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人気のある微分積分 >

integral of 1/(cos(x)sqrt(sin(x)))

解

\int \frac{1}{cos ( x ) sin ( x )} d x

解

arctan (sin (x)) + \frac{1}{2} ln sin (x) + 1 - \frac{1}{2} ln sin (x) - 1 + C

解答ステップ

\int \frac{1}{cos ( x ) sin ( x )} d x

u置換積分法を適用する

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

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{1 - v ^{4}} : \frac{1}{2 ( v ^{2} + 1 )} + \frac{1}{4 ( v + 1 )} - \frac{1}{4 ( v - 1 )}

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

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

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

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

\int \frac{1}{4 ( v + 1 )} d v = \frac{1}{4} ln ∣ v + 1 ∣

\int \frac{1}{4 ( v - 1 )} d v = \frac{1}{4} ln ∣ v - 1 ∣

= 2 (\frac{1}{2} arctan (v) + \frac{1}{4} ln ∣ v + 1 ∣ - \frac{1}{4} ln ∣ v - 1 ∣)

代用を戻す

= 2 (\frac{1}{2} arctan (sin (x)) + \frac{1}{4} ln sin (x) + 1 - \frac{1}{4} ln sin (x) - 1)

簡素化

2 (\frac{1}{2} arctan (sin (x)) + \frac{1}{4} ln sin (x) + 1 - \frac{1}{4} ln sin (x) - 1) : arctan (sin (x)) + \frac{1}{2} ln sin (x) + 1 - \frac{1}{2} ln sin (x) - 1

= arctan (sin (x)) + \frac{1}{2} ln sin (x) + 1 - \frac{1}{2} ln sin (x) - 1

定数を解答に追加する

= arctan (sin (x)) + \frac{1}{2} ln sin (x) + 1 - \frac{1}{2} ln sin (x) - 1 + C

グラフ

人気の例

integral of 1/(16x^2+1)

integral of (7^{2/3}-x^{2/3})^3

integral of sin(3x-1)

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

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