What is the integral of 4sec^2(8x+3) ?

The integral of 4sec^2(8x+3) is 4(-3/8+1/8 tan(8x+3))+C

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integral of 4sec^2(8x+3)

\int 4 sec^{2} (8 x + 3) d x

4 (- \frac{3}{8} + \frac{1}{8} tan (8 x + 3)) + C

Solution steps

\int 4 sec^{2} (8 x + 3) d x

Take the constant out:

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

= 4 \cdot \int sec^{2} (8 x + 3) d x

Rewrite using trig identities

= 4 \cdot \int 1 + tan^{2} (8 x + 3) 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

= 4 (\int 1 d x + \int tan^{2} (8 x + 3) d x)

\int 1 d x = x

\int tan^{2} (8 x + 3) d x = - x - \frac{3}{8} + \frac{1}{8} tan (8 x + 3)

= 4 (x - x - \frac{3}{8} + \frac{1}{8} tan (8 x + 3))

Simplify

= 4 (- \frac{3}{8} + \frac{1}{8} tan (8 x + 3))

Add a constant to the solution

= 4 (- \frac{3}{8} + \frac{1}{8} tan (8 x + 3)) + C

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What is the integral of 4sec^2(8x+3) ?
The integral of 4sec^2(8x+3) is 4(-3/8+1/8 tan(8x+3))+C