What is the integral from 0 to pi/(26) of tan(13x) ?

The integral from 0 to pi/(26) of tan(13x) is diverges

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integral from 0 to pi/(26) of tan(13x)

\int_{0}^{\frac{π}{26}} tan (13 x) d x

diverges

Solution steps

\int_{0}^{\frac{π}{26}} tan (13 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{26}} \frac{sin ( 13 x )}{cos ( 13 x )} d x

Apply u-substitution

: \int_{1}^{0} - \frac{1}{13 u} d u

= \int_{1}^{0} - \frac{1}{13 u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{1}{13 u} d u

Take the constant out:

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

= - (- \frac{1}{13} \cdot \int_{0}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{13} [ln ∣ u ∣]_{0}^{1})

Simplify

= \frac{1}{13} [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

5 x^{2} y^{^{'}} = y^{^{'}} + 2 x e^{- y}

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 7, y (0) = 7, y^{^{'}} (0) = 7

\frac{\partial}{\partial x} (2 y (y^{2} + e^{2 x}))

\int_{0}^{N} e^{(- s t)} sin (2 t) d t

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

What is the integral from 0 to pi/(26) of tan(13x) ?
The integral from 0 to pi/(26) of tan(13x) is diverges