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

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

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

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

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

Solution steps

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

Factor

\frac{sin ( 3 x )}{( cos ( 3 x ) ) ^{3}}

= \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

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