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

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

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integral of-tan(x)sin(x)

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

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

Solution steps

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

Take the constant out:

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

= - \int tan (x) sin (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

4 x y^{^{'}} - 16 y = x^{- 7}, y (1) = 7

\frac{\partial}{\partial y} (3 x^{2} - 2 x y - y)

x \to 0 lim (\frac{5}{1 + e ^{\frac{1}{x}}})

\frac{\partial}{\partial x} (48 x^{5} y^{7} - 35 x^{4} y^{6})

t an g e n t f (x) = \frac{6}{( 1 + e ^{- x} ) , at ( 0 , 3 )}

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