What is the integral of sec^2(2xta)n^52x ?

The integral of sec^2(2xta)n^52x is (n^5)/(at)xtan(2atx)+n^5 1/(2a^2t^2)ln|cos(2atx)|+C

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integral of sec^2(2xta)n^52x

\int sec^{2} (2 x t a) n^{5} 2 x d x

\frac{n ^{5}}{a t} x tan (2 a t x) + n^{5} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣ + C

Solution steps

\int sec^{2} (2 x t a) n^{5} \cdot 2 x d x

Take the constant out:

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

= 2 n^{5} \cdot \int sec^{2} (2 a t x) x d x

Apply u-substitution

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

= 2 n^{5} \cdot \int \frac{u sec ^{2} ( u )}{4 a ^{2} t ^{2}} d u

Take the constant out:

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

= 2 n^{5} \frac{1}{4 a ^{2} t ^{2}} \cdot \int u sec^{2} (u) d u

Apply Integration By Parts

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

= 2 n^{5} \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^{5} \frac{1}{4 a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

Substitute back

u = 2 a t x

= 2 n^{5} \frac{1}{4 a ^{2} t ^{2}} (2 a t x tan (2 a t x) - (- ln ∣ cos (2 a t x) ∣))

Simplify

2 n^{5} \frac{1}{4 a ^{2} t ^{2}} (2 a t x tan (2 a t x) - (- ln ∣ cos (2 a t x) ∣)) : \frac{n ^{5}}{a t} x tan (2 a t x) + n^{5} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣

= \frac{n ^{5}}{a t} x tan (2 a t x) + n^{5} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣

Add a constant to the solution

= \frac{n ^{5}}{a t} x tan (2 a t x) + n^{5} \frac{1}{2 a ^{2} t ^{2}} ln ∣ cos (2 a t x) ∣ + C

\int \frac{x}{3 - x} d x

\frac{e ^{x}}{3 + e ^{x}}

(p \cdot m - p^{2})^{- 1} \cdot d p = k \cdot d t

\frac{d}{d x} (9 x^{5} cos (x))

\frac{d}{d x} (x + x^{5})

What is the integral of sec^2(2xta)n^52x ?
The integral of sec^2(2xta)n^52x is (n^5)/(at)xtan(2atx)+n^5 1/(2a^2t^2)ln|cos(2atx)|+C