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

The integral of x*sec^2(3x) is 1/2 x^2sec^2(3x)-1/18 (9x^2tan^2(3x)-6xtan(3x)-2ln|cos(3x)|+9x^2)+C

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integral of x*sec^2(3x)

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

\frac{1}{2} x^{2} sec^{2} (3 x) - \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2}) + C

Solution steps

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

Rewrite using trig identities

= \int x (1 + tan^{2} (3 x)) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} sec^{2} (3 x) - \int 3 x^{2} sec^{2} (3 x) tan (3 x) d x

= \frac{1}{2} x^{2} sec^{2} (3 x) - \int 3 x^{2} sec^{2} (3 x) tan (3 x) d x

\int 3 x^{2} sec^{2} (3 x) tan (3 x) d x = \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2})

= \frac{1}{2} x^{2} sec^{2} (3 x) - \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2})

Add a constant to the solution

= \frac{1}{2} x^{2} sec^{2} (3 x) - \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2}) + C

\frac{\partial}{\partial x} (cos (t))

8 (sin (x))^{x}

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

n = 1 \sum \infty 3 (\frac{7}{2})^{n - 1}

x \to 0 + lim (tan (3 x))

What is the integral of x*sec^2(3x) ?
The integral of x*sec^2(3x) is 1/2 x^2sec^2(3x)-1/18 (9x^2tan^2(3x)-6xtan(3x)-2ln|cos(3x)|+9x^2)+C