integral from 0 to 2pi of (4-cos(3x))^2

Lösung

\int_{0}^{2 π} (4 - cos (3 x))^{2} d x

33 π

Dezimale

103.67255 \dots

Schritte zur Lösung

\int_{0}^{2 π} (4 - cos (3 x))^{2} d x

Faktorisiere

4 - cos (3 x) : - (- 4 + cos (3 x))

= \int_{0}^{2 π} (- (- 4 + cos (3 x)))^{2} d x

Vereinfache

(- (- 4 + cos (3 x)))^{2} : (- 4 + cos (3 x))^{2}

= \int_{0}^{2 π} (- 4 + cos (3 x))^{2} d x

Multipliziere aus

(- 4 + cos (3 x))^{2} : 16 - 8 cos (3 x) + cos^{2} (3 x)

= \int_{0}^{2 π} 16 - 8 cos (3 x) + cos^{2} (3 x) d x

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2 π} 16 d x - \int_{0}^{2 π} 8 cos (3 x) d x + \int_{0}^{2 π} cos^{2} (3 x) d x

\int_{0}^{2 π} 16 d x = 32 π

\int_{0}^{2 π} 8 cos (3 x) d x = 0

\int_{0}^{2 π} cos^{2} (3 x) d x = π

= 32 π - 0 + π

Vereinfache

= 33 π

\int_{5}^{8} f (4 x^{3} + 1) x^{2} d x

\int_{- 1}^{1} 3 x^{4} d x

\int_{0}^{\frac{π}{2}} 3 sin (x) cos (x) d x

\int_{1}^{4} 2 x^{2} + 3 x - 1 d x

\int_{0}^{8} 2 b x^{- \frac{1}{3}} d x