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

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

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integral of (sin^3(3x))/(cos^4(3x))

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{tan ^{3} ( 3 x )}{cos ( 3 x )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

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

Rewrite using trig identities

= \int (- 1 + sec^{2} (3 x)) tan (3 x) sec (3 x) d x

Apply u-substitution

: \int \frac{- 1 + u ^{2}}{3} d u

= \int \frac{- 1 + u ^{2}}{3} d u

Expand

\frac{- 1 + u ^{2}}{3} : - \frac{1}{3} + \frac{u ^{2}}{3}

= \int - \frac{1}{3} + \frac{u ^{2}}{3} 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}{3} d u + \int \frac{u ^{2}}{3} d u

\int \frac{1}{3} d u = \frac{1}{3} u

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

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

Substitute back

u = sec (3 x)

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

Add a constant to the solution

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

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