What is the integral of tan^2(2y) ?

The integral of tan^2(2y) is -y+1/2 tan(2y)+C

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integral of tan^2(2y)

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

- y + \frac{1}{2} tan (2 y) + C

Solution steps

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

Rewrite using trig identities

= \int - 1 + sec^{2} (2 y) d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d y + \int sec^{2} (2 y) d y

\int 1 d y = y

\int sec^{2} (2 y) d y = \frac{1}{2} tan (2 y)

= - y + \frac{1}{2} tan (2 y)

Add a constant to the solution

= - y + \frac{1}{2} tan (2 y) + C

\frac{d}{d x} ((x + 5)^{2})

x \to \infty lim (3 x - 1)

t \to 0 lim (\frac{sin ( 3 t )}{t} + 2)

\int 4 x e^{5 x} d x

(0, - 5), (- 2, 1)

What is the integral of tan^2(2y) ?
The integral of tan^2(2y) is -y+1/2 tan(2y)+C