What is the integral of sec^4(xta)n^5x ?

The integral of sec^4(xta)n^5x is

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integral of sec^4(xta)n^5x

\int sec^{4} (x t a) n^{5} x d x

n^{5} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x)) + C

Solution steps

\int sec^{4} (x t a) n^{5} x d x

Take the constant out:

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

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

Apply u-substitution

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

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

Apply Integration By Parts

: u (tan (u) + \frac{1}{3} tan^{3} (u)) - \int tan (u) + \frac{1}{3} tan^{3} (u) d u

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

\int tan (u) + \frac{1}{3} tan^{3} (u) d u = - ln ∣ cos (u) ∣ + \frac{1}{3} (- ln ∣ sec (u) ∣ + \frac{sec ^{2} ( u )}{2})

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

Substitute back

u = a t x

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

Simplify

n^{5} \frac{1}{a ^{2} t ^{2}} (a t x (tan (a t x) + \frac{1}{3} tan^{3} (a t x)) - (- ln ∣ cos (a t x) ∣ + \frac{1}{3} (- ln ∣ sec (a t x) ∣ + \frac{sec ^{2} ( a t x )}{2}))) : n^{5} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x))

= n^{5} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x))

Add a constant to the solution

= n^{5} \frac{1}{a ^{2} t ^{2}} (a t x tan (a t x) + \frac{1}{3} a t x tan^{3} (a t x) + ln ∣ cos (a t x) ∣ + \frac{1}{3} ln ∣ sec (a t x) ∣ - \frac{1}{6} sec^{2} (a t x)) + C

n = 1 \sum \infty \frac{4}{n ( n + 7 )}

n = 2 \sum \infty \frac{n}{n + 2}

f (x) = \frac{x ^{2} + 3 x - 2}{x ^{2} - 5 x + 6}

f (x) = e^{x} - x^{3}

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (1) = 0, y^{^{'}} (1) = - 1