integral of ((1))/((sin(x)-2csc(x)))

Lösung

\int \frac{( 1 )}{( sin ( x ) - 2 csc ( x ) )} d x

arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1) + C

Schritte zur Lösung

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

Wende U-Substitution an

= \int \frac{2 u}{- u ^{4} - 1} d u

Entferne die Konstante:

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

= 2 \cdot \int \frac{u}{- u ^{4} - 1} d u

Ermittle den Partialbruch von

\frac{u}{- u ^{4} - 1} : \frac{1}{2 2 ( u ^{2} + 2 u + 1 )} - \frac{1}{2 2 ( u ^{2} - 2 u + 1 )}

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

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

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

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

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

Setze in

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

ein

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

Vereinfache

2 (\frac{1}{2} arctan (2 tan (\frac{x}{2}) + 1) - \frac{1}{2} arctan (2 tan (\frac{x}{2}) - 1)) : arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1)

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

Füge eine Konstante zur Lösung hinzu

= arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1) + C

\int - 4 e^{5 x} d x

\int sin (2 x) sin (8 x) d x

y^{^{''}} + 5 y = 0, y^{^{'}} (1) = 3

x \to 0 + lim (tan (x))

\int \frac{1}{cos ( 2 y ) - cos ^{2} ( y )} d y