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人気のある微分積分 >

integral of 5sec(3x^2+4)tan(3x^2+4)x

解

\int 5 sec (3 x^{2} + 4) tan (3 x^{2} + 4) x d x

解

\frac{5}{6} sec (3 x^{2} + 4) + C

解答ステップ

\int 5 sec (3 x^{2} + 4) tan (3 x^{2} + 4) x d x

定数を除く:

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

= 5 \cdot \int sec (3 x^{2} + 4) tan (3 x^{2} + 4) x d x

u置換積分法を適用する

= 5 \cdot \int \frac{sec ( u ) tan ( u )}{6} d u

定数を除く:

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

= 5 \cdot \frac{1}{6} \cdot \int sec (u) tan (u) d u

u置換積分法を適用する

= 5 \cdot \frac{1}{6} \cdot \int 1 d v

定数の積分:

\int a d x = a x

= 5 \cdot \frac{1}{6} \cdot 1 \cdot v

代用を戻す

= 5 \cdot \frac{1}{6} \cdot 1 \cdot sec (3 x^{2} + 4)

簡素化

5 \cdot \frac{1}{6} \cdot 1 \cdot sec (3 x^{2} + 4) : \frac{5}{6} sec (3 x^{2} + 4)

= \frac{5}{6} sec (3 x^{2} + 4)

定数を解答に追加する

= \frac{5}{6} sec (3 x^{2} + 4) + C

グラフ

人気の例

integral of x(3x^2+2)^5

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

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\int \frac{4 x ^{2}}{x - 1} d x

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integral of ((2x^2-5x+1))/((x+2)(x^2+8))

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