What is the integral of sec^6(3x) ?

The integral of sec^6(3x) is 1/3 (tan(3x)+(2tan^3(3x))/3+(tan^5(3x))/5)+C

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integral of sec^6(3x)

\int sec^{6} (3 x) d x

\frac{1}{3} (tan (3 x) + \frac{2 tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5}) + C

Solution steps

\int sec^{6} (3 x) d x

Apply u-substitution

: \int sec^{6} (u) \frac{1}{3} d u

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

(1 + v^{2})^{2} : 1 + 2 v^{2} + v^{4}

= \frac{1}{3} \cdot \int 1 + 2 v^{2} + v^{4} 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 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

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

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (v + \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= \frac{1}{3} (tan (3 x) + \frac{2 tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5})

Add a constant to the solution

= \frac{1}{3} (tan (3 x) + \frac{2 tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5}) + C

y^{^{'}} - \frac{4 y}{t} = 5

\frac{6 x ^{2} + 1}{x ^{2} + 9}

\frac{d}{d x} (\frac{- cos ( x )}{x})

\int \frac{5 x e ^{2 x}}{( 1 + 2 x ) ^{2}} d x

\int sin^{5} (3 x) cos^{3} (3 x) d x

What is the integral of sec^6(3x) ?
The integral of sec^6(3x) is 1/3 (tan(3x)+(2tan^3(3x))/3+(tan^5(3x))/5)+C