What is the integral of-2tan(t) ?

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

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integral of-2tan(t)

\int - 2 tan (t) d t

2 ln ∣ cos (t) ∣ + C

Solution steps

\int - 2 tan (t) d t

Take the constant out:

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

= - 2 \cdot \int tan (t) d t

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 (- ln ∣ u ∣)

Substitute back

u = cos (t)

= - 2 (- ln ∣ cos (t) ∣)

Simplify

= 2 ln ∣ cos (t) ∣

Add a constant to the solution

= 2 ln ∣ cos (t) ∣ + C

\int tan^{5} (θ) sec^{3} (θ) d θ

\frac{\partial}{\partial x} (x^{3} z + z^{3} x - 2 yz)

f (x) = x^{2} x^{3} + 1, at x = 1

x \to 1 lim (x^{2} + 3 x - 2)

\int_{0}^{1} x \cdot e^{- x} d x