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

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

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

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

\frac{3}{2} ln (2)

Decimal

1.03972 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{3 π}{4}} \frac{sin ( \frac{x}{3} )}{cos ( \frac{x}{3} )} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

= 3 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

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

= 3 \cdot \frac{1}{2} ln (2)

Simplify

= \frac{3}{2} ln (2)

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\int_{1}^{12} \frac{4}{x ^{2}} ln (x) d x

\int_{- 7}^{8} x d x

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