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 y}{d t} - \frac{4 t}{( 1 + 4 t ^{2} )} y = 4 t

y^{^{'}} = x^{2} sec (y)

\int_{0}^{1} x e^{- 3 x^{2}} d x

n = 0 \sum \infty arctan (2 n)

(2, 2), (0, - 1)

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