What is the integral of sin^3(3x)cos^{-2}(3x) ?

The integral of sin^3(3x)cos^{-2}(3x) is 1/3 (sec(3x)+cos(3x))+C

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integral of sin^3(3x)cos^{-2}(3x)

\int sin^{3} (3 x) cos^{- 2} (3 x) d x

\frac{1}{3} (sec (3 x) + cos (3 x)) + C

Solution steps

\int sin^{3} (3 x) cos^{- 2} (3 x) d x

Apply u-substitution

: \int \frac{sin ^{3} ( u )}{3 cos ^{2} ( u )} d u

= \int \frac{sin ^{3} ( u )}{3 cos ^{2} ( u )} 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 \frac{sin ^{3} ( u )}{cos ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{( 1 - cos ^{2} ( u ) ) sin ( u )}{cos ^{2} ( u )} d u

Apply u-substitution

: \int - \frac{1 - v ^{2}}{v ^{2}} d v

= \frac{1}{3} \cdot \int - \frac{1 - v ^{2}}{v ^{2}} d v

Expand

- \frac{1 - v ^{2}}{v ^{2}} : - \frac{1}{v ^{2}} + 1

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

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

\int 1 d v = v

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

Substitute back

= \frac{1}{3} (- (- \frac{1}{cos ( 3 x )}) + cos (3 x))

Simplify

\frac{1}{3} (- (- \frac{1}{cos ( 3 x )}) + cos (3 x)) : \frac{1}{3} (sec (3 x) + cos (3 x))

= \frac{1}{3} (sec (3 x) + cos (3 x))

Add a constant to the solution

= \frac{1}{3} (sec (3 x) + cos (3 x)) + C

\frac{d}{d x} ((\frac{x}{7} + \frac{7}{x})^{7})

\int \frac{1}{3 x ^{2} - 1} d x

i = 2 \sum \infty \frac{5}{i ^{2} + i - 2}

x \to 5 lim (2 x + 3)

3 \cdot cos^{2} (t)

What is the integral of sin^3(3x)cos^{-2}(3x) ?
The integral of sin^3(3x)cos^{-2}(3x) is 1/3 (sec(3x)+cos(3x))+C