integral of 1/(sin(x)cos^3(x))

Lösung

\int \frac{1}{sin ( x ) cos ^{3} ( x )} d x

\frac{tan ^{2} ( x )}{2} + ln ∣ tan (x) ∣ + C

Schritte zur Lösung

\int \frac{1}{sin ( x ) cos ^{3} ( x )} d x

Trigonometrische Substitution anwenden

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

Vereinfache

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

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

Multipliziere aus

\frac{u ^{2} + 1}{u} : u + \frac{1}{u}

= \int u + \frac{1}{u} 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

= \int u d u + \int \frac{1}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + ln ∣ u ∣

Setze in

u = tan (x)

ein

= \frac{tan ^{2} ( x )}{2} + ln ∣ tan (x) ∣

Füge eine Konstante zur Lösung hinzu

= \frac{tan ^{2} ( x )}{2} + ln ∣ tan (x) ∣ + C

\int_{3}^{4} \frac{( 2 x - 3 )}{6} d x

1 - 9 x^{2} y^{^{'}} = x

\frac{d}{d x} (\frac{∣ x ∣}{2 - x ^{2}})

3 (1 + t^{2}) (\frac{d y}{d t}) = 2 t y (y^{3} - 1)

\int \frac{7 ( 4 - x ^{2} )}{( x - 4 ) ( x - 2 ) ( x + 3 )} d x