integral de pi/2 a pi de csc(x)

Solución

\int_{\frac{π}{2}}^{π} csc (x) d x

Es divergente

Pasos de solución

\int_{\frac{π}{2}}^{π} csc (x) d x

Re-escribir usando identidades trigonométricas

= \int_{\frac{π}{2}}^{π} \frac{1}{sin ( x )} d x

Aplicar integración por sustitución

: \int_{1}^{0} \frac{1}{u 1 - u ^{2}} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Aplicar integración por sustitución

: \int_{1}^{0} - \frac{1}{- v ^{2} + 1} d v

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Sacar la constante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Aplicar la regla de integración:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - (- (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1}))

Simplificar

- (- (- [\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}]_{0}^{1})) : - [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{0}^{1}

= - [\frac{1}{2} (ln ∣ v + 1 ∣ - ln ∣ v - 1 ∣)]_{0}^{1}

Calcular los limites:

Es divergente

= Es divergente

\int_{0}^{1} (x^{2} + 3) e^{5 x} d x

\int_{0}^{\frac{π}{2}} sin (x) - cos (x) d x

\int_{- 2}^{0} x^{3} d x

\int_{e}^{e^{4}} \frac{1}{x ln ( x )} d x

\int_{1}^{2} \frac{2 ln ( x )}{x} d x