integral from 0 to 1 of xcos((npi)/2 x)

Lösung

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

\frac{2 ( πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 )}{π ^{2} n ^{2}}

Schritte zur Lösung

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

\frac{nπ}{2} = nπ \frac{1}{2}

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

Wende U-Substitution an

= \int_{0}^{\frac{πn}{2}} \frac{4 u cos ( u )}{π ^{2} n ^{2}} d u

Entferne die Konstante:

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

= \frac{4}{π ^{2} n ^{2}} \cdot \int_{0}^{\frac{πn}{2}} u cos (u) d u

Wende die partielle Integration an

= \frac{4}{π ^{2} n ^{2}} [u sin (u) - \int sin (u) d u]_{0}^{\frac{πn}{2}}

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

= \frac{4}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{0}^{\frac{πn}{2}}

Vereinfache

= \frac{4}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{0}^{\frac{πn}{2}}

Berechne die Grenzen:

\frac{πn}{2} sin (\frac{πn}{2}) + cos (\frac{πn}{2}) - 1

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

Vereinfache

= \frac{2 ( πn sin ( \frac{πn}{2} ) + 2 cos ( \frac{πn}{2} ) - 2 )}{π ^{2} n ^{2}}

\int_{2}^{x} - \frac{2}{t ^{3}} d t

\int_{4}^{6} x^{2} + 2 x + 2 d x

\int_{0}^{1} x e^{- \frac{x}{2}} d x

\int_{0}^{\frac{π}{2}} cos (4 θ) d θ

\int_{0}^{25} π (x)^{2} d x