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

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

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

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

- \frac{1}{6} tan^{2} (3 x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Rewrite using trig identities

= - \int \frac{tan ( 3 x )}{cos ^{2} ( 3 x )} d x

Use the basic trigonometric identity:

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

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

Apply u-substitution

: \int \frac{u}{3} d u

= - \int \frac{u}{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 u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - \frac{1}{3} \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (3 x)

= - \frac{1}{3} \cdot \frac{tan ^{2} ( 3 x )}{2}

Simplify

- \frac{1}{3} \cdot \frac{tan ^{2} ( 3 x )}{2} : - \frac{1}{6} tan^{2} (3 x)

= - \frac{1}{6} tan^{2} (3 x)

Add a constant to the solution

= - \frac{1}{6} tan^{2} (3 x) + C

sin (2 x)

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

x \to - 3 lim (4 x^{2})

\frac{d}{d x} (4 csc (x))

\frac{d}{d x} (\frac{x ^{\frac{3}{2}} + 2}{x})

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