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

The integral of xsec^2(6x) is 1/2 x^2sec^2(6x)-1/36 (18x^2tan^2(6x)-6xtan(6x)-ln|cos(6x)|+18x^2)+C

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integral of xsec^2(6x)

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

\frac{1}{2} x^{2} sec^{2} (6 x) - \frac{1}{36} (18 x^{2} tan^{2} (6 x) - 6 x tan (6 x) - ln ∣ cos (6 x) ∣ + 18 x^{2}) + C

Solution steps

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

Rewrite using trig identities

= \int x (1 + tan^{2} (6 x)) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} sec^{2} (6 x) - \int 6 x^{2} sec^{2} (6 x) tan (6 x) d x

= \frac{1}{2} x^{2} sec^{2} (6 x) - \int 6 x^{2} sec^{2} (6 x) tan (6 x) d x

\int 6 x^{2} sec^{2} (6 x) tan (6 x) d x = \frac{1}{36} (18 x^{2} tan^{2} (6 x) - 6 x tan (6 x) - ln ∣ cos (6 x) ∣ + 18 x^{2})

= \frac{1}{2} x^{2} sec^{2} (6 x) - \frac{1}{36} (18 x^{2} tan^{2} (6 x) - 6 x tan (6 x) - ln ∣ cos (6 x) ∣ + 18 x^{2})

Add a constant to the solution

= \frac{1}{2} x^{2} sec^{2} (6 x) - \frac{1}{36} (18 x^{2} tan^{2} (6 x) - 6 x tan (6 x) - ln ∣ cos (6 x) ∣ + 18 x^{2}) + C

\int (6 x^{5} - \frac{5}{x ^{5}}) d x

\frac{d y}{d x} = \frac{e ^{x}}{9 + e ^{x}}

(1 + x) y^{^{'}} = (x + 2) (y - 1)

x \to \frac{x}{6} lim (cot (x))

y = x^{2} + 5 x

What is the integral of xsec^2(6x) ?
The integral of xsec^2(6x) is 1/2 x^2sec^2(6x)-1/36 (18x^2tan^2(6x)-6xtan(6x)-ln|cos(6x)|+18x^2)+C