What is the integral from 0 to (5pi)/4 of 3tan(x/5) ?

The integral from 0 to (5pi)/4 of 3tan(x/5) is 15/2 ln(2)

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integral from 0 to (5pi)/4 of 3tan(x/5)

\int_{0}^{\frac{5 π}{4}} 3 tan (\frac{x}{5}) d x

\frac{15}{2} ln (2)

Decimal Notation

5.19860 \dots

Solution steps

\int_{0}^{\frac{5 π}{4}} 3 tan (\frac{x}{5}) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{\frac{5 π}{4}} tan (\frac{x}{5}) d x

Rewrite using trig identities

= 3 \cdot \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

= 3 \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

= 3 (- \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

= 3 (- (- 5 \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1}))

Simplify

3 (- (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1})) : 15 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 15 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= 15 \cdot \frac{1}{2} ln (2)

Simplify

= \frac{15}{2} ln (2)

n = 0 \sum \infty \frac{1}{n ^{2} + 1}

\frac{d y}{d x}, y = 7^{x}

n = 0 \sum \infty \frac{1 ^{n}}{4 ^{n}}

(3 x - 2)^{^{'}}

x \to 3 + lim (\frac{5 x}{x - 3})

What is the integral from 0 to (5pi)/4 of 3tan(x/5) ?
The integral from 0 to (5pi)/4 of 3tan(x/5) is 15/2 ln(2)