What is the integral of sec^2(3x+5) ?

The integral of sec^2(3x+5) is -5/3+1/3 tan(3x+5)+C

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integral of sec^2(3x+5)

\int sec^{2} (3 x + 5) d x

- \frac{5}{3} + \frac{1}{3} tan (3 x + 5) + C

Solution steps

\int sec^{2} (3 x + 5) d x

Rewrite using trig identities

= \int 1 + tan^{2} (3 x + 5) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d x + \int tan^{2} (3 x + 5) d x

\int 1 d x = x

\int tan^{2} (3 x + 5) d x = - x - \frac{5}{3} + \frac{1}{3} tan (3 x + 5)

= x - x - \frac{5}{3} + \frac{1}{3} tan (3 x + 5)

Simplify

= - \frac{5}{3} + \frac{1}{3} tan (3 x + 5)

Add a constant to the solution

= - \frac{5}{3} + \frac{1}{3} tan (3 x + 5) + C

x \to 0 lim (2 + 3 e^{- \frac{3}{x}})

\frac{d}{d x} (2 sin (\frac{1}{2} x))

\int - 2 x d x

y = e^{x^{2} + 9 x + 4}

y = 3 x^{2}, y = 16 - x^{2}

What is the integral of sec^2(3x+5) ?
The integral of sec^2(3x+5) is -5/3+1/3 tan(3x+5)+C