0-pi-4-sin-x-sin-x-cos-x-dx-

Question Number 127528 by bramlexs22 last updated on 30/Dec/20

\int_{0}^{π / 4} \frac{\sin x}{\sin x + \cos x} dx = ?

Answered by liberty last updated on 30/Dec/20

L e t L = \int_{0}^{π / 4} \frac{\sin x}{\sin x + \cos x} d x

a n d l e t Y = \int_{0}^{π / 4} \frac{\cos x}{\sin x + \cos x} d x

(1) L + Y = \int_{0}^{π / 4} \frac{\sin x + \cos x}{\sin x + \cos x} d x = \frac{π}{4}

(2) L - Y = \int_{0}^{π / 4} \frac{\sin x - \cos x}{\sin x + \cos x} d x

= - \int_{0}^{π / 4} \frac{d (\sin x + \cos x)}{\sin x + \cos x}

= - {[\ln (\sin x + \cos x)]}_{0}^{π / 4}

= - \frac{1}{2} \ln (2)

(3) L = \frac{(L + Y) + (L - Y)}{2} = \frac{π}{8} - \frac{1}{4} \ln (2)

Commented by bramlexs22 last updated on 30/Dec/20

amazing \dots

Answered by Dwaipayan Shikari last updated on 30/Dec/20

\int_{0}^{1} \frac{t}{t + 1} . \frac{1}{t^{2} + 1} d t t a n x = t

= \frac{1}{2} \int_{0}^{1} t (\frac{1}{t + 1} - \frac{t - 1}{t^{2} + 1}) d t

= \frac{1}{2} \int_{0}^{1} - \frac{1}{t + 1} + \frac{t}{t^{2} + 1} + \frac{1}{1 + t^{2}} d t = {[\frac{1}{2} l o g (\frac{\sqrt{t^{2} + 1}}{t + 1})]}_{0}^{1} + {[\frac{1}{2} {t a n}^{- 1} t]}_{0}^{1}

= \frac{π}{8} - \frac{1}{4} l o g (2)

Answered by mathmax by abdo last updated on 31/Dec/20

I = \int_{0}^{\frac{π}{4}} \frac{sinx}{sinx + cosx} dx we do the changement \tan (\frac{x}{2}) = t \Rightarrow

I = \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 t}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}} \times \frac{2 dt}{1 + t^{2}} = \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(1 + t^{2}) (- t^{2} + 2 t + 1)} dt

= - 4 \int_{0}^{\sqrt{2} - 1} \frac{tdt}{(t^{2} + 1) (t^{2} - 2 t - 1)} let decompose F (t) = \frac{t}{(t^{2} + 1) (t^{2} - 2 t - 1)}

t^{2} - 2 t - 1 = 0 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} and t_{2} = 1 - \sqrt{2}

F (t) = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{ct + d}{t^{2} + 1}

a = \frac{t_{1}}{(t_{1}^{2} + 1) (t_{1} - t_{2})} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (1 + 3 + 2 \sqrt{2})} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (4 + 2 \sqrt{2})}

b = \frac{t_{2}}{(t_{2}^{2} + 1) (t_{2} - t_{1})} = \frac{1 - \sqrt{2}}{(- 2 \sqrt{2}) (3 - 2 \sqrt{2}) + 1)} = \frac{\sqrt{2} - 1}{2 \sqrt{2} (4 - 2 \sqrt{2})}

\lim_{t \to + \infty} tF (t) = 0 = a + b + c \Rightarrow c = - a - b

F (0) = 0 = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + d \Rightarrow d = \frac{a}{t_{1}} + \frac{b}{t_{2}} \Rightarrow

\int_{0}^{\sqrt{2} - 1} F (t) dt = a \int_{0}^{\sqrt{2} - 1} \frac{dt}{t - t_{1}} + b \int_{0}^{\sqrt{2} - 1} \frac{dt}{t - t_{2}} + \frac{c}{2} \int_{0}^{\sqrt{2} - 1} \frac{2 t}{t^{2} + 1} dt

+ d \int_{0}^{\sqrt{2} - 1} \frac{dt}{t^{2} + 1} = a {[\ln ∣ t - t_{1} ∣]}_{0}^{\sqrt{2} - 1} + b {[\ln ∣ t - t_{2} ∣]}_{0}^{\sqrt{2} - 1} + \frac{c}{2} {[\ln (t^{2} + 1)]}_{0}^{\sqrt{2} - 1}

+ d {[arctant]}_{0}^{\sqrt{2} - 1} rest to finish calculus \dots