integral from pi to x of t^3sin^2(t)

Lösung

\int_{π}^{x} t^{3} sin^{2} (t) d t

\frac{x ^{4} sin ^{2} ( x ) - x ^{3} sin ( 2 x )}{4} + \frac{x ^{4} cos ( 2 x ) - 3 x ^{2} cos ( 2 x ) + 3 x sin ( 2 x )}{8} + \frac{3}{16} cos (2 x) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

Schritte zur Lösung

\int_{π}^{x} t^{3} sin^{2} (t) d t

Wende die partielle Integration an

= [\frac{1}{4} t^{4} sin^{2} (t) - \int \frac{1}{4} t^{4} sin (2 t) d t]_{π}^{x}

\int \frac{1}{4} t^{4} sin (2 t) d t = \frac{1}{16} (- 2 t^{4} cos (2 t) + 4 t^{3} sin (2 t) - 3 (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t)))

= [\frac{1}{4} t^{4} sin^{2} (t) - \frac{1}{16} (- 2 t^{4} cos (2 t) + 4 t^{3} sin (2 t) - 3 (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t)))]_{π}^{x}

Berechne die Grenzen:

\frac{1}{4} x^{4} sin^{2} (x) - \frac{1}{16} (- 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

= \frac{1}{4} x^{4} sin^{2} (x) - \frac{1}{16} (- 2 x^{4} cos (2 x) + 4 x^{3} sin (2 x) - 3 (- 2 x^{2} cos (2 x) + 2 x sin (2 x) + cos (2 x))) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

Vereinfache

= \frac{x ^{4} sin ^{2} ( x ) - x ^{3} sin ( 2 x )}{4} + \frac{x ^{4} cos ( 2 x ) - 3 x ^{2} cos ( 2 x ) + 3 x sin ( 2 x )}{8} + \frac{3}{16} cos (2 x) + \frac{- 2 π ^{4} - 3 ( - 2 π ^{2} + 1 )}{16}

\int_{0}^{2} \frac{y ^{3}}{1 + y ^{4}} d y

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

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

\int_{0}^{\frac{π}{4}} tan (x) + sin (x) d x

\int_{3}^{6} \frac{1}{8 x + 1} d x