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

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

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

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

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

Solution steps

\int 2 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

= 2 \cdot \int sec (2 x) tan (2 x) d x

Rewrite using trig identities

= 2 \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)

= 2 (- \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)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + (\frac{1}{x}) y = 8 x^{2}

\int_{2}^{4} x^{2} y + y^{2} x + x + y d x

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

y^{^{''}} - 5 y^{^{'}} = 0, y (0) = 6, y^{^{'}} (0) = - 9

\frac{\partial}{\partial x} (ln (x^{4} - y^{4}))

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