(df)/(dx)=tan^3(x)-sec^3(x)

Lösung

\frac{df}{d x} = tan^{3} (x) - sec^{3} (x)

f (x) = - ln (sec (x)) + \frac{1}{2} sec^{2} (x) - \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln (tan (x) + sec (x)) + c_{1}

Schritte zur Lösung

\frac{df}{d x} = tan^{3} (x) - sec^{3} (x)

Ersetze

\frac{df}{d x}

mit

f^{'} (x)

f^{^{'}} (x) = tan^{3} (x) - sec^{3} (x)

Wenn

f^{^{'}} (x) = g (x)

dann

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

f (x) = \int tan^{3} (x) - sec^{3} (x) d x

\int tan^{3} (x) - sec^{3} (x) d x = - ln (sec (x)) + \frac{1}{2} sec^{2} (x) - \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln (tan (x) + sec (x)) + c_{1}

f (x) = - ln (sec (x)) + \frac{1}{2} sec^{2} (x) - \frac{1}{2} sec (x) tan (x) - \frac{1}{2} ln (tan (x) + sec (x)) + c_{1}

n = 1 \sum \infty \frac{80}{8 ^{n}}

\int 4 e^{x + y}

t an g e n t y = x^{2} - 5, (- 5, 20)

d er i v a t i v e f (x) = \frac{1}{t ^{2}}

\int 43 tan^{5} (x) sec^{4} (x) d x