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

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

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integral of 2sec^4(x/2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Substitute back

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

Simplify

2 \cdot 2 (tan (\frac{x}{2}) + \frac{tan ^{3} ( \frac{x}{2} )}{3}) : 4 (tan (\frac{x}{2}) + \frac{1}{3} tan^{3} (\frac{x}{2}))

= 4 (tan (\frac{x}{2}) + \frac{1}{3} tan^{3} (\frac{x}{2}))

Add a constant to the solution

= 4 (tan (\frac{x}{2}) + \frac{1}{3} tan^{3} (\frac{x}{2})) + C

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

\int - e^{- t} + 1 d t

f (x) = x^{4} - 10 x^{2} + 9, at x = 1

h (g (x) k (x))

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

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