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 (sin^2(x))/(sin(x)+2cos(x))

解

\int \frac{sin ^{2} ( x )}{sin ( x ) + 2 cos ( x )} d x

解

4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})))) + C

解答ステップ

\int \frac{sin ^{2} ( x )}{sin ( x ) + 2 cos ( x )} d x

u置換積分法を適用する

= \int \frac{4 u ^{2}}{( - u ^{2} + u + 1 ) ( u ^{2} + 1 ) ^{2}} d u

定数を除く:

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

= 4 \cdot \int \frac{u ^{2}}{( - u ^{2} + u + 1 ) ( u ^{2} + 1 ) ^{2}} d u

以下の部分分数を得る:

\frac{u ^{2}}{( - u ^{2} + u + 1 ) ( u ^{2} + 1 ) ^{2}} : - \frac{1}{5 ( u ^{2} - u - 1 )} + \frac{1}{5 ( u ^{2} + 1 )} + \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}}

= 4 \cdot \int - \frac{1}{5 ( u ^{2} - u - 1 )} + \frac{1}{5 ( u ^{2} + 1 )} + \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}} 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

= 4 (- \int \frac{1}{5 ( u ^{2} - u - 1 )} d u + \int \frac{1}{5 ( u ^{2} + 1 )} d u + \int \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}} d u)

\int \frac{1}{5 ( u ^{2} - u - 1 )} d u = - \frac{1}{5 5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1)

\int \frac{1}{5 ( u ^{2} + 1 )} d u = \frac{1}{5} arctan (u)

\int \frac{u - 2}{5 ( u ^{2} + 1 ) ^{2}} d u = \frac{- 2 u ^{2} arctan ( u ) - 2 u - 2 arctan ( u ) - 1}{10 ( u ^{2} + 1 )}

= 4 (- (- \frac{1}{5 5} (ln \frac{2 u - 1}{5} + 1 - ln \frac{2 u - 1}{5} - 1)) + \frac{1}{5} arctan (u) + \frac{- 2 u ^{2} arctan ( u ) - 2 u - 2 arctan ( u ) - 1}{10 ( u ^{2} + 1 )})

代用を戻す

u = tan (\frac{x}{2})

= 4 (- (- \frac{1}{5 5} (ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} - 1)) + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{- 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 arctan ( tan ( \frac{x}{2} ) ) - 1}{10 ( tan ^{2} ( \frac{x}{2} ) + 1 )})

簡素化

4 (- (- \frac{1}{5 5} (ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} + 1 - ln \frac{2 tan ( \frac{x}{2} ) - 1}{5} - 1)) + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{- 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 arctan ( tan ( \frac{x}{2} ) ) - 1}{10 ( tan ^{2} ( \frac{x}{2} ) + 1 )}) : 4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2}))))

= 4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2}))))

定数を解答に追加する

= 4 (\frac{- 1 - 2 arctan ( tan ( \frac{x}{2} ) ) - 2 tan ( \frac{x}{2} ) - 2 tan ^{2} ( \frac{x}{2} ) arctan ( tan ( \frac{x}{2} ) )}{10 sec ^{2} ( \frac{x}{2} )} + \frac{1}{5} arctan (tan (\frac{x}{2})) + \frac{1}{5 5} (- ln - 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})) + ln 1 + \frac{1}{5} (- 1 + 2 tan (\frac{x}{2})))) + C

グラフ

人気の例

integral of sin(cx)

\int sin (c x) d x

integral of 2x*e^{3x}

\int 2 x \cdot e^{3 x} d x

integral of (6x^2+x)/((16x^3+4x^2)^2)

\int \frac{6 x ^{2} + x}{( 16 x ^{3} + 4 x ^{2} ) ^{2}} d x

integral of 2/(1+t^2)

\int \frac{2}{1 + t ^{2}} d t

integral of t/((2t^2-3)^{2/3)}

\int \frac{t}{( 2 t ^{2} - 3 ) ^{\frac{2}{3}}} d t