Evaluate the integral:

(cos(2-3x) - tan(5-3x))dx

we know that

∫cos(u) du = sin(u)
∫tan(u) du = -log cos(u)

So, what we get here is
(-1/3)sin(2-3x) - (-1/3)(-log cos(5-3x))
= -1/3 (sin(2-3x) + log cos(5-3x)) + C

the correct answer I was given has a negative 1/3 before the log. Also, how did you get the 1/3 value?

To evaluate the integral ∫ (cos(2-3x) - tan(5-3x)) dx, we can use some basic integration techniques.

Let's first evaluate the integral of the first term, cos(2-3x). To do this, we can use the substitution method.

Let u = 2-3x, then du = -3 dx. Rearrange this equation to solve for dx: dx = -du/3.

Now substitute the value of u and dx into the original integral:

∫ (cos(2-3x)) dx = ∫ (cos(u)) (-du/3) = (-1/3) ∫ cos(u) du

Integrating the cosine function, we get:

= (-1/3) sin(u) + C

Where C is the constant of integration.

Now, let's evaluate the integral of the second term, tan(5-3x). To do this, we can use the substitution method as well.

Let v = 5-3x, then dv = -3 dx. Similarly, rearrange this equation to solve for dx: dx = -dv/3.

Now substitute the value of v and dx into the original integral:

∫ (tan(5-3x)) dx = ∫ (tan(v)) (-dv/3) = (-1/3) ∫ tan(v) dv

Integrating the tangent function, we get:

= (-1/3) ln|sec(v)| + C

Finally, to evaluate the entire integral, we can combine the results from the two integrals:

∫ (cos(2-3x) - tan(5-3x)) dx = (∫ cos(u) du) - (∫ tan(v) dv)

= (-1/3) sin(u) + C - (-1/3) ln|sec(v)| + C

= (-1/3) sin(2-3x) + (-1/3) ln|sec(5-3x)| + C

Therefore, the integral evaluates to (-1/3) sin(2-3x) - (1/3) ln|sec(5-3x)| + C, where C is the constant of integration.