What is the integral from 0 to (5pi)/4 of tan(x/5) ?

The integral from 0 to (5pi)/4 of tan(x/5) is 5/2 ln(2)

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integral from 0 to (5pi)/4 of tan(x/5)

\int_{0}^{\frac{5 π}{4}} tan (\frac{x}{5}) d x

\frac{5}{2} ln (2)

Solution steps

\int_{0}^{\frac{5 π}{4}} tan (\frac{x}{5}) d x

Rewrite using trig identities

= \int_{0}^{\frac{5 π}{4}} \frac{sin ( \frac{x}{5} )}{cos ( \frac{x}{5} )} d x

Apply u-substitution

: \int_{1}^{\frac{2}{2}} - \frac{5}{u} d u

= \int_{1}^{\frac{2}{2}} - \frac{5}{u} d u

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

= - \int_{\frac{2}{2}}^{1} - \frac{5}{u} d u

Take the constant out:

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

= - (- 5 \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1})

Simplify

- (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1}) : 5 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 5 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= 5 \cdot \frac{1}{2} ln (2)

Simplify

= \frac{5}{2} ln (2)

\int_{5}^{4} sin (x) d x

\int_{0}^{2} x cos (x^{2}) d x

\int_{2}^{4} (- x^{2} + 6 x - 8)^{2} d x

\int_{1}^{2} (x^{2} + x - 1) d x

\int_{2}^{\infty} \frac{1}{x - 1} d x

What is the integral from 0 to (5pi)/4 of tan(x/5) ?
The integral from 0 to (5pi)/4 of tan(x/5) is 5/2 ln(2)