What is the integral of sec^4(2y) ?

The integral of sec^4(2y) is 1/2 (tan(2y)+(tan^3(2y))/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(2y)

\int sec^{4} (2 y) d y

\frac{1}{2} (tan (2 y) + \frac{tan ^{3} ( 2 y )}{3}) + C

Solution steps

\int sec^{4} (2 y) d y

Apply u-substitution

: \int sec^{4} (u) \frac{1}{2} d u

= \int sec^{4} (u) \frac{1}{2} d u

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= \frac{1}{2} \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}{2} (\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}{2} (v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{2} (tan (2 y) + \frac{tan ^{3} ( 2 y )}{3})

Add a constant to the solution

= \frac{1}{2} (tan (2 y) + \frac{tan ^{3} ( 2 y )}{3}) + C

\int ln^{2} (\frac{4 a}{x}) d x

\int x^{3} (2 + x^{2})^{\frac{5}{2}} d x

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

\int \frac{1}{3 x ^{2} - 5 x + 1} d x

\int \frac{3}{x ^{2} 4 - x ^{2}} d x

What is the integral of sec^4(2y) ?
The integral of sec^4(2y) is 1/2 (tan(2y)+(tan^3(2y))/3)+C