What is the integral of 8tan(x)sin(y)cos(y) ?

The integral of 8tan(x)sin(y)cos(y) is -8cos(y)sin(y)ln|cos(x)|+C

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integral of 8tan(x)sin(y)cos(y)

\int 8 tan (x) sin (y) cos (y) d x

- 8 cos (y) sin (y) ln ∣ cos (x) ∣ + C

Solution steps

\int 8 tan (x) sin (y) cos (y) d x

Take the constant out:

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

= 8 cos (y) sin (y) \cdot \int tan (x) d x

Rewrite using trig identities

= 8 cos (y) sin (y) \cdot \int \frac{sin ( x )}{cos ( x )} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= 8 cos (y) sin (y) \cdot \int - \frac{1}{u} d u

Take the constant out:

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

= 8 cos (y) sin (y) (- \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 cos (y) sin (y) (- ln ∣ u ∣)

Substitute back

u = cos (x)

= 8 cos (y) sin (y) (- ln ∣ cos (x) ∣)

Simplify

= - 8 cos (y) sin (y) ln ∣ cos (x) ∣

Add a constant to the solution

= - 8 cos (y) sin (y) ln ∣ cos (x) ∣ + C

\frac{d}{d x} ((2 x^{2} - 7)^{5} - (1 + 4 x^{3})^{5})

\frac{3 x}{e ^{x}}

(x \cdot e^{x})^{^{'}}

\frac{d}{d x} (- x cos (3 x))

\frac{d}{d x} (arcsin (3 x + 1))

What is the integral of 8tan(x)sin(y)cos(y) ?
The integral of 8tan(x)sin(y)cos(y) is -8cos(y)sin(y)ln|cos(x)|+C