What is the integral of-sin(2x)tan(2x) ?

The integral of-sin(2x)tan(2x) is -1/2 (ln|tan(2x)+sec(2x)|-sin(2x))+C

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integral of-sin(2x)tan(2x)

\int - sin (2 x) tan (2 x) d x

- \frac{1}{2} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x)) + C

Solution steps

\int - sin (2 x) tan (2 x) d x

Take the constant out:

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

= - \int sin (2 x) tan (2 x) d x

Rewrite using trig identities

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

Apply u-substitution

= - \int \frac{sin ( u ) tan ( u )}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int sin (u) tan (u) d u

Rewrite using trig identities

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

Expand

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

= - \frac{1}{2} \cdot \int \frac{1}{cos ( u )} - cos (u) d u

Apply the Sum Rule:

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

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

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

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

= - \frac{1}{2} (ln ∣ tan (u) + sec (u) ∣ - sin (u))

Substitute back

u = 2 x

= - \frac{1}{2} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x))

Add a constant to the solution

= - \frac{1}{2} (ln ∣ tan (2 x) + sec (2 x) ∣ - sin (2 x)) + C

What is the integral of-sin(2x)tan(2x) ?
The integral of-sin(2x)tan(2x) is -1/2 (ln|tan(2x)+sec(2x)|-sin(2x))+C