integral of tan(x)sqrt(1+sec^4(x))

Lösung

\int tan (x) 1 + sec^{4} (x) d x

- (- \frac{1}{2} sec^{2} (x) cos^{4} (x) + 1 + \frac{1}{2} ln cos^{4} (x) + 1 + cos^{2} (x)) + C

Schritte zur Lösung

\int tan (x) 1 + sec^{4} (x) d x

Trigonometrische Substitution anwenden

= \int - \frac{u ^{4} + 1}{u ^{3}} d u

Entferne die Konstante:

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

= - \int \frac{u ^{4} + 1}{u ^{3}} d u

Wende die partielle Integration an

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

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

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

Setze in

u = cos (x)

ein

= - (- \frac{cos ^{4} ( x ) + 1}{2 cos ^{2} ( x )} - (- \frac{1}{2} ln cos^{4} (x) + 1 + cos^{2} (x)))

Vereinfache

- (- \frac{cos ^{4} ( x ) + 1}{2 cos ^{2} ( x )} - (- \frac{1}{2} ln cos^{4} (x) + 1 + cos^{2} (x))) : - (- \frac{1}{2} sec^{2} (x) cos^{4} (x) + 1 + \frac{1}{2} ln cos^{4} (x) + 1 + cos^{2} (x))

= - (- \frac{1}{2} sec^{2} (x) cos^{4} (x) + 1 + \frac{1}{2} ln cos^{4} (x) + 1 + cos^{2} (x))

Füge eine Konstante zur Lösung hinzu

= - (- \frac{1}{2} sec^{2} (x) cos^{4} (x) + 1 + \frac{1}{2} ln cos^{4} (x) + 1 + cos^{2} (x)) + C

\int tan^{5} (x) sec^{4} (x) d x

d y + e^{3 y} d x = 0

\frac{d}{d x} (4 sec (x) - 3 x)

d er i v a t i v e 20 ln (3 x + 1) - \frac{x}{3}

x (x - 2) \frac{d y}{d x} + 2 y = 0