integral of sec^4(xta)n^2x

解

\int sec^{4} (x t a) n^{2} x d x

n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x)) + C

解答ステップ

\int sec^{4} (x t a) n^{2} x d x

定数を除く:

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

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

u置換積分法を適用する

: \int \frac{u sec ^{4} ( u )}{a ^{2} t ^{2}} d u

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

定数を除く:

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

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

部分積分を適用する

: u (tan (u) + \frac{1}{3} tan^{3} (u)) - \int tan (u) + \frac{1}{3} tan^{3} (u) d u

= n^{2} \frac{1}{a ^{2} t ^{2}} (u (tan (u) + \frac{1}{3} tan^{3} (u)) - \int tan (u) + \frac{1}{3} tan^{3} (u) d u)

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

= n^{2} \frac{1}{a ^{2} t ^{2}} (u (tan (u) + \frac{1}{3} tan^{3} (u)) - (- ln ∣ cos (u) ∣ + \frac{1}{3} (- ln ∣ sec (u) ∣ + \frac{sec ^{2} ( u )}{2})))

代用を戻す

u = a t x

= n^{2} \frac{1}{a ^{2} t ^{2}} (a t x (tan (a t x) + \frac{1}{3} tan^{3} (a t x)) - (- ln ∣ cos (a t x) ∣ + \frac{1}{3} (- ln ∣ sec (a t x) ∣ + \frac{sec ^{2} ( a t x )}{2})))

簡素化

n^{2} \frac{1}{a ^{2} t ^{2}} (a t x (tan (a t x) + \frac{1}{3} tan^{3} (a t x)) - (- ln ∣ cos (a t x) ∣ + \frac{1}{3} (- ln ∣ sec (a t x) ∣ + \frac{sec ^{2} ( a t x )}{2}))) : n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x))

= n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x))

定数を解答に追加する

= n^{2} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x)) + C