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人気のある微分積分 >

integral of (cos(x))/(3cos(x)-5)

解

\int \frac{cos ( x )}{3 cos ( x ) - 5} d x

解

- \frac{5}{6} arctan (2 tan (\frac{x}{2})) + \frac{2}{3} arctan (tan (\frac{x}{2})) + C

解答ステップ

\int \frac{cos ( x )}{3 cos ( x ) - 5} d x

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

\frac{- u ^{2} + 1}{( 4 u ^{2} + 1 ) ( u ^{2} + 1 )} : \frac{5}{3 ( 4 u ^{2} + 1 )} - \frac{2}{3 ( u ^{2} + 1 )}

= - \int \frac{5}{3 ( 4 u ^{2} + 1 )} - \frac{2}{3 ( 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

= - (\int \frac{5}{3 ( 4 u ^{2} + 1 )} d u - \int \frac{2}{3 ( u ^{2} + 1 )} d u)

\int \frac{5}{3 ( 4 u ^{2} + 1 )} d u = \frac{5}{6} arctan (2 u)

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

= - (\frac{5}{6} arctan (2 u) - \frac{2}{3} arctan (u))

代用を戻す

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

= - (\frac{5}{6} arctan (2 tan (\frac{x}{2})) - \frac{2}{3} arctan (tan (\frac{x}{2})))

簡素化

= - \frac{5}{6} arctan (2 tan (\frac{x}{2})) + \frac{2}{3} arctan (tan (\frac{x}{2}))

定数を解答に追加する

= - \frac{5}{6} arctan (2 tan (\frac{x}{2})) + \frac{2}{3} arctan (tan (\frac{x}{2})) + C

グラフ

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