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人気のある微分積分 >

integral of 1/(64sec^2(x)tan(x))

解

\int \frac{1}{64 sec ^{2} ( x ) tan ( x )} d x

解

\frac{1}{64} (ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2}) + C

解答ステップ

\int \frac{1}{64 sec ^{2} ( x ) tan ( x )} d x

定数を除く:

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

= \frac{1}{64} \cdot \int \frac{1}{sec ^{2} ( x ) tan ( x )} d x

三角関数の公式を使用して書き換える

= \frac{1}{64} \cdot \int \frac{( 1 - sin ^{2} ( x ) ) cos ( x )}{sin ( x )} d x

u置換積分法を適用する

= \frac{1}{64} \cdot \int \frac{1 - u ^{2}}{u} d u

拡張

\frac{1 - u ^{2}}{u} : \frac{1}{u} - u

= \frac{1}{64} \cdot \int \frac{1}{u} - u 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

= \frac{1}{64} (\int \frac{1}{u} d u - \int u d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= \frac{1}{64} (ln ∣ u ∣ - \frac{u ^{2}}{2})

代用を戻す

u = sin (x)

= \frac{1}{64} (ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2})

定数を解答に追加する

= \frac{1}{64} (ln ∣ sin (x) ∣ - \frac{sin ^{2} ( x )}{2}) + C

グラフ

人気の例

integral of x^9ln(3x)

\int x^{9} ln (3 x) d x

integral of cos(x)*ln(sin(x))

\int cos (x) \cdot ln (sin (x)) d x

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

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

integral of ((y^3-1)/y)^2

\int (\frac{y ^{3} - 1}{y})^{2} d y

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

\int \frac{cos ( x )}{1 - ( cos ( x ) ) ^{2}} d x