What is the integral of sec^4(1/2 x) ?

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

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{cos ^{4} ( \frac{x}{2} )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

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

Apply u-substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= 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

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

\int 1 d v = v

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

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

Substitute back

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

Add a constant to the solution

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

\int ln ((x + 1)^{2}) d x

\int x \cdot ln^{2} (x) d x

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

\int \frac{4 y}{- y ^{2} - 8 y - 7} d y

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