0-pi-4-sin-d-cos-sin-d-where-zeta-

Question Number 124963 by 676597498 last updated on 07/Dec/20

\int_{0}^{\frac{π}{4}} \frac{\sin (ς) d ς}{\cos (ς) + \sin (ς)} d ς

where ς : zeta

Commented by MJS_new last updated on 07/Dec/20

syntax error .

Answered by Dwaipayan Shikari last updated on 07/Dec/20

\int_{0}^{\frac{π}{4}} \frac{1}{t a n (ζ) + 1} d ζ = \int_{0}^{1} \frac{1}{t + 1} . \frac{1}{t^{2} + 1} d t t a n ζ = t \Rightarrow {s e c}^{2} ζ = \frac{d t}{d ζ}

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

= \frac{1}{2} l o g (2) - \frac{1}{4} \int_{0}^{1} \frac{2 t}{t^{2} + 1} + \frac{1}{2} \int_{0}^{1} \frac{1}{1 + t^{2}} d t

= - \frac{1}{4} l o g (2) + \frac{π}{8}

Commented by 676597498 last updated on 07/Dec/20

thanks

but there are 2 (d ς)

look again

it is \int \frac{\sin (ς) d ς}{\cos (ς) + \sin (ς)} d ς

Commented by Dwaipayan Shikari last updated on 07/Dec/20

S o r r y b u t i t {d o e s n}^{'} t m a k e a n y s e n s e t o m e

Commented by 676597498 last updated on 07/Dec/20

it makes sense

Commented by mr W last updated on 07/Dec/20

it makes \boldsymbol N o sense!

Commented by MJS_new last updated on 07/Dec/20

if it makes sense, what sense does it make ?

Answered by mathmax by abdo last updated on 07/Dec/20

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

A = \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 t}{1 + t^{2}}}{\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}}} \frac{2 dt}{1 + t^{2}} = \int_{0}^{\sqrt{2} - 1} \frac{4 t}{(t^{2} + 1) (- 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 \to Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} and t_{2} = 1 - \sqrt{2} \Rightarrow

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

a = \frac{t_{1}}{2 \sqrt{2} (t_{1}^{2} + 1)} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (3 + 2 \sqrt{2} + 1)} = \frac{1 + \sqrt{2}}{2 \sqrt{2} (4 + 2 \sqrt{2})} = \frac{1 + \sqrt{2}}{8 \sqrt{2} + 8} = \frac{1}{8}

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

= \frac{1 - \sqrt{2}}{- 8 \sqrt{2} + 8} = \frac{1}{8}

\lim_{t \to + \infty} tF (t) = 0 = a + b + m \Rightarrow m = - \frac{1}{4}

F (o) = 0 = - \frac{a}{t_{1}} - \frac{b}{t_{2}} + n \Rightarrow n = \frac{a}{t_{1}} + \frac{b}{t_{2}} = \frac{1}{8 (1 + \sqrt{2})} + \frac{1}{8 (1 - \sqrt{2})}

= \frac{1}{8} (\frac{2}{(- 1)}) = - \frac{1}{4} \Rightarrow

F (t) = \frac{1}{8 (t - (1 + \sqrt{2})} + \frac{1}{8 (t - (1 - \sqrt{2}))} - \frac{1}{4} \times \frac{t + 1}{t^{2} + 1} \Rightarrow

\int_{0}^{\sqrt{2} - 1} F (t) dt = \frac{1}{8} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t - 1 - \sqrt{2}} + \frac{1}{8} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t - 1 + \sqrt{2}}

- \frac{1}{8} \int_{0}^{\sqrt{2} - 1} \frac{2 t}{t^{2} + 1} - \frac{1}{4} \int_{0}^{\sqrt{2} - 1} \frac{dt}{t^{2} + 1}

= \frac{1}{8} {[\ln ∣ {(t - 1)}^{2} - 2 ∣]}_{0}^{\sqrt{2} - 1} - \frac{1}{8} {[\ln (t^{2} + 1)]}_{0}^{\sqrt{2} - 1} - \frac{1}{4} {[arctant]}_{0}^{\sqrt{2} - 1}

= \frac{1}{8} {\ln ∣ {(\sqrt{2} - 2)}^{2} - 2 ∣ - \frac{1}{8} \ln ({(\sqrt{2} - 1)}^{2} + 1) - \frac{1}{4} \times \frac{π}{8}

= \frac{1}{8} \ln ∣ 2 - 4 \sqrt{2} + 4 - 2 ∣ - \frac{1}{8} \ln (3 - 2 \sqrt{2} + 1) - \frac{π}{32}

= \frac{1}{8} \ln (4 \sqrt{2} - 4) - \frac{1}{8} \ln (4 - 2 \sqrt{2}) - \frac{π}{32} \Rightarrow

I = - \frac{1}{2} \ln (4 \sqrt{2} - 4) + \frac{1}{2} \ln (4 - 2 \sqrt{2}) + \frac{π}{8}

Commented by talminator2856791 last updated on 07/Dec/20

abdo to the rescue!!!!!

abdo to the rescue!!!!!