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

The integral from 0 to pi/3 of yztan(x) is yzln(2)

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integral from 0 to pi/3 of yztan(x)

\int_{0}^{\frac{π}{3}} yz tan (x) d x

yz ln (2)

Solution steps

\int_{0}^{\frac{π}{3}} yz tan (x) d x

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= yz \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

= yz (- \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

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

Simplify

= yz [ln ∣ u ∣]_{\frac{1}{2}}^{1}

Compute the boundaries:

ln (2)

= yz ln (2)

\int_{2}^{3} \frac{1}{x ^{2} + 3 x + 2} d x

\int_{y^{2}}^{y} \frac{y}{x} d x

\int_{0}^{8} 2 x d x

\int_{- 1}^{1} (x^{2} + x) d x

\int_{0}^{2} x^{2} + 5 d x

What is the integral from 0 to pi/3 of yztan(x) ?
The integral from 0 to pi/3 of yztan(x) is yzln(2)