What is the integral from 0 to pi/(21) of tan(7x) ?

The integral from 0 to pi/(21) of tan(7x) is 1/7 ln(2)

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integral from 0 to pi/(21) of tan(7x)

\int_{0}^{\frac{π}{21}} tan (7 x) d x

\frac{1}{7} ln (2)

Decimal Notation

0.09902 \dots

Solution steps

\int_{0}^{\frac{π}{21}} tan (7 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{21}} \frac{sin ( 7 x )}{cos ( 7 x )} d x

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{1}{7 u} d u

= \int_{1}^{\frac{1}{2}} - \frac{1}{7 u} d u

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

= - \int_{\frac{1}{2}}^{1} - \frac{1}{7 u} d u

Take the constant out:

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

= - (- \frac{1}{7} \cdot \int_{\frac{1}{2}}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- \frac{1}{7} [ln ∣ u ∣]_{\frac{1}{2}}^{1})

Simplify

= \frac{1}{7} [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

ln (2)

= \frac{1}{7} ln (2)

y^{^{'}} + 2 y = 1

16 y^{^{''}} + 40 y^{^{'}} + 25 y = 0

\int (x + 4) e^{x^{2} + 8 x} d x

(6 x + 1) y^{2} (\frac{d y}{d x}) + 3 x^{2} + 2 y^{3} = 0

y = x^{2} cos (x)

What is the integral from 0 to pi/(21) of tan(7x) ?
The integral from 0 to pi/(21) of tan(7x) is 1/7 ln(2)