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

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

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

\int_{0}^{\frac{π}{3}} y ln (z) tan (x) d x

y ln (z) ln (2)

Solution steps

\int_{0}^{\frac{π}{3}} y ln (z) tan (x) d x

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

= ln (z) y \cdot \int_{1}^{\frac{1}{2}} - \frac{1}{u} d u

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

= ln (z) y (- \int_{\frac{1}{2}}^{1} - \frac{1}{u} d u)

Take the constant out:

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

= ln (z) y (- (- \int_{\frac{1}{2}}^{1} \frac{1}{u} d u))

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

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

Compute the boundaries:

ln (2)

= y ln (z) ln (2)

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