Evaluar

Question

$$\int_{-\pi/2}^{\pi/2} \frac{\cos(x)}{1+e^x} dx$$

Usando integración por partes

$$\int u v = u \int v - \int u' \int v$$

$u(x) = \cos(x)$ y $v(x) = \frac{1}{1+e^x}$,

$$ u \int{-\pi/2}^{\pi/2} v dx = \left[ \cos(x) \int{-\pi/2}^{\pi/2} \frac{1}{1+e^x} dx \right]{-\pi/2}^{\pi/2} = \left[ \cos(x) ( x - log(1+e^x) \right]{-\pi/2}^{\pi/2} $$

$$ \int{-\pi/2}^{\pi/2} u' \left( \int{-\pi/2}^{\pi/2} v dx \right) dx = \int{-\pi/2}^{\pi/2} -\sin(x) \int{-\pi/2}^{\pi/2} \frac{1}{1+e^x} dx $$

$$ = \int{-\pi/2}^{\pi/2} -\sin(x) \left[ ( x - log(1+e^x) \right]{-\pi/2}^{\pi/2} $$

¿Corta de seguir ad nauseam, hay una mejor manera para determinar la respuesta (dada como $1$)?

Answer 1

$\displaystyle\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

Si $\displaystyle f(x)=\frac{\cos x}{1+e^x}, f\left[\frac\pi2+\left(-\frac\pi2\right)-x\right]=\frac{\cos(-x)}{1+e^{-x}}=\frac{e^x\cos x}{1+e^x}$

$$I=\int{-\frac\pi2}^{\frac\pi2}\frac{\cos x}{1+e^x}dx=\int{-\frac\pi2}^{\frac\pi2}\frac{e^x\cos x}{1+e^x}dx$$

$$\implies I+I=\int_{-\frac\pi2}^{\frac\pi2}\cos x\ dx$$

Answer 2

$$I=\int_{-\frac\pi2}^{\frac\pi2}\frac{\cos x}{1+e^x}dx$$

$$=\int_{-\frac\pi2}^0\frac{\cos x}{1+e^x}+\int_0^{\frac\pi2}\frac{\cos x}{1+e^x}dx$$

Ajuste $-x=y$ $$I1=\int{-\frac\pi2}^0\frac{\cos x}{1+e^x}=\int_\frac\pi2^0\frac{\cos(-y)}{1+e^{-y}}(-dy)$$I_1=\int_0^{\frac\pi2}\frac{e^y\cos y}{1+e^y}dy\text{ as }\int_c^df(y)\ dy=-\int_d^cf(y)\ dy

$$\implies I_1=\int_0^{\frac\pi2}\frac{e^x\cos x}{1+e^x}dx$$

$$\implies I=\int_0^{\frac\pi2}\frac{e^x\cos x}{1+e^x}dx+\int_0^{\frac\pi2}\frac{\cos x}{1+e^x}dx=\cdots$$