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

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

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integral of sec^2(6x)

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

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

Solution steps

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

Apply u-substitution

: \int sec^{2} (u) \frac{1}{6} d u

= \int sec^{2} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{6} tan (u)

Substitute back

u = 6 x

= \frac{1}{6} tan (6 x)

Add a constant to the solution

= \frac{1}{6} tan (6 x) + C

\int \frac{t}{e ^{t}} d t

\int \frac{x + 3}{x ^{2} - x + 4} d x

y^{^{''}} + 2 y^{^{'}} + y = 4 e - t

1 y^{2} - 7 y^{4} (y + 9 y^{3})

x \to 36 lim (\frac{x ^{2}}{9} - \frac{4}{x})