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人気のある微分積分 >

integral of 13tan^4(xse)c^6x

解

\int 13 tan^{4} (x se) c^{6} x d x

解

\frac{13 c ^{6}}{6 e ^{2} s ^{2}} (3 e^{2} s^{2} x^{2} + 2 es x tan^{3} (es x) - 6 es x tan (es x) + 2 ln ∣ sec (es x) ∣ - sec^{2} (es x) - 6 ln ∣ cos (es x) ∣) + C

解答ステップ

\int 13 tan^{4} (x se) c^{6} x d x

定数を除く:

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

= 13 c^{6} \cdot \int tan^{4} (se x) x d x

u置換積分法を適用する

= 13 c^{6} \cdot \int \frac{u tan ^{4} ( u )}{e ^{2} s ^{2}} d u

定数を除く:

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

= 13 c^{6} \frac{1}{e ^{2} s ^{2}} \cdot \int u tan^{4} (u) d u

部分積分を適用する

= 13 c^{6} \frac{1}{e ^{2} s ^{2}} (u (\frac{1}{3} tan^{3} (u) + u - tan (u)) - \int \frac{1}{3} tan^{3} (u) + u - tan (u) d u)

\int \frac{1}{3} tan^{3} (u) + u - tan (u) d u = \frac{1}{3} (- ln ∣ sec (u) ∣ + \frac{1}{2} sec^{2} (u)) + \frac{u ^{2}}{2} + ln ∣ cos (u) ∣

= 13 c^{6} \frac{1}{e ^{2} s ^{2}} (u (\frac{1}{3} tan^{3} (u) + u - tan (u)) - (\frac{1}{3} (- ln ∣ sec (u) ∣ + \frac{1}{2} sec^{2} (u)) + \frac{u ^{2}}{2} + ln ∣ cos (u) ∣))

代用を戻す

u = se x

= 13 c^{6} \frac{1}{e ^{2} s ^{2}} (se x (\frac{1}{3} tan^{3} (se x) + se x - tan (se x)) - (\frac{1}{3} (- ln ∣ sec (se x) ∣ + \frac{1}{2} sec^{2} (se x)) + \frac{( se x ) ^{2}}{2} + ln ∣ cos (se x) ∣))

簡素化

13 c^{6} \frac{1}{e ^{2} s ^{2}} (se x (\frac{1}{3} tan^{3} (se x) + se x - tan (se x)) - (\frac{1}{3} (- ln ∣ sec (se x) ∣ + \frac{1}{2} sec^{2} (se x)) + \frac{( se x ) ^{2}}{2} + ln ∣ cos (se x) ∣)) : \frac{13 c ^{6}}{6 e ^{2} s ^{2}} (3 e^{2} s^{2} x^{2} + 2 es x tan^{3} (es x) - 6 es x tan (es x) + 2 ln ∣ sec (es x) ∣ - sec^{2} (es x) - 6 ln ∣ cos (es x) ∣)

= \frac{13 c ^{6}}{6 e ^{2} s ^{2}} (3 e^{2} s^{2} x^{2} + 2 es x tan^{3} (es x) - 6 es x tan (es x) + 2 ln ∣ sec (es x) ∣ - sec^{2} (es x) - 6 ln ∣ cos (es x) ∣)

定数を解答に追加する

= \frac{13 c ^{6}}{6 e ^{2} s ^{2}} (3 e^{2} s^{2} x^{2} + 2 es x tan^{3} (es x) - 6 es x tan (es x) + 2 ln ∣ sec (es x) ∣ - sec^{2} (es x) - 6 ln ∣ cos (es x) ∣) + C

グラフ

人気の例

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