What is the integral of-16tan(x)cos(x) ?

The integral of-16tan(x)cos(x) is 16cos(x)+C

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integral of-16tan(x)cos(x)

\int - 16 tan (x) cos (x) d x

16 cos (x) + C

Solution steps

\int - 16 tan (x) cos (x) d x

Take the constant out:

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

= - 16 \cdot \int tan (x) cos (x) d x

Express with sin, cos

= - 16 \cdot \int sin (x) d x

Use the common integral:

\int sin (x) d x = - cos (x)

= - 16 (- cos (x))

Simplify

= 16 cos (x)

Add a constant to the solution

= 16 cos (x) + C

x \to 0 lim (\frac{\frac{1}{1 - x} + 1}{x})

x^{^{''}} + \frac{9}{4} x^{^{'}} + \frac{9}{64} x = 0, x^{^{'}} (0) = 1, x (0) = 2

n = 0 \sum \infty \frac{1}{n ^{0.51}}

\frac{d v}{d t} = 10 - 2 v

What is the integral of-16tan(x)cos(x) ?
The integral of-16tan(x)cos(x) is 16cos(x)+C