ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

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

解

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

解

\frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x)) + C

解答ステップ

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

u置換積分法を適用する

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

拡張

(tan (u) + sec (u))^{2} : tan^{2} (u) + 2 tan (u) sec (u) + sec^{2} (u)

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

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

\int 2 tan (u) sec (u) d u = 2 sec (u)

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

= \frac{1}{5} (- u + tan (u) + 2 sec (u) + tan (u))

代用を戻す

u = 5 x

= \frac{1}{5} (- 5 x + tan (5 x) + 2 sec (5 x) + tan (5 x))

簡素化

\frac{1}{5} (- 5 x + tan (5 x) + 2 sec (5 x) + tan (5 x)) : \frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x))

= \frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x))

定数を解答に追加する

= \frac{1}{5} (- 5 x + 2 tan (5 x) + 2 sec (5 x)) + C

グラフ

人気の例

(dy)/(dx)=(x^2-3x)^3

\frac{d y}{d x} = (x^{2} - 3 x)^{3}

tangent 1/(x^3)

t an g e n t \frac{1}{x ^{3}}

integral from 0 to 2 of-5x^3sqrt(x^2+4)

\int_{0}^{2} - 5 x^{3} x^{2} + 4 d x

integral of 5θcos(θ)

\int 5 θ cos (θ) d θ

inverselaplace ax+1

in v erse l a pl a ce a x + 1