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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

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

u置換積分法を適用する

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

因数

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

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

定数を除く:

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

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

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

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

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

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

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

\int 1 d v = v

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

代用を戻す

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of (2x+3)e^x

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

integral of (x+5)/(x^2-6x+34)

\int \frac{x + 5}{x ^{2} - 6 x + 34} d x

integral of (2-sqrt(tan(x)))^2sec^2(x)

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

integral of (3+3tan^2(2x))

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

integral of (2x^2-1)cos(3x)

\int (2 x^{2} - 1) cos (3 x) d x