show-that-0-pi-2-cos-x-cos-x-sin-x-1-dx-1-4-pi-2ln-2-

Question Number 151220 by peter frank last updated on 19/Aug/21

show that

\int_{0}^{\frac{π}{2}} \frac{\cos x}{\cos x + \sin x + 1} dx = \frac{1}{4} (π - 2 \ln 2)

Answered by ArielVyny last updated on 19/Aug/21

t = t a n (\frac{x}{2}) \to d t = \frac{1}{2} (1 + t^{2}) d x

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

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

= \int_{0}^{1} \frac{1}{1 + t^{2}} d t - \int_{0}^{1} \frac{t}{1 + t^{2}} d t = \frac{π}{4} - Σ {(- 1)}^{n} \int_{0}^{1} t^{2 n + 1}

\frac{π}{4} - Σ {(- 1)}^{n} \frac{1}{2 n + 2} = \frac{π}{4} - \frac{1}{2} l n (2)

\int_{0}^{\frac{π}{2}} \frac{c o s x}{c o s x + s i n x + 1} = \frac{π}{4} - \frac{1}{2} l n (2 +

Commented by peter frank last updated on 19/Aug/21

thank you

Answered by Lordose last updated on 19/Aug/21

Ω = \int_{0}^{\frac{π}{2}} \frac{\cos (x)}{\cos (x) + \sin (x) + 1} dx \overset{t = \tan (\frac{x}{2})}{=} \int_{0}^{1} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{(1 + t^{2}) (\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}} + 1)} dt

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

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

Ω = \frac{π}{4} - \frac{1}{2} \ln (2)

Commented by peter frank last updated on 19/Aug/21

thank you

Answered by Lordose last updated on 19/Aug/21

Ω = \int_{0}^{\frac{π}{2}} \frac{\cos (x)}{\cos (x) + \sin (x) + 1} dx \overset{t = \tan (\frac{x}{2})}{=} \int_{0}^{1} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{(1 + t^{2}) (\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}} + 1)} dt

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

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

Ω = \frac{π}{4} - \frac{1}{2} \ln (2)

Facebook Tweet Pin