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

The integral of sec^4(5x) is 1/5 (tan(5x)+(tan^3(5x))/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sec^4(5x)

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

\frac{1}{5} (tan (5 x) + \frac{tan ^{3} ( 5 x )}{3}) + C

Solution steps

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

Apply u-substitution

: \int sec^{4} (u) \frac{1}{5} d u

= \int sec^{4} (u) \frac{1}{5} d u

Take the constant out:

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

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

Rewrite using trig identities

= \frac{1}{5} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int 1 + v^{2} d v

= \frac{1}{5} \cdot \int 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{5} (v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{5} (tan (5 x) + \frac{tan ^{3} ( 5 x )}{3})

Add a constant to the solution

= \frac{1}{5} (tan (5 x) + \frac{tan ^{3} ( 5 x )}{3}) + C

f (x) = 2^{x}, at x = 0

\int \frac{1}{9 + x ^{2}} d x

y = \frac{- 5}{3 x}

f (x) = 2 x^{2} - 4 x - 5, at x = 2

\int sin (e^{x}) e^{x} d x

What is the integral of sec^4(5x) ?
The integral of sec^4(5x) is 1/5 (tan(5x)+(tan^3(5x))/3)+C