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

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

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integral of sec^4(3x)

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

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

Solution steps

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

Apply u-substitution

= \int sec^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

= \frac{1}{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

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

\int 1 d v = v

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

= \frac{1}{3} (v + \frac{v ^{3}}{3})

Substitute back

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

Add a constant to the solution

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

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

\int x^{2} (x - 1)^{2} d x

t a y l or \frac{2}{( 3 x + 1 )}, x = 2

\int - x e^{- l n (x)} d x

t an g e n t f (x) = 2 x^{3} + 3 x^{2}, at x = 1

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