What is the integral of sec^2(4xta)n^34x ?

The integral of sec^2(4xta)n^34x is (n^3)/(at)xtan(4atx)+n^3 1/(4a^2t^2)ln|cos(4atx)|+C

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integral of sec^2(4xta)n^34x

\int sec^{2} (4 x t a) n^{3} 4 x d x

\frac{n ^{3}}{a t} x tan (4 a t x) + n^{3} \frac{1}{4 a ^{2} t ^{2}} ln ∣ cos (4 a t x) ∣ + C

Solution steps

\int sec^{2} (4 x t a) n^{3} \cdot 4 x d x

Take the constant out:

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

= 4 n^{3} \cdot \int sec^{2} (4 a t x) x d x

Apply u-substitution

: \int \frac{u sec ^{2} ( u )}{16 a ^{2} t ^{2}} d u

= 4 n^{3} \cdot \int \frac{u sec ^{2} ( u )}{16 a ^{2} t ^{2}} d u

Take the constant out:

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

= 4 n^{3} \frac{1}{16 a ^{2} t ^{2}} \cdot \int u sec^{2} (u) d u

Apply Integration By Parts

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

= 4 n^{3} \frac{1}{16 a ^{2} t ^{2}} (u tan (u) - \int tan (u) d u)

\int tan (u) d u = - ln ∣ cos (u) ∣

= 4 n^{3} \frac{1}{16 a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

Substitute back

u = 4 a t x

= 4 n^{3} \frac{1}{16 a ^{2} t ^{2}} (4 a t x tan (4 a t x) - (- ln ∣ cos (4 a t x) ∣))

Simplify

4 n^{3} \frac{1}{16 a ^{2} t ^{2}} (4 a t x tan (4 a t x) - (- ln ∣ cos (4 a t x) ∣)) : \frac{n ^{3}}{a t} x tan (4 a t x) + n^{3} \frac{1}{4 a ^{2} t ^{2}} ln ∣ cos (4 a t x) ∣

= \frac{n ^{3}}{a t} x tan (4 a t x) + n^{3} \frac{1}{4 a ^{2} t ^{2}} ln ∣ cos (4 a t x) ∣

Add a constant to the solution

= \frac{n ^{3}}{a t} x tan (4 a t x) + n^{3} \frac{1}{4 a ^{2} t ^{2}} ln ∣ cos (4 a t x) ∣ + C

5 e^{2 t}

\frac{d}{d x} (\frac{1}{2 1 + x})

\frac{\partial}{\partial x} (x^{2} - 2 x y + 2 y^{2} + x - 2 y)

\frac{d}{d x} (cos (π \cdot x))

x^{2} \cdot e^{x} + 4 \cdot y \cdot e^{(- y)} \cdot y^{^{'}} = 0

What is the integral of sec^2(4xta)n^34x ?
The integral of sec^2(4xta)n^34x is (n^3)/(at)xtan(4atx)+n^3 1/(4a^2t^2)ln|cos(4atx)|+C