What is the integral of sec^2(4x+9) ?

The integral of sec^2(4x+9) is -9/4+1/4 tan(4x+9)+C

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integral of sec^2(4x+9)

\int sec^{2} (4 x + 9) d x

- \frac{9}{4} + \frac{1}{4} tan (4 x + 9) + C

Solution steps

\int sec^{2} (4 x + 9) d x

Rewrite using trig identities

= \int 1 + tan^{2} (4 x + 9) 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} (4 x + 9) d x

\int 1 d x = x

\int tan^{2} (4 x + 9) d x = - x - \frac{9}{4} + \frac{1}{4} tan (4 x + 9)

= x - x - \frac{9}{4} + \frac{1}{4} tan (4 x + 9)

Simplify

= - \frac{9}{4} + \frac{1}{4} tan (4 x + 9)

Add a constant to the solution

= - \frac{9}{4} + \frac{1}{4} tan (4 x + 9) + C

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What is the integral of sec^2(4x+9) ?
The integral of sec^2(4x+9) is -9/4+1/4 tan(4x+9)+C