (dt/(cost +sint)) =∫_0 ^1 (1/(((1−u^2 )/(1+u^2 ))+((2u)/(1+u^2 ))))×((2du)/(1+u^2 )) =∫_0 ^1 ((2du)/(1−u^2 +2u)) =−2 ∫_0 ^1 (du/(u^2 −2u−1)) =−2 ∫_0 ^1 (du/((u−1)^2 −2)) =−2 ∫_0 ^1 (du/((u−1−(√2))(u−1+(√2)))) =−(2/(2(√2))) ∫_0 ^1 {(1/(u−1−(√2)))−(1/(u−1+(√2)))}du =−(1/(√2))[ln∣((u−1−(√2))/(u−1+(√2)))∣]_0 ^1 =−(1/(√2))(−ln(((1+(√2))/((√2)−1))) =(1/(√2))ln(((√2)+1)^2 ) =(√2)ln(1+(√2)) ⇒A =((π(√2))/4)ln(1+(√2))$
$A = \int_{0}^{\frac{π}{2}} \frac{x d x}{s i n x + c o s x} c h a n g e m e n t x = \frac{π}{2} - t g i v e$ $A = - \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - t) (- d t)}{c o s t + s i n t} = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{d t}{c o s t + s i n t} - A \Rightarrow$ $2 A = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{d t}{c o s t + s i n t} \Rightarrow A = \frac{π}{4} \int_{0}^{\frac{π}{2}} \frac{d t}{c o s t + s i n t}$ $c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $\int_{0}^{\frac{π}{2}} \frac{d t}{c o s t + s i n t} = \int_{0}^{1} \frac{1}{\frac{1 - u^{2}}{1 + u^{2}} + \frac{2 u}{1 + u^{2}}} \times \frac{2 d u}{1 + u^{2}} = \int_{0}^{1} \frac{2 d u}{1 - u^{2} + 2 u}$ $= - 2 \int_{0}^{1} \frac{d u}{u^{2} - 2 u - 1} = - 2 \int_{0}^{1} \frac{d u}{{(u - 1)}^{2} - 2} = - 2 \int_{0}^{1} \frac{d u}{(u - 1 - \sqrt{2}) (u - 1 + \sqrt{2})}$ $= - \frac{2}{2 \sqrt{2}} \int_{0}^{1} {\frac{1}{u - 1 - \sqrt{2}} - \frac{1}{u - 1 + \sqrt{2}}} d u$ $= - \frac{1}{\sqrt{2}} {[l n ∣ \frac{u - 1 - \sqrt{2}}{u - 1 + \sqrt{2}} ∣]}_{0}^{1} = - \frac{1}{\sqrt{2}} (- l n (\frac{1 + \sqrt{2}}{\sqrt{2} - 1}) = \frac{1}{\sqrt{2}} l n ({(\sqrt{2} + 1)}^{2})$ $= \sqrt{2} l n (1 + \sqrt{2}) \Rightarrow A = \frac{π \sqrt{2}}{4} l n (1 + \sqrt{2})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum