What is the integral of sec^2(xta)n^4x ?

The integral of sec^2(xta)n^4x is n^4 1/(a^2t^2)(atxtan(atx)+ln|cos(atx)|)+C

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

: u tan (u) - \int tan (u) d u

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

\int tan (u) d u = - ln ∣ cos (u) ∣

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

Substitute back

u = a t x

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

Simplify

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

Add a constant to the solution

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

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What is the integral of sec^2(xta)n^4x ?
The integral of sec^2(xta)n^4x is n^4 1/(a^2t^2)(atxtan(atx)+ln|cos(atx)|)+C