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人気のある微分積分 >

integral of (tan(x))/(sqrt(4+3sec(x)))

解

\int \frac{tan ( x )}{4 + 3 sec ( x )} d x

解

- \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} + 1 + \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} - 1 + C

解答ステップ

\int \frac{tan ( x )}{4 + 3 sec ( x )} d x

u置換積分法を適用する

= \int \frac{1}{u 4 + 3 u} d u

u置換積分法を適用する

= \int \frac{2}{v ^{2} - 4} d v

定数を除く:

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

= 2 \cdot \int \frac{1}{v ^{2} - 4} d v

置換積分法を適用する

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

定数を除く:

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

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

因数

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

= 2 \cdot \frac{1}{2} \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

= 2 \cdot \frac{1}{2} (- \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}

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

代用を戻す

= 2 \cdot \frac{1}{2} - \frac{ln \frac{4 + 3 s e c ( x )}{2} + 1}{2} - \frac{ln \frac{4 + 3 s e c ( x )}{2} - 1}{2}

簡素化

2 \cdot \frac{1}{2} - \frac{ln \frac{4 + 3 s e c ( x )}{2} + 1}{2} - \frac{ln \frac{4 + 3 s e c ( x )}{2} - 1}{2} : - \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} + 1 + \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} - 1

= - \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} + 1 + \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} - 1

定数を解答に追加する

= - \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} + 1 + \frac{1}{2} ln \frac{4 + 3 sec ( x )}{2} - 1 + C

グラフ

人気の例

integral of 5xtan^2(3x^2-5)sec^2(3x^2-5)

\int 5 x tan^{2} (3 x^{2} - 5) sec^{2} (3 x^{2} - 5) d x

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\int sin (15 x) sin (16 x) d x

integral of a/((x^2+a^2)^{3/2)}

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\int (x^{3} - 2)^{2} d x

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\int \frac{12 x ^{2} - 6 x + 2}{( 4 x + 1 ) ( x ^{2} + 1 )} d x