What is the integral from 1 to 2 of tan(xy) ?

The integral from 1 to 2 of tan(xy) is -(ln|cos(2x)|-ln|cos(x)|)/x

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integral from 1 to 2 of tan(xy)

\int_{1}^{2} tan (x y) d y

- \frac{ln ∣ cos ( 2 x ) ∣ - ln ∣ cos ( x ) ∣}{x}

Solution steps

\int_{1}^{2} tan (x y) d y

Rewrite using trig identities

= \int_{1}^{2} \frac{sin ( x y )}{cos ( x y )} d y

Apply u-substitution

: \int_{c o s (x)}^{c o s (x \cdot 2)} - \frac{1}{xu} d u

= \int_{c o s (x)}^{c o s (x \cdot 2)} - \frac{1}{xu} d u

Take the constant out:

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

= - \frac{1}{x} \cdot \int_{c o s (x)}^{c o s (x \cdot 2)} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{x} [ln ∣ u ∣]_{c o s (x)}^{c o s (x \cdot 2)}

Compute the boundaries:

ln ∣ cos (x \cdot 2) ∣ - ln ∣ cos (x) ∣

= - \frac{1}{x} (ln ∣ cos (x \cdot 2) ∣ - ln ∣ cos (x) ∣)

Simplify

= - \frac{ln ∣ cos ( 2 x ) ∣ - ln ∣ cos ( x ) ∣}{x}

\frac{d}{d x} (8 x^{4} + 5 x^{3} - x^{2} + 4 x - 1)

- 7 (t - 4)

\frac{d}{d x} (x (2 x + 1)^{5})

\frac{d y}{d x} = 5 y + y^{2}

f (x) = sin (x) + \frac{1}{2} e^{x}

What is the integral from 1 to 2 of tan(xy) ?
The integral from 1 to 2 of tan(xy) is -(ln|cos(2x)|-ln|cos(x)|)/x