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 pi/4 of 2/(1+sin(2x))

解

\int_{0}^{\frac{π}{4}} \frac{2}{1 + sin ( 2 x )} d x

解

1

解答ステップ

\int_{0}^{\frac{π}{4}} \frac{2}{1 + sin ( 2 x )} d x

定数を除く:

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

= 2 \cdot \int_{0}^{\frac{π}{4}} \frac{1}{1 + sin ( 2 x )} d x

u置換積分法を適用する

= 2 \cdot \int_{0}^{\frac{π}{2}} \frac{1}{2 ( 1 + sin ( u ) )} d u

定数を除く:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1}{1 + sin ( u )} d u

u置換積分法を適用する

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{1} \frac{2}{1 + v ^{2} + 2 v} d v

定数を除く:

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

= 2 \cdot \frac{1}{2} \cdot 2 \cdot \int_{0}^{1} \frac{1}{1 + v ^{2} + 2 v} d v

平方完成する

1 + v^{2} + 2 v : (v + 1)^{2}

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

u置換積分法を適用する

= 2 \cdot \frac{1}{2} \cdot 2 \cdot \int_{1}^{2} \frac{1}{w ^{2}} d w

べき乗則を適用する

= 2 \cdot \frac{1}{2} \cdot 2 [- \frac{1}{w}]_{1}^{2}

簡素化

2 \cdot \frac{1}{2} \cdot 2 [- \frac{1}{w}]_{1}^{2} : 2 [- \frac{1}{w}]_{1}^{2}

= 2 [- \frac{1}{w}]_{1}^{2}

限度を計算する:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

簡素化

= 1

グラフ

人気の例

integral from 3 to 5 of 2

\int_{3}^{5} 2 d x

integral from 0 to 9 of 2x-9

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

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

\int_{- 1}^{\infty} \frac{x}{1 + x ^{2}} d x

integral from-pi/4 to 0 of xsec^2(x)

\int_{- \frac{π}{4}}^{0} x sec^{2} (x) d x

integral from 1 to 4 of (x^2-x+1)

\int_{1}^{4} (x^{2} - x + 1) d x