What is the integral of-sec^2((9x)/8) ?

The integral of-sec^2((9x)/8) is -8/9 tan((9x)/8)+C

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integral of-sec^2((9x)/8)

\int - sec^{2} (\frac{9 x}{8}) d x

- \frac{8}{9} tan (\frac{9 x}{8}) + C

Solution steps

\int - sec^{2} (\frac{9 x}{8}) d x

Take the constant out:

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

= - \int sec^{2} (\frac{9 x}{8}) d x

Rewrite using trig identities

= - \int 1 + tan^{2} (\frac{9 x}{8}) 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} (\frac{9 x}{8}) d x)

\int 1 d x = x

\int tan^{2} (\frac{9 x}{8}) d x = - x + \frac{8}{9} tan (\frac{9 x}{8})

= - (x - x + \frac{8}{9} tan (\frac{9 x}{8}))

Simplify

= - \frac{8}{9} tan (\frac{9 x}{8})

Add a constant to the solution

= - \frac{8}{9} tan (\frac{9 x}{8}) + C

What is the integral of-sec^2((9x)/8) ?
The integral of-sec^2((9x)/8) is -8/9 tan((9x)/8)+C