integral of (cos(2x)-1)/(cos(2x)+1)

Lösung

\int \frac{cos ( 2 x ) - 1}{cos ( 2 x ) + 1} d x

- tan (x) + x + C

Schritte zur Lösung

\int \frac{cos ( 2 x ) - 1}{cos ( 2 x ) + 1} d x

\frac{cos ( 2 x ) - 1}{cos ( 2 x ) + 1} = \frac{cos ( 2 x ) - 1 - 2 cos ^{2} ( x )}{2 cos ^{2} ( x )} + 1

= \int \frac{cos ( 2 x ) - 1 - 2 cos ^{2} ( x )}{2 cos ^{2} ( x )} + 1 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 \frac{cos ( 2 x ) - 1 - 2 cos ^{2} ( x )}{2 cos ^{2} ( x )} d x + \int 1 d x

\int \frac{cos ( 2 x ) - 1 - 2 cos ^{2} ( x )}{2 cos ^{2} ( x )} d x = - tan (x)

\int 1 d x = x

= - tan (x) + x

Füge eine Konstante zur Lösung hinzu

= - tan (x) + x + C

\frac{d}{d x} ((x^{5} - 9 x)^{2})

d er i v a t i v e f (x) = 4 x^{3} - 5 x + 1

\frac{\partial}{\partial z} (2 x y + z^{2})

\frac{\partial}{\partial y} (y x^{2})

\int x (x - 2021) d x