integral of 1/(sqrt(sin(x)(cos(x))^2))

Lösung

\int \frac{1}{sin ( x ) ( cos ( x ) ) ^{2}} d x

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

Schritte zur Lösung

\int \frac{1}{sin ( x ) ( cos ( x ) ) ^{2}} d x

sin (x) cos^{2} (x) = cos (x) sin (x)

= \int \frac{1}{cos ( x ) sin ( x )} d x

Wende U-Substitution an

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Ermittle den Partialbruch von

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

= 2 \cdot \int \frac{1}{2 ( v ^{2} + 1 )} + \frac{1}{4 ( v + 1 )} - \frac{1}{4 ( v - 1 )} d v

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 ( v ^{2} + 1 )} d v + \int \frac{1}{4 ( v + 1 )} d v - \int \frac{1}{4 ( v - 1 )} d v)

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

\int \frac{1}{4 ( v + 1 )} d v = \frac{1}{4} ln ∣ v + 1 ∣

\int \frac{1}{4 ( v - 1 )} d v = \frac{1}{4} ln ∣ v - 1 ∣

= 2 (\frac{1}{2} arctan (v) + \frac{1}{4} ln ∣ v + 1 ∣ - \frac{1}{4} ln ∣ v - 1 ∣)

Ersetze zurück

= 2 (\frac{1}{2} arctan (sin (x)) + \frac{1}{4} ln sin (x) + 1 - \frac{1}{4} ln sin (x) - 1)

Vereinfache

2 (\frac{1}{2} arctan (sin (x)) + \frac{1}{4} ln sin (x) + 1 - \frac{1}{4} ln sin (x) - 1) : arctan (sin (x)) + \frac{1}{2} ln sin (x) + 1 - \frac{1}{2} ln sin (x) - 1

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

Füge eine Konstante zur Lösung hinzu

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

\int ln (x^{0}) d x

\int \frac{3 x ^{2} + 3 x}{2 x ^{3} + 3 x ^{2} - 5} d x

\int \frac{5 - 3 y ^{2}}{y} d y

\int cos^{3} (x) csc^{3} (x) d x

\int x 25 x - 8 d x