integral de 0 a 1/2 de xcos((pikx)/2)

Solución

\int_{0}^{\frac{1}{2}} x cos (\frac{πk x}{2}) d x

\frac{πk sin ( \frac{πk}{4} ) + 4 cos ( \frac{πk}{4} ) - 4}{π ^{2} k ^{2}}

Pasos de solución

\int_{0}^{\frac{1}{2}} x cos (\frac{πk x}{2}) d x

Aplicar integración por sustitución

= \int_{0}^{\frac{πk}{2}} \frac{u cos ( \frac{u}{2} )}{π ^{2} k ^{2}} d u

Sacar la constante:

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

= \frac{1}{π ^{2} k ^{2}} \cdot \int_{0}^{\frac{πk}{2}} u cos (\frac{u}{2}) d u

Aplicar integración por sustitución

= \frac{1}{π ^{2} k ^{2}} \cdot \int_{0}^{\frac{πk}{4}} 4 v cos (v) d v

Sacar la constante:

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

= \frac{1}{π ^{2} k ^{2}} \cdot 4 \cdot \int_{0}^{\frac{πk}{4}} v cos (v) d v

Aplicar integración por partes

= \frac{1}{π ^{2} k ^{2}} \cdot 4 [v sin (v) - \int sin (v) d v]_{0}^{\frac{πk}{4}}

\int sin (v) d v = - cos (v)

= \frac{1}{π ^{2} k ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{0}^{\frac{πk}{4}}

Simplificar

\frac{1}{π ^{2} k ^{2}} \cdot 4 [v sin (v) - (- cos (v))]_{0}^{\frac{πk}{4}} : \frac{4}{π ^{2} k ^{2}} [v sin (v) + cos (v)]_{0}^{\frac{πk}{4}}

= \frac{4}{π ^{2} k ^{2}} [v sin (v) + cos (v)]_{0}^{\frac{πk}{4}}

Calcular los limites:

\frac{πk}{4} sin (\frac{πk}{4}) + cos (\frac{πk}{4}) - 1

= \frac{4}{π ^{2} k ^{2}} (\frac{πk}{4} sin (\frac{πk}{4}) + cos (\frac{πk}{4}) - 1)

Simplificar

= \frac{πk sin ( \frac{πk}{4} ) + 4 cos ( \frac{πk}{4} ) - 4}{π ^{2} k ^{2}}