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受欢迎的微积分 >

integral of sec^2(2xta)n^62x

解答

\int sec^{2} (2 x t a) n^{6} 2 x d x

解答

\frac{n ^{6}}{a t} x tan (2 a t x) + n^{6} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣ + C

求解步骤

\int sec^{2} (2 x t a) n^{6} \cdot 2 x d x

提出常数:

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

= 2 n^{6} \cdot \int sec^{2} (2 a t x) x d x

使用换元积分法

= 2 n^{6} \cdot \int \frac{u sec ^{2} ( u )}{4 a ^{2} t ^{2}} d u

提出常数:

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

= 2 n^{6} \frac{1}{4 a ^{2} t ^{2}} \cdot \int u sec^{2} (u) d u

使用分布积分法

= 2 n^{6} \frac{1}{4 a ^{2} t ^{2}} (u tan (u) - \int tan (u) d u)

\int tan (u) d u = - ln ∣ cos (u) ∣

= 2 n^{6} \frac{1}{4 a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

u = 2 a t x

代回

= 2 n^{6} \frac{1}{4 a ^{2} t ^{2}} (2 a t x tan (2 a t x) - (- ln ∣ cos (2 a t x) ∣))

化简

2 n^{6} \frac{1}{4 a ^{2} t ^{2}} (2 a t x tan (2 a t x) - (- ln ∣ cos (2 a t x) ∣)) : \frac{n ^{6}}{a t} x tan (2 a t x) + n^{6} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣

= \frac{n ^{6}}{a t} x tan (2 a t x) + n^{6} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣

解答补常数

= \frac{n ^{6}}{a t} x tan (2 a t x) + n^{6} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣ + C

作图

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