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人気のある微分積分 >

integral of 17tan^2(x)sec^3(x)

解

\int 17 tan^{2} (x) sec^{3} (x) d x

解

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

解答ステップ

\int 17 tan^{2} (x) sec^{3} (x) d x

定数を除く:

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

= 17 \cdot \int tan^{2} (x) sec^{3} (x) d x

三角関数の公式を使用して書き換える

= 17 \cdot \int (- 1 + sec^{2} (x)) sec^{3} (x) d x

拡張

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

= 17 \cdot \int - sec^{3} (x) + sec^{5} (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

= 17 (- \int sec^{3} (x) d x + \int sec^{5} (x) d x)

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

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

= 17 (- (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣))

簡素化

17 (- (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)) : 17 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x))

= 17 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x))

定数を解答に追加する

= 17 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x)) + C

グラフ

人気の例

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