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

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

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integral of 2sec^4(1/2 x)

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

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

Solution steps

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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