What is the integral of sin^3(6x)cos^{-2}(6x) ?

The integral of sin^3(6x)cos^{-2}(6x) is 1/6 (sec(6x)+cos(6x))+C

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integral of sin^3(6x)cos^{-2}(6x)

\int sin^{3} (6 x) cos^{- 2} (6 x) d x

\frac{1}{6} (sec (6 x) + cos (6 x)) + C

Solution steps

\int sin^{3} (6 x) cos^{- 2} (6 x) d x

Apply u-substitution

: \int \frac{sin ^{3} ( u )}{6 cos ^{2} ( u )} d u

= \int \frac{sin ^{3} ( u )}{6 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}{6} \cdot \int \frac{sin ^{3} ( u )}{cos ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{6} \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}{6} \cdot \int - \frac{1 - v ^{2}}{v ^{2}} d v

Expand

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

= \frac{1}{6} \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}{6} (- \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}{6} (- (- \frac{1}{v}) + v)

Substitute back

= \frac{1}{6} (- (- \frac{1}{cos ( 6 x )}) + cos (6 x))

Simplify

\frac{1}{6} (- (- \frac{1}{cos ( 6 x )}) + cos (6 x)) : \frac{1}{6} (sec (6 x) + cos (6 x))

= \frac{1}{6} (sec (6 x) + cos (6 x))

Add a constant to the solution

= \frac{1}{6} (sec (6 x) + cos (6 x)) + C

\int \frac{I n x}{x} d x

y^{^{''}} + 9 y^{^{'}} + 8 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\int_{1}^{\infty} x e^{- 9 x} d x

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

2^{2 - 3 x}

What is the integral of sin^3(6x)cos^{-2}(6x) ?
The integral of sin^3(6x)cos^{-2}(6x) is 1/6 (sec(6x)+cos(6x))+C