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

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

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

\int_{0}^{π} 5 tan (\frac{x}{3}) d x

15 ln (2)

Decimal

10.39720 \dots

Solution steps

\int_{0}^{π} 5 tan (\frac{x}{3}) d x

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int_{1}^{\frac{1}{2}} - \frac{3}{u} d u

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

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

= 15 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

ln (2)

= 15 ln (2)

\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{1}{sin ( x )} d x

\int_{0}^{x} \frac{1}{3} d x

\int_{0}^{0.5} 2 x + 3 - x e^{3 x} d x

\int_{0}^{1} \frac{9}{1 + p ^{2}} d p

\int_{- 4}^{5} (25 - x^{2}) - (5 - x) d x

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