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

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

\int \frac{x ^{2} - 5}{( 1 + x ) ^{3}} d x

\frac{1}{4 x}

\int 6 x + 12 x^{2} d x

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