integral of 1/(sec(x)+1)

Lösung

\int \frac{1}{sec ( x ) + 1} d x

2 arctan (tan (\frac{x}{2})) - tan (\frac{x}{2}) + C

Schritte zur Lösung

\int \frac{1}{sec ( x ) + 1} d x

Wende U-Substitution an

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

\frac{1 - u ^{2}}{1 + u ^{2}} = \frac{2}{1 + u ^{2}} - 1

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

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

\int 1 d u = u

= 2 arctan (u) - u

Setze in

u = tan (\frac{x}{2})

ein

= 2 arctan (tan (\frac{x}{2})) - tan (\frac{x}{2})

Füge eine Konstante zur Lösung hinzu

= 2 arctan (tan (\frac{x}{2})) - tan (\frac{x}{2}) + C

t an g e n t y = x, (49, 7)

\int \frac{25}{x ^{\frac{3}{2}}} d x

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f^{'} (x)

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