integral of (cos(x))/(5+4cos(x))

Lösung

\int \frac{cos ( x )}{5 + 4 cos ( x )} d x

2 (\frac{1}{4} arctan (tan (\frac{x}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{x}{2}))) + C

Schritte zur Lösung

\int \frac{cos ( x )}{5 + 4 cos ( x )} d x

Wende U-Substitution an

= \int \frac{2 ( 1 - u ^{2} )}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{1 - u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u

Multipliziere aus

\frac{1 - u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} : \frac{1}{( u ^{2} + 9 ) ( 1 + u ^{2} )} - \frac{u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )}

Wende die Summenregel an:

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

= 2 (\int \frac{1}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u - \int \frac{u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u)

\int \frac{1}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u = - \frac{1}{24} arctan (\frac{u}{3}) + \frac{1}{8} arctan (u)

\int \frac{u ^{2}}{( u ^{2} + 9 ) ( 1 + u ^{2} )} d u = \frac{3}{8} arctan (\frac{u}{3}) - \frac{1}{8} arctan (u)

= 2 (- \frac{1}{24} arctan (\frac{u}{3}) + \frac{1}{8} arctan (u) - (\frac{3}{8} arctan (\frac{u}{3}) - \frac{1}{8} arctan (u)))

Setze in

u = tan (\frac{x}{2})

ein

= 2 (- \frac{1}{24} arctan (\frac{tan ( \frac{x}{2} )}{3}) + \frac{1}{8} arctan (tan (\frac{x}{2})) - (\frac{3}{8} arctan (\frac{tan ( \frac{x}{2} )}{3}) - \frac{1}{8} arctan (tan (\frac{x}{2}))))

Vereinfache

2 (- \frac{1}{24} arctan (\frac{tan ( \frac{x}{2} )}{3}) + \frac{1}{8} arctan (tan (\frac{x}{2})) - (\frac{3}{8} arctan (\frac{tan ( \frac{x}{2} )}{3}) - \frac{1}{8} arctan (tan (\frac{x}{2})))) : 2 (\frac{1}{4} arctan (tan (\frac{x}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{x}{2})))

= 2 (\frac{1}{4} arctan (tan (\frac{x}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{x}{2})))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{4} arctan (tan (\frac{x}{2})) - \frac{5}{12} arctan (\frac{1}{3} tan (\frac{x}{2}))) + C

\frac{d}{d x} (3 \frac{x ^{2} + 1}{x ^{2} - 1})

\int_{0}^{\frac{π}{2}} 6 sec^{4} (\frac{t}{2}) d t

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

\int \frac{ln ( x )}{x ( ln ^{2} ( x ) + 4 )} d x

\frac{d y}{d x} = \frac{x - 3}{y - 3}