What is the integral of sec^2(4xta)n^54x ?

The integral of sec^2(4xta)n^54x is (n^5)/(at)xtan(4atx)+n^5 1/(4a^2t^2)ln|cos(4atx)|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^2(4xta)n^54x

\int sec^{2} (4 x t a) n^{5} 4 x d x

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

Solution steps

\int sec^{2} (4 x t a) n^{5} \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^{5} \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^{5} \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^{5} \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^{5} \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^{5} \frac{1}{16 a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

Substitute back

u = 4 a t x

= 4 n^{5} \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^{5} \frac{1}{16 a ^{2} t ^{2}} (4 a t x tan (4 a t x) - (- ln ∣ cos (4 a t x) ∣)) : \frac{n ^{5}}{a t} x tan (4 a t x) + n^{5} \frac{1}{4 a ^{2} t ^{2}} ln ∣ cos (4 a t x) ∣

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

Add a constant to the solution

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

\int \frac{x ^{3} + 2 x ^{2}}{x} d x

\frac{d}{d x} ((1 + 3 x)^{2})

6 y^{^{''}} - 5 y^{^{'}} + y = 0

x \to 0 lim (e^{s i n (x)})

sec^{2} (x), 8 cos (x)

What is the integral of sec^2(4xta)n^54x ?
The integral of sec^2(4xta)n^54x is (n^5)/(at)xtan(4atx)+n^5 1/(4a^2t^2)ln|cos(4atx)|+C