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人気のある微分積分 >

integral of sin(ln(13x))

解

\int sin (ln (13 x)) d x

解

\frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x))) + C

解答ステップ

\int sin (ln (13 x)) d x

u置換積分法を適用する

= \int sin (ln (u)) \frac{1}{13} d u

定数を除く:

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

= \frac{1}{13} \cdot \int sin (ln (u)) d u

部分積分を適用する

= \frac{1}{13} (u sin (ln (u)) - \int cos (ln (u)) d u)

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

= \frac{1}{13} (u sin (ln (u)) - (\frac{1}{2} u sin (ln (u)) + \frac{1}{2} u cos (ln (u))))

代用を戻す

u = 13 x

= \frac{1}{13} (13 x sin (ln (13 x)) - (\frac{1}{2} \cdot 13 x sin (ln (13 x)) + \frac{1}{2} \cdot 13 x cos (ln (13 x))))

簡素化

\frac{1}{13} (13 x sin (ln (13 x)) - (\frac{1}{2} \cdot 13 x sin (ln (13 x)) + \frac{1}{2} \cdot 13 x cos (ln (13 x)))) : \frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x)))

= \frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x)))

定数を解答に追加する

= \frac{1}{2} x (sin (ln (13 x)) - cos (ln (13 x))) + C

グラフ

人気の例

tangent f(x)=2x^2+5x,\at x=-3

t an g e n t f (x) = 2 x^{2} + 5 x, at x = - 3

(d^2h)/(dt^2)=-kh

\frac{d ^{2} h}{d t ^{2}} = - kh

limit as x approaches 2 of 2x^2-3x+1

x \to 2 lim (2 x^{2} - 3 x + 1)

integral of (x^2)/4-1

\int \frac{x ^{2}}{4} - 1 d x

y^{''}+36y=-36sec(6t)

y^{^{''}} + 36 y = - 36 sec (6 t)