What is the integral of sec^4(x/3) ?

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

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integral of sec^4(x/3)

\int sec^{4} (\frac{x}{3}) d x

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

Solution steps

\int sec^{4} (\frac{x}{3}) d x

Apply u-substitution

: \int sec^{4} (u) \cdot 3 d u

= \int sec^{4} (u) \cdot 3 d u

Take the constant out:

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

= 3 \cdot \int sec^{4} (u) d u

Rewrite using trig identities

= 3 \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

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

= 3 \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

= 3 (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

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

Substitute back

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

Simplify

3 (tan (\frac{x}{3}) + \frac{tan ^{3} ( \frac{x}{3} )}{3}) : 3 tan (\frac{x}{3}) + tan^{3} (\frac{x}{3})

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

Add a constant to the solution

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

y = x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

\int_{- π}^{π} cos^{2} (x) d x

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

y = x^{2} + 2 x + 6

x^{2}, x + 2, 0

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