integral of sec^2(3x)+tan(6x)

Lösung

\int sec^{2} (3 x) + tan (6 x) d x

\frac{1}{3} tan (3 x) - \frac{1}{6} ln ∣ cos (6 x) ∣ + C

Schritte zur Lösung

\int sec^{2} (3 x) + tan (6 x) d x

Wende die Summenregel an:

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

= \int sec^{2} (3 x) d x + \int tan (6 x) d x

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

\int tan (6 x) d x = - \frac{1}{6} ln ∣ cos (6 x) ∣

= \frac{1}{3} tan (3 x) - \frac{1}{6} ln ∣ cos (6 x) ∣

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} tan (3 x) - \frac{1}{6} ln ∣ cos (6 x) ∣ + C

\int \frac{x ^{2}}{( x ^{2} + 9 ) ^{3}} d x

\int sin^{7} (2 x) cos^{3} (2 x) d x

\int \frac{1}{2 + 9 x ^{2}} d x

\int x cos (6 π x^{2}) d x

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