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

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

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integral of 4sec(2x)tan(2x)

\int 4 sec (2 x) tan (2 x) d x

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

Solution steps

\int 4 sec (2 x) tan (2 x) 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 x) tan (2 x) d x

Rewrite using trig identities

= 4 \cdot \int \frac{sin ( 2 x )}{cos ^{2} ( 2 x )} d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

y = 2 x + 4

\frac{48}{x}

\frac{d}{d x} (1 + 2 x^{\frac{3}{2}})

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

\frac{d}{d x} (lo g_{2} (7 - 8 x))

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