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人気のある微分積分 >

integral of 1/(1+tan(x))

解

\int \frac{1}{1 + tan ( x )} d x

解

\frac{1}{2} (- \frac{1}{2} ln (sec^{2} (x)) + arctan (tan (x))) + \frac{1}{2} ln ∣ tan (x) + 1 ∣ + C

解答ステップ

\int \frac{1}{1 + tan ( x )} d x

u置換積分法を適用する

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{2 ( u + 1 )} d u = \frac{1}{2} ln ∣ u + 1 ∣

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

代用を戻す

u = tan (x)

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

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

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

定数を解答に追加する

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

グラフ

人気の例

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