What is the integral of (-sin(x)tan(x))/1 ?

The integral of (-sin(x)tan(x))/1 is -ln|tan(x)+sec(x)|+sin(x)+C

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integral of (-sin(x)tan(x))/1

\int \frac{- sin ( x ) tan ( x )}{1} d x

- ln ∣ tan (x) + sec (x) ∣ + sin (x) + C

Solution steps

\int \frac{- sin ( x ) tan ( x )}{1} d x

Take the constant out:

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

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

Rewrite using trig identities

= - \int \frac{1 - cos ^{2} ( x )}{cos ( x )} d x

Expand

\frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{cos ( x )} - cos (x)

= - \int \frac{1}{cos ( x )} - cos (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int \frac{1}{cos ( x )} d x - \int cos (x) d x)

\int \frac{1}{cos ( x )} d x = ln ∣ tan (x) + sec (x) ∣

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

= - (ln ∣ tan (x) + sec (x) ∣ - sin (x))

Simplify

= - ln ∣ tan (x) + sec (x) ∣ + sin (x)

Add a constant to the solution

= - ln ∣ tan (x) + sec (x) ∣ + sin (x) + C

\frac{d}{d x} (\frac{14}{x})

\frac{6}{3 x + 1}, at x = - 1

\frac{\partial}{\partial x} (- x - 6)

\frac{\partial}{\partial x} (4 - x^{2} - y^{2})

121 y^{^{''}} - 154 y^{^{'}} + 85 y = 0

What is the integral of (-sin(x)tan(x))/1 ?
The integral of (-sin(x)tan(x))/1 is -ln|tan(x)+sec(x)|+sin(x)+C