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

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

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

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

- x + tan (2 x) - cos (2 x) - sin (2 x) tan (2 x) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{( sin ( 2 x ) - 1 ) ^{2}}{cos ^{2} ( 2 x )} d x

Expand

\frac{( sin ( 2 x ) - 1 ) ^{2}}{cos ^{2} ( 2 x )} : \frac{sin ^{2} ( 2 x )}{cos ^{2} ( 2 x )} - \frac{2 sin ( 2 x )}{cos ^{2} ( 2 x )} + \frac{1}{cos ^{2} ( 2 x )}

= \int \frac{sin ^{2} ( 2 x )}{cos ^{2} ( 2 x )} - \frac{2 sin ( 2 x )}{cos ^{2} ( 2 x )} + \frac{1}{cos ^{2} ( 2 x )} 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 \frac{sin ^{2} ( 2 x )}{cos ^{2} ( 2 x )} d x - \int \frac{2 sin ( 2 x )}{cos ^{2} ( 2 x )} d x + \int \frac{1}{cos ^{2} ( 2 x )} d x

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

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

\int \frac{1}{cos ^{2} ( 2 x )} d x = \frac{1}{2} tan (2 x)

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

Simplify

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

= - x + tan (2 x) - cos (2 x) - sin (2 x) tan (2 x)

Add a constant to the solution

= - x + tan (2 x) - cos (2 x) - sin (2 x) tan (2 x) + C

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\int 5^{2 x + 1} d x

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