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 from 0 to 1 of sqrt(tan(θ))

解

\int_{0}^{1} tan (θ) d θ

解

\frac{4 arctan ( 2 tan ( 1 ) + 1 ) + 4 arctan ( 2 tan ( 1 ) - 1 ) - 2 ln ( tan ( 1 ) + 2 tan ( 1 ) + 1 ) - 2 ln ( \frac{1}{t a n ( 1 ) - 2 t a n ( 1 ) + 1} )}{4 2}

+1

十進法表記

0.72729 \dots

解答ステップ

\int_{0}^{1} tan (θ) d θ

三角関数による置換を適用する

= \int_{0}^{t a n (1)} \frac{u}{1 + u ^{2}} d u

置換積分法を適用する

= \int_{0}^{2 t a n (1)} \frac{2 v}{4 + v ^{2}} d v

定数を除く:

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

= 2 \cdot \int_{0}^{2 t a n (1)} \frac{v}{4 + v ^{2}} d v

u置換積分法を適用する

= 2 \cdot \int_{0}^{2 t a n (1)} \frac{2 w ^{2}}{4 + w ^{4}} d w

定数を除く:

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

= 2 \cdot 2 \cdot \int_{0}^{2 t a n (1)} \frac{w ^{2}}{4 + w ^{4}} d w

以下の部分分数を得る:

\frac{w ^{2}}{4 + w ^{4}} : - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )}

= 2 \cdot 2 \cdot \int_{0}^{2 t a n (1)} - \frac{w}{4 ( w ^{2} + 2 w + 2 )} + \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w

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

= 2 \cdot 2 (- \int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w + \int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w)

\int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} + 2 w + 2 )} d w = \frac{1}{4} (\frac{1}{2} ln (tan (1) + 2 tan (1) + 1) - arctan (2 tan (1) + 1) + \frac{π}{4})

\int_{0}^{2 t a n (1)} \frac{w}{4 ( w ^{2} - 2 w + 2 )} d w = \frac{1}{4} (- \frac{1}{2} ln (\frac{1}{tan ( 1 ) - 2 tan ( 1 ) + 1}) + arctan (2 tan (1) - 1) + \frac{π}{4})

= 2 \cdot 2 (- \frac{1}{4} (\frac{1}{2} ln (tan (1) + 2 tan (1) + 1) - arctan (2 tan (1) + 1) + \frac{π}{4}) + \frac{1}{4} (- \frac{1}{2} ln (\frac{1}{tan ( 1 ) - 2 tan ( 1 ) + 1}) + arctan (2 tan (1) - 1) + \frac{π}{4}))

簡素化

2 \cdot 2 (- \frac{1}{4} (\frac{1}{2} ln (tan (1) + 2 tan (1) + 1) - arctan (2 tan (1) + 1) + \frac{π}{4}) + \frac{1}{4} (- \frac{1}{2} ln (\frac{1}{tan ( 1 ) - 2 tan ( 1 ) + 1}) + arctan (2 tan (1) - 1) + \frac{π}{4})) : \frac{4 arctan ( 2 tan ( 1 ) + 1 ) + 4 arctan ( 2 tan ( 1 ) - 1 ) - 2 ln ( tan ( 1 ) + 2 tan ( 1 ) + 1 ) - 2 ln ( \frac{1}{t a n ( 1 ) - 2 t a n ( 1 ) + 1} )}{4 2}

= \frac{4 arctan ( 2 tan ( 1 ) + 1 ) + 4 arctan ( 2 tan ( 1 ) - 1 ) - 2 ln ( tan ( 1 ) + 2 tan ( 1 ) + 1 ) - 2 ln ( \frac{1}{t a n ( 1 ) - 2 t a n ( 1 ) + 1} )}{4 2}

グラフ

人気の例

integral from 0 to ln(10) of cosh(x)

\int_{0}^{l n (10)} cosh (x) d x

integral from 0 to 1 of 2/(sqrt(9-2x^2))

\int_{0}^{1} \frac{2}{9 - 2 x ^{2}} d x

integral from 0 to 4 of (x^3-7x)

\int_{0}^{4} (x^{3} - 7 x) d x

integral from 2 to 3 of (2x^2)/(x^2-1)

\int_{2}^{3} \frac{2 x ^{2}}{x ^{2} - 1} d x

integral from 0 to 1 of e^{1-x}cos(2x)

\int_{0}^{1} e^{1 - x} cos (2 x) d x