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人気のある微分積分 >

integral of 1/(4+5cos(x))

解

\int \frac{1}{4 + 5 cos ( x )} d x

解

\frac{1}{3} (ln \frac{tan ( \frac{x}{2} )}{3} + 1 - ln \frac{tan ( \frac{x}{2} )}{3} - 1) + C

解答ステップ

\int \frac{1}{4 + 5 cos ( x )} d x

u置換積分法を適用する

= \int \frac{2}{- u ^{2} + 9} 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 ^{2} + 9} d u

置換積分法を適用する

= 2 \cdot \int \frac{1}{3 ( - 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 \cdot \frac{1}{3} \cdot \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 \cdot \frac{1}{3} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

代用を戻す

= 2 \cdot \frac{1}{3} \frac{ln \frac{t a n ( \frac{x}{2} )}{3} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} )}{3} - 1}{2}

簡素化

2 \cdot \frac{1}{3} \frac{ln \frac{t a n ( \frac{x}{2} )}{3} + 1}{2} - \frac{ln \frac{t a n ( \frac{x}{2} )}{3} - 1}{2} : \frac{1}{3} (ln \frac{tan ( \frac{x}{2} )}{3} + 1 - ln \frac{tan ( \frac{x}{2} )}{3} - 1)

= \frac{1}{3} (ln \frac{tan ( \frac{x}{2} )}{3} + 1 - ln \frac{tan ( \frac{x}{2} )}{3} - 1)

定数を解答に追加する

= \frac{1}{3} (ln \frac{tan ( \frac{x}{2} )}{3} + 1 - ln \frac{tan ( \frac{x}{2} )}{3} - 1) + C

グラフ

人気の例

integral of cos(x)e^{ax}

\int cos (x) e^{a x} d x

derivative t(ln(2t))^2

d er i v a t i v e t (ln (2 t))^{2}

derivative 3x(a^2-x^2)

d er i v a t i v e 3 x (a^{2} - x^{2})

derivative log_{8}(x^2-3x)

d er i v a t i v e lo g_{8} (x^{2} - 3 x)

integral of (ln(x+1))

\int (ln (x + 1)) d x