What is the integral from 0 to x of 4tan(t) ?

The integral from 0 to x of 4tan(t) is -4ln|cos(x)|

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integral from 0 to x of 4tan(t)

\int_{0}^{x} 4 tan (t) d t

- 4 ln ∣ cos (x) ∣

Solution steps

\int_{0}^{x} 4 tan (t) d t

Take the constant out:

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

= 4 \cdot \int_{0}^{x} tan (t) d t

Rewrite using trig identities

= 4 \cdot \int_{0}^{x} \frac{sin ( t )}{cos ( t )} d t

Apply u-substitution

: \int_{1}^{c o s (x)} - \frac{1}{u} d u

= 4 \cdot \int_{1}^{c o s (x)} - \frac{1}{u} d u

Take the constant out:

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

= 4 (- \int_{1}^{c o s (x)} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- [ln ∣ u ∣]_{1}^{c o s (x)})

Simplify

= - 4 [ln ∣ u ∣]_{1}^{c o s (x)}

Compute the boundaries:

ln ∣ cos (x) ∣

= - 4 ln ∣ cos (x) ∣

y^{^{''}} + 9 y = e^{3 x}

\frac{\partial}{\partial x} (sin^{2} (x y))

121 y^{^{''}} + 44 y^{^{'}} + 4 y = 0

\frac{1}{s ( s ^{2} + 4 )}

x \to \frac{7}{8} lim ((x)^{4})

What is the integral from 0 to x of 4tan(t) ?
The integral from 0 to x of 4tan(t) is -4ln|cos(x)|