What is the integral of tan(2t) ?

The integral of tan(2t) is -1/2 ln|cos(2t)|+C

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integral of tan(2t)

\int tan (2 t) d t

- \frac{1}{2} ln ∣ cos (2 t) ∣ + C

Solution steps

\int tan (2 t) d t

Rewrite using trig identities

= \int \frac{sin ( 2 t )}{cos ( 2 t )} d t

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = cos (2 t)

= - \frac{1}{2} ln ∣ cos (2 t) ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ cos (2 t) ∣ + C

y^{^{''}} + 16 y = t e^{4 t}

\frac{y}{x}

(\frac{100}{lo g _{10} ( 2 x + 1 )})^{^{'}}

f (x) = x^{3} e^{2 x}

x \to 1 lim (x^{\frac{3}{1 - x}})