What is the integral of sec^4(9x) ?

The integral of sec^4(9x) is 1/9 (tan(9x)+(tan^3(9x))/3)+C

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integral of sec^4(9x)

\int sec^{4} (9 x) d x

\frac{1}{9} (tan (9 x) + \frac{tan ^{3} ( 9 x )}{3}) + C

Solution steps

\int sec^{4} (9 x) d x

Apply u-substitution

: \int sec^{4} (u) \frac{1}{9} d u

= \int sec^{4} (u) \frac{1}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int sec^{4} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

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

= \frac{1}{9} \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}{9} (\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}{9} (v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{9} (tan (9 x) + \frac{tan ^{3} ( 9 x )}{3})

Add a constant to the solution

= \frac{1}{9} (tan (9 x) + \frac{tan ^{3} ( 9 x )}{3}) + C

y^{^{'}} = cos (y - x + 3)

\int 7 tan (7 x) d x

\int 3 tan (x) d x

\frac{d}{d t} (\frac{t}{t ^{2} + 16})

- x^{3} + x^{2} + 16 x, 4 x

What is the integral of sec^4(9x) ?
The integral of sec^4(9x) is 1/9 (tan(9x)+(tan^3(9x))/3)+C