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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

u置換積分法を適用する

= 2 \cdot \int \frac{1}{v ^{2} - 1} d v

因数

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

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

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of (2e^x+3/x)

\int (2 e^{x} + \frac{3}{x}) d x

integral of 4xsec^2(x)-4xtan^2(x)

\int 4 x sec^{2} (x) - 4 x tan^{2} (x) d x

integral of 1/(4+49x^2)

\int \frac{1}{4 + 49 x ^{2}} d x

integral of (e^x)/(sqrt(e^{2x))}

\int \frac{e ^{x}}{e ^{2 x}} d x

integral of 1/((x^2-6x+10)^{5/2)}

\int \frac{1}{( x ^{2} - 6 x + 10 ) ^{\frac{5}{2}}} d x