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

The integral from 0 to (5pi)/4 of 5tan(x/5) is 25/2 ln(2)

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

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

\frac{25}{2} ln (2)

Decimal

8.66433 \dots

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

= 5 \cdot \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

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

Simplify

5 (- (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : 25 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 25 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= 25 \cdot \frac{1}{2} ln (2)

Simplify

= \frac{25}{2} ln (2)

What is the integral from 0 to (5pi)/4 of 5tan(x/5) ?
The integral from 0 to (5pi)/4 of 5tan(x/5) is 25/2 ln(2)