What is the integral of sec^4(x/6) ?

The integral of sec^4(x/6) is 6(tan(x/6)+(tan^3(x/6))/3)+C

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integral of sec^4(x/6)

\int sec^{4} (\frac{x}{6}) d x

6 (tan (\frac{x}{6}) + \frac{tan ^{3} ( \frac{x}{6} )}{3}) + C

Solution steps

\int sec^{4} (\frac{x}{6}) d x

Apply u-substitution

: \int sec^{4} (u) \cdot 6 d u

= \int sec^{4} (u) \cdot 6 d u

Take the constant out:

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

= 6 \cdot \int sec^{4} (u) d u

Rewrite using trig identities

= 6 \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

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

= 6 \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

= 6 (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= 6 (v + \frac{v ^{3}}{3})

Substitute back

= 6 (tan (\frac{x}{6}) + \frac{tan ^{3} ( \frac{x}{6} )}{3})

Add a constant to the solution

= 6 (tan (\frac{x}{6}) + \frac{tan ^{3} ( \frac{x}{6} )}{3}) + C

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What is the integral of sec^4(x/6) ?
The integral of sec^4(x/6) is 6(tan(x/6)+(tan^3(x/6))/3)+C