What is the integral of (cos(9x))/(sec(9x)-tan(9x)) ?

The integral of (cos(9x))/(sec(9x)-tan(9x)) is -1/9 cos(9x)+x+C

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integral of (cos(9x))/(sec(9x)-tan(9x))

\int \frac{cos ( 9 x )}{sec ( 9 x ) - tan ( 9 x )} d x

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

Solution steps

\int \frac{cos ( 9 x )}{sec ( 9 x ) - tan ( 9 x )} d x

Rewrite using trig identities

= \int sin (9 x) + 1 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 sin (9 x) d x + \int 1 d x

\int sin (9 x) d x = - \frac{1}{9} cos (9 x)

\int 1 d x = x

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

Add a constant to the solution

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

\int \frac{tan ( y )}{sin ^{2} ( y )} d y

\int \frac{x ^{3} - 1}{( 4 x ^{2} - 1 ) ^{2}} d x

\int e^{- x} sin (2 x + 1) d x

\int \frac{1}{2 - 3 e ^{- x}} d x

\frac{1}{x ^{2} - 1}

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