What is the integral of sqrt(tan(6x))(sec(6x))^2 ?

The integral of sqrt(tan(6x))(sec(6x))^2 is 1/9 tan^{3/2}(6x)+C

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integral of sqrt(tan(6x))(sec(6x))^2

\int tan (6 x) (sec (6 x))^{2} d x

\frac{1}{9} tan^{\frac{3}{2}} (6 x) + C

Solution steps

\int tan (6 x) (sec (6 x))^{2} d x

Rewrite using trig identities

= \int \frac{sin ( 6 x )}{cos ( 6 x )} sec^{2} (6 x) d x

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= \int sec^{2} (6 x) tan (6 x) d x

Apply u-substitution

: \int \frac{u ^{2}}{3} d u

= \int \frac{u ^{2}}{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^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

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

Substitute back

u = tan (6 x)

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

Simplify

\frac{1}{3} \cdot \frac{( tan ( 6 x ) ) ^{3}}{3} : \frac{1}{9} tan^{\frac{3}{2}} (6 x)

= \frac{1}{9} tan^{\frac{3}{2}} (6 x)

Add a constant to the solution

= \frac{1}{9} tan^{\frac{3}{2}} (6 x) + C

\int 8 sin (x) d x

\frac{d}{d y} (1 - \frac{3}{y} + x)

\frac{d y}{d x} = - y^{2} sin (x)

\int \frac{1}{x ( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (x \cdot ln (x) - 1)

What is the integral of sqrt(tan(6x))(sec(6x))^2 ?
The integral of sqrt(tan(6x))(sec(6x))^2 is 1/9 tan^{3/2}(6x)+C