integral of 4+8cos(θ)+4cos^2(θ)

Lösung

\int 4 + 8 cos (θ) + 4 cos^{2} (θ) d θ

6 θ + 8 sin (θ) + sin (2 θ) + C

Schritte zur Lösung

\int 4 + 8 cos (θ) + 4 cos^{2} (θ) d θ

Wende die Summenregel an:

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

= \int 4 d θ + \int 8 cos (θ) d θ + \int 4 cos^{2} (θ) d θ

\int 4 d θ = 4 θ

\int 8 cos (θ) d θ = 8 sin (θ)

\int 4 cos^{2} (θ) d θ = 2 θ + sin (2 θ)

= 4 θ + 8 sin (θ) + 2 θ + sin (2 θ)

Vereinfache

= 6 θ + 8 sin (θ) + sin (2 θ)

Füge eine Konstante zur Lösung hinzu

= 6 θ + 8 sin (θ) + sin (2 θ) + C

x^{2} + y^{^{'}} = x^{2}

\frac{\partial}{\partial x} (6 x e^{y})

\int \frac{- 4 ( ln ( x ) ) ^{4}}{x} d x

d er i v a t i v e f (x) = \frac{x ^{2} + 9}{x}

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