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人気のある微分積分 >

integral of (sin(x))/(1+sin(x)+cos(x))

解

\int \frac{sin ( x )}{1 + sin ( x ) + cos ( x )} d x

解

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

解答ステップ

\int \frac{sin ( x )}{1 + sin ( x ) + cos ( x )} d x

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

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

= 2 (- \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{2} (\frac{1}{2} ln u^{2} + 1 + arctan (u)))

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of 5/(x(ln(x))^4)

\int \frac{5}{x ( ln ( x ) ) ^{4}} d x

integral of 1/(x(x^7+1))

\int \frac{1}{x ( x ^{7} + 1 )} d x

integral of (125)/((5x-300)^2)

\int \frac{125}{( 5 x - 300 ) ^{2}} d x

integral of 3/(1+9x^2)

\int \frac{3}{1 + 9 x ^{2}} d x

integral of xsin(3x-1)

\int x sin (3 x - 1) d x