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人気のある微分積分 >

integral of 1/((3sin(x)-4cos(x)))

解

\int \frac{1}{( 3 sin ( x ) - 4 cos ( x ) )} d x

解

- \frac{1}{5} (ln \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2})) + C

解答ステップ

\int \frac{1}{3 sin ( x ) - 4 cos ( x )} d x

u置換積分法を適用する

= \int \frac{1}{3 u - 2 ( - u ^{2} + 1 )} d u

平方完成する

3 u - 2 (- u^{2} + 1) : 2 (u + \frac{3}{4})^{2} - \frac{25}{8}

= \int \frac{1}{2 ( u + \frac{3}{4} ) ^{2} - \frac{25}{8}} d u

u置換積分法を適用する

= \int \frac{8}{16 v ^{2} - 25} d v

定数を除く:

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

= 8 \cdot \int \frac{1}{16 v ^{2} - 25} d v

置換積分法を適用する

= 8 \cdot \int \frac{1}{20 ( w ^{2} - 1 )} d w

定数を除く:

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

= 8 \cdot \frac{1}{20} \cdot \int \frac{1}{w ^{2} - 1} d w

因数

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

= 8 \cdot \frac{1}{20} \cdot \int \frac{1}{- ( - w ^{2} + 1 )} d w

定数を除く:

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

= 8 \cdot \frac{1}{20} (- \int \frac{1}{- w ^{2} + 1} d w)

共通積分を使用する:

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

= 8 \cdot \frac{1}{20} (- (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}))

代用を戻す

= 8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) - 1}{2}))

簡素化

8 \cdot \frac{1}{20} (- (\frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) + 1}{2} - \frac{ln \frac{4}{5} ( tan ( \frac{x}{2} ) + \frac{3}{4} ) - 1}{2})) : - \frac{1}{5} (ln \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2}))

= - \frac{1}{5} (ln \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2}))

定数を解答に追加する

= - \frac{1}{5} (ln \frac{8}{5} + \frac{4}{5} tan (\frac{x}{2}) - ln - \frac{2}{5} + \frac{4}{5} tan (\frac{x}{2})) + C

グラフ

人気の例

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