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人気のある微分積分 >

integral of-(sec^2(x)-1)/(tan(x))

解

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

解

- ln ∣ tan (x) ∣ + ln ∣ sin (x) ∣ + C

解答ステップ

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

定数を除く:

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

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

拡張

\frac{sec ^{2} ( x ) - 1}{tan ( x )} : \frac{sec ^{2} ( x )}{tan ( x )} - \frac{1}{tan ( x )}

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

総和規則を適用する:

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

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

\int \frac{sec ^{2} ( x )}{tan ( x )} d x = ln ∣ tan (x) ∣

\int \frac{1}{tan ( x )} d x = ln ∣ sin (x) ∣

= - (ln ∣ tan (x) ∣ - ln ∣ sin (x) ∣)

簡素化

= - ln ∣ tan (x) ∣ + ln ∣ sin (x) ∣

定数を解答に追加する

= - ln ∣ tan (x) ∣ + ln ∣ sin (x) ∣ + C

グラフ

人気の例

inverselaplace (s+3)/(s^2+s+1)

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