What is the integral of tan(3x)sec^2(3x) ?

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

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integral of tan(3x)sec^2(3x)

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

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

Solution steps

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

\frac{d}{d x} (\frac{40 x ^{5}}{e ^{x}})

x \to 0 + lim (1 + csc (x))

\frac{d}{d x} (3 sin (x) - 4 cos (x))

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

\frac{e ^{(- s)}}{s ^{2} - 4}

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