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人気のある微分積分 >

integral of (sec(5x)+csc(5x))^2

解

\int (sec (5 x) + csc (5 x))^{2} d x

解

\frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x)) + C

解答ステップ

\int (sec (5 x) + csc (5 x))^{2} d x

拡張

(sec (5 x) + csc (5 x))^{2} : sec^{2} (5 x) + 2 sec (5 x) csc (5 x) + csc^{2} (5 x)

= \int sec^{2} (5 x) + 2 sec (5 x) csc (5 x) + csc^{2} (5 x) d x

u置換積分法を適用する

= \int (sec^{2} (u) + 2 sec (u) csc (u) + csc^{2} (u)) \frac{1}{5} d u

定数を除く:

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

= \frac{1}{5} \cdot \int sec^{2} (u) + 2 sec (u) csc (u) + csc^{2} (u) d u

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

= \frac{1}{5} (\int sec^{2} (u) d u + \int 2 sec (u) csc (u) d u + \int csc^{2} (u) d u)

\int sec^{2} (u) d u = tan (u)

\int 2 sec (u) csc (u) d u = ln (tan^{2} (u))

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{5} (tan (u) + ln (tan^{2} (u)) - cot (u))

代用を戻す

u = 5 x

= \frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x))

定数を解答に追加する

= \frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x)) + C

グラフ

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