What is the integral of sec^2(3xta)n^53x ?

The integral of sec^2(3xta)n^53x is (n^5)/(at)xtan(3atx)+n^5 1/(3a^2t^2)ln|cos(3atx)|+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(3xta)n^53x

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u sec ^{2} ( u )}{9 a ^{2} t ^{2}} d u

= 3 n^{5} \cdot \int \frac{u sec ^{2} ( u )}{9 a ^{2} t ^{2}} d u

Take the constant out:

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

= 3 n^{5} \frac{1}{9 a ^{2} t ^{2}} \cdot \int u sec^{2} (u) d u

Apply Integration By Parts

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

= 3 n^{5} \frac{1}{9 a ^{2} t ^{2}} (u tan (u) - \int tan (u) d u)

\int tan (u) d u = - ln ∣ cos (u) ∣

= 3 n^{5} \frac{1}{9 a ^{2} t ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

Substitute back

u = 3 a t x

= 3 n^{5} \frac{1}{9 a ^{2} t ^{2}} (3 a t x tan (3 a t x) - (- ln ∣ cos (3 a t x) ∣))

Simplify

3 n^{5} \frac{1}{9 a ^{2} t ^{2}} (3 a t x tan (3 a t x) - (- ln ∣ cos (3 a t x) ∣)) : \frac{n ^{5}}{a t} x tan (3 a t x) + n^{5} \frac{1}{3 a ^{2} t ^{2}} ln ∣ cos (3 a t x) ∣

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

Add a constant to the solution

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

y ln (x) \frac{d y}{d x} = (\frac{( y + 1 )}{x})^{2}

(x^{2} - 7)^{2}

\frac{\partial}{\partial x} (ln (cos^{2} (x) + 1))

f (x) = \frac{x}{( 5 x - 4 ) ^{8}}, at x = 1

(- 2.2) (5.3)

What is the integral of sec^2(3xta)n^53x ?
The integral of sec^2(3xta)n^53x is (n^5)/(at)xtan(3atx)+n^5 1/(3a^2t^2)ln|cos(3atx)|+C