What is the integral of sec^2(5xta)n^35x ?

The integral of sec^2(5xta)n^35x is (n^3)/(at)xtan(5atx)+n^3 1/(5a^2t^2)ln|cos(5atx)|+C

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integral of sec^2(5xta)n^35x

\int sec^{2} (5 x t a) n^{3} 5 x d x

\frac{n ^{3}}{a t} x tan (5 a t x) + n^{3} \frac{1}{5 a ^{2} t ^{2}} ln ∣ cos (5 a t x) ∣ + C

Solution steps

\int sec^{2} (5 x t a) n^{3} \cdot 5 x d x

Take the constant out:

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

= 5 n^{3} \cdot \int sec^{2} (5 a t x) x d x

Apply u-substitution

: \int \frac{u sec ^{2} ( u )}{25 a ^{2} t ^{2}} d u

= 5 n^{3} \cdot \int \frac{u sec ^{2} ( u )}{25 a ^{2} t ^{2}} d u

Take the constant out:

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

= 5 n^{3} \frac{1}{25 a ^{2} t ^{2}} \cdot \int u sec^{2} (u) d u

Apply Integration By Parts

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

= 5 n^{3} \frac{1}{25 a ^{2} t ^{2}} (u tan (u) - \int tan (u) d u)

\int tan (u) d u = - ln ∣ cos (u) ∣

= 5 n^{3} \frac{1}{25 a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

Substitute back

u = 5 a t x

= 5 n^{3} \frac{1}{25 a ^{2} t ^{2}} (5 a t x tan (5 a t x) - (- ln ∣ cos (5 a t x) ∣))

Simplify

5 n^{3} \frac{1}{25 a ^{2} t ^{2}} (5 a t x tan (5 a t x) - (- ln ∣ cos (5 a t x) ∣)) : \frac{n ^{3}}{a t} x tan (5 a t x) + n^{3} \frac{1}{5 a ^{2} t ^{2}} ln ∣ cos (5 a t x) ∣

= \frac{n ^{3}}{a t} x tan (5 a t x) + n^{3} \frac{1}{5 a ^{2} t ^{2}} ln ∣ cos (5 a t x) ∣

Add a constant to the solution

= \frac{n ^{3}}{a t} x tan (5 a t x) + n^{3} \frac{1}{5 a ^{2} t ^{2}} ln ∣ cos (5 a t x) ∣ + C

\int cos (x) cos (5 x) d x

f (x) = x^{2}, (1, 5), 5

\frac{d}{d x} (7^{x} ln^{2} (7))

arccsc (2 x^{4} - 5)

\frac{d}{d x} (\frac{2}{3} x^{3} - 8 x - 15)

What is the integral of sec^2(5xta)n^35x ?
The integral of sec^2(5xta)n^35x is (n^3)/(at)xtan(5atx)+n^3 1/(5a^2t^2)ln|cos(5atx)|+C