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

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

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

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

n^{2} \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^{2} x d x

Take the constant out:

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

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

\int tan (u) d u = - ln ∣ cos (u) ∣

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

Substitute back

u = a t x

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

Simplify

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

f (x) = x^{3} - \frac{3}{2} x^{2} + x, at x = 1

n = 2 \sum \infty \frac{( - 1 ) ^{n}}{n ln ( n )}

\int 4 + y^{2} d y

\int \frac{14}{( x - 1 ) ^{2} ( x - 2 ) ^{2}} d x

f (x) = arcsin (4 x)

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