What is the integral of tan^3(9x)sec(9x) ?

The integral of tan^3(9x)sec(9x) is -1/9 sec(9x)+(sec^3(9x))/(27)+C

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integral of tan^3(9x)sec(9x)

\int tan^{3} (9 x) sec (9 x) d x

- \frac{1}{9} sec (9 x) + \frac{sec ^{3} ( 9 x )}{27} + C

Solution steps

\int tan^{3} (9 x) sec (9 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (9 x)) tan (9 x) sec (9 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{9} d u

= \int \frac{- 1 + u ^{2}}{9} d u

Expand

\frac{- 1 + u ^{2}}{9} : - \frac{1}{9} + \frac{u ^{2}}{9}

= \int - \frac{1}{9} + \frac{u ^{2}}{9} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{9} d u + \int \frac{u ^{2}}{9} d u

\int \frac{1}{9} d u = \frac{1}{9} u

\int \frac{u ^{2}}{9} d u = \frac{u ^{3}}{27}

= - \frac{1}{9} u + \frac{u ^{3}}{27}

Substitute back

u = sec (9 x)

= - \frac{1}{9} sec (9 x) + \frac{sec ^{3} ( 9 x )}{27}

Add a constant to the solution

= - \frac{1}{9} sec (9 x) + \frac{sec ^{3} ( 9 x )}{27} + C

y = cos (a^{9} + x^{9})

y = 6 x^{2} - 3 x, at 1, 5

(8.1) (4.1)

\frac{\partial}{\partial y} (x^{4} + 2 x^{2} y)

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

What is the integral of tan^3(9x)sec(9x) ?
The integral of tan^3(9x)sec(9x) is -1/9 sec(9x)+(sec^3(9x))/(27)+C