What is the integral of sec(2x)tan(2x)-4/(x^2) ?

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

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integral of sec(2x)tan(2x)-4/(x^2)

\int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} d x

\frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + C

Solution steps

\int sec (2 x) tan (2 x) - \frac{4}{x ^{2}} 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 sec (2 x) tan (2 x) d x - \int \frac{4}{x ^{2}} d x

\int sec (2 x) tan (2 x) d x = \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x)

\int \frac{4}{x ^{2}} d x = - \frac{4}{x}

= \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) - (- \frac{4}{x})

Simplify

= \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x}

Add a constant to the solution

= \frac{1}{2} cos (2 x) + \frac{1}{2} sin (2 x) tan (2 x) + \frac{4}{x} + C

\int (cos (y)) d y

(- 3, 5), (7, - 9)

\frac{8}{1 - x}

\int \frac{x ^{5} - x ^{3} + 2 x}{x ^{4}} d x

\int e^{16 x} d x

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