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人気のある微分積分 >

integral of 1/(tan(x)-sin(x))

解

\int \frac{1}{tan ( x ) - sin ( x )} d x

解

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

解答ステップ

\int \frac{1}{tan ( x ) - sin ( x )} d x

u置換積分法を適用する

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

部分積分を適用する

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

\int - \frac{1}{u} d u = - ln ∣ u ∣

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

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of e^{5x+1}

\int e^{5 x + 1} d x

integral of x^2e^{3x^3}

\int x^{2} e^{3 x^{3}} d x

integral of x/(sqrt(2x-1))

\int \frac{x}{2 x - 1} d x

derivative f(x)=1-5x

d er i v a t i v e f (x) = 1 - 5 x

integral from 0 to 8 of 1/(sqrt(64+t^2))

\int_{0}^{8} \frac{1}{64 + t ^{2}} d t