What is the integral of sec^3(x/2) ?

The integral of sec^3(x/2) is sec(x/2)tan(x/2)+ln|tan(x/2)+sec(x/2)|+C

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integral of sec^3(x/2)

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

sec (\frac{x}{2}) tan (\frac{x}{2}) + ln tan (\frac{x}{2}) + sec (\frac{x}{2}) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{cos ^{3} ( \frac{x}{2} )} d x

Apply u-substitution

= \int 2 sec^{3} (u) d u

Take the constant out:

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

= 2 \cdot \int sec^{3} (u) d u

Apply Integral Reduction

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

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = \frac{x}{2}

= 2 (\frac{sec ^{2} ( \frac{x}{2} ) sin ( \frac{x}{2} )}{2} + \frac{1}{2} ln tan (\frac{x}{2}) + sec (\frac{x}{2}))

Simplify

2 (\frac{sec ^{2} ( \frac{x}{2} ) sin ( \frac{x}{2} )}{2} + \frac{1}{2} ln tan (\frac{x}{2}) + sec (\frac{x}{2})) : sec (\frac{x}{2}) tan (\frac{x}{2}) + ln tan (\frac{x}{2}) + sec (\frac{x}{2})

= sec (\frac{x}{2}) tan (\frac{x}{2}) + ln tan (\frac{x}{2}) + sec (\frac{x}{2})

Add a constant to the solution

= sec (\frac{x}{2}) tan (\frac{x}{2}) + ln tan (\frac{x}{2}) + sec (\frac{x}{2}) + C

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