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人気のある微分積分 >

integral of sec^3(5x)tan^2(5x)

解

\int sec^{3} (5 x) tan^{2} (5 x) d x

解

\frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x)) + C

解答ステップ

\int sec^{3} (5 x) tan^{2} (5 x) d x

u置換積分法を適用する

= \int sec^{3} (u) tan^{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^{3} (u) tan^{2} (u) d u

三角関数の公式を使用して書き換える

= \frac{1}{5} \cdot \int sec^{3} (u) (- 1 + sec^{2} (u)) d u

拡張

sec^{3} (u) (- 1 + sec^{2} (u)) : - sec^{3} (u) + sec^{5} (u)

= \frac{1}{5} \cdot \int - sec^{3} (u) + sec^{5} (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^{3} (u) d u + \int sec^{5} (u) d u)

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

\int sec^{5} (u) d u = \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= \frac{1}{5} (- (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣) + \frac{sec ^{4} ( u ) sin ( u )}{4} + \frac{3}{4} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣))

代用を戻す

u = 5 x

= \frac{1}{5} (- (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) + \frac{sec ^{4} ( 5 x ) sin ( 5 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣))

簡素化

\frac{1}{5} (- (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣) + \frac{sec ^{4} ( 5 x ) sin ( 5 x )}{4} + \frac{3}{4} (\frac{1}{2} sec (5 x) tan (5 x) + \frac{1}{2} ln ∣ tan (5 x) + sec (5 x) ∣)) : \frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x))

= \frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x))

定数を解答に追加する

= \frac{1}{40} (- sec (5 x) tan (5 x) - ln ∣ tan (5 x) + sec (5 x) ∣ + 2 sec^{4} (5 x) sin (5 x)) + C

グラフ

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