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人気のある微分積分 >

integral of cot^4(xcs)c^2x

解

\int cot^{4} (x cs) c^{2} x d x

解

- \frac{c}{3 s} x cot^{3} (cs x) + c^{2} x^{2} + \frac{c}{s} x cot (cs x) - \frac{c ^{2} x ^{2}}{2} - \frac{4}{3 s ^{2}} ln ∣ sin (cs x) ∣ - \frac{1}{6 s ^{2}} cot^{2} (cs x) + C

解答ステップ

\int cot^{4} (x cs) c^{2} x d x

定数を除く:

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

= c^{2} \cdot \int cot^{4} (cs x) x d x

u置換積分法を適用する

= c^{2} \cdot \int \frac{u cot ^{4} ( u )}{c ^{2} s ^{2}} d u

定数を除く:

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

= c^{2} \frac{1}{c ^{2} s ^{2}} \cdot \int u cot^{4} (u) d u

部分積分を適用する

= c^{2} \frac{1}{c ^{2} s ^{2}} (u (- \frac{1}{3} cot^{3} (u) + u + cot (u)) - \int - \frac{1}{3} cot^{3} (u) + u + cot (u) d u)

\int - \frac{1}{3} cot^{3} (u) + u + cot (u) d u = \frac{u ^{2}}{2} + \frac{4}{3} ln ∣ sin (u) ∣ + \frac{1}{6} cot^{2} (u)

= c^{2} \frac{1}{c ^{2} s ^{2}} (u (- \frac{1}{3} cot^{3} (u) + u + cot (u)) - (\frac{u ^{2}}{2} + \frac{4}{3} ln ∣ sin (u) ∣ + \frac{1}{6} cot^{2} (u)))

代用を戻す

u = cs x

= c^{2} \frac{1}{c ^{2} s ^{2}} (cs x (- \frac{1}{3} cot^{3} (cs x) + cs x + cot (cs x)) - (\frac{( cs x ) ^{2}}{2} + \frac{4}{3} ln ∣ sin (cs x) ∣ + \frac{1}{6} cot^{2} (cs x)))

簡素化

c^{2} \frac{1}{c ^{2} s ^{2}} (cs x (- \frac{1}{3} cot^{3} (cs x) + cs x + cot (cs x)) - (\frac{( cs x ) ^{2}}{2} + \frac{4}{3} ln ∣ sin (cs x) ∣ + \frac{1}{6} cot^{2} (cs x))) : - \frac{c}{3 s} x cot^{3} (cs x) + c^{2} x^{2} + \frac{c}{s} x cot (cs x) - \frac{c ^{2} x ^{2}}{2} - \frac{4}{3 s ^{2}} ln ∣ sin (cs x) ∣ - \frac{1}{6 s ^{2}} cot^{2} (cs x)

= - \frac{c}{3 s} x cot^{3} (cs x) + c^{2} x^{2} + \frac{c}{s} x cot (cs x) - \frac{c ^{2} x ^{2}}{2} - \frac{4}{3 s ^{2}} ln ∣ sin (cs x) ∣ - \frac{1}{6 s ^{2}} cot^{2} (cs x)

定数を解答に追加する

= - \frac{c}{3 s} x cot^{3} (cs x) + c^{2} x^{2} + \frac{c}{s} x cot (cs x) - \frac{c ^{2} x ^{2}}{2} - \frac{4}{3 s ^{2}} ln ∣ sin (cs x) ∣ - \frac{1}{6 s ^{2}} cot^{2} (cs x) + C

グラフ

人気の例

integral of cot^3(θ)csc^2(θ)

\int cot^{3} (θ) csc^{2} (θ) d θ

(d^2y)/(dt^2)+3((dy)/(dt))+2y=1

\frac{d ^{2} y}{d t ^{2}} + 3 (\frac{d y}{d t}) + 2 y = 1

integral from 1 to infinity of x/(6+x^2)

\int_{1}^{\infty} \frac{x}{6 + x ^{2}} d x

derivative of 0.001x^3-0.14x^2+7x+160

\frac{d}{d x} (0.001 x^{3} - 0.14 x^{2} + 7 x + 160)

テイラー [(z-1)^3(z+1)^2]-1

t a y l or [(z - 1)^{3} (z + 1)^{2}] - 1