(dy)/(dx)= 1/(2cos^2(x))

Lösung

\frac{d y}{d x} = \frac{1}{2 cos ^{2} ( x )}

y = \frac{1}{2} tan (x) + c_{1}

Schritte zur Lösung

\frac{d y}{d x} = \frac{1}{2 cos ^{2} ( x )}

Ersetze

\frac{d y}{d x}

mit

y^{'} (x)

y^{^{'}} (x) = \frac{1}{2 cos ^{2} ( x )}

Wenn

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

dann

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

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

\int \frac{1}{2 cos ^{2} ( x )} d x = \frac{1}{2} tan (x) + c_{1}

y = \frac{1}{2} tan (x) + c_{1}

\int_{0}^{1} x (x^{2} - 1) e^{- x^{2}} d x

\int_{0}^{3} (x) d x

\int 8 x (4 x^{2} + 5)^{6} d x

\int \frac{4}{7 ^{l n (x)}} d x

x \to 0 + lim (ln (x^{x}))