Evaluate the integral.

S= integral sign
I= absolute value

S ((cos x)/(2 + sin x))dx

Not sure if I'm doing this right:

u= 2 + sin x
du= 0 + cos x dx

= S du/u = ln IuI + C
= ln I 2 + sin x I + C
= ln (2 + sin x) + C

Another problem:

S ((sin (ln x))/(x)) dx

I don't even know what to put as u and du?

Please help with explanations.

The first problem is correct.

As for the second problem, take:
u = ln x
du = (1/x) dx

So:
S (ln u) du = ?
Try it yourself

Why is it S (ln u) du = ?

What about the sin? And to multiply them are you reciprocating?

Oh sorry. It should be:

S sin(u) du

So I get this:

= ln Isin(u)I + C
= ln Isin(ln x)I + C
= ln (sin(ln x)) + C
OR
= ln (sin(1/x)) + C

Is it correct or incorrect?

To evaluate the first integral, let's use a substitution.

Let u = 2 + sin(x). Taking the derivative of both sides, we get du = cos(x) dx.

Now, substitute u and du into the integral:

∫ (cos(x))/(2 + sin(x)) dx = ∫ du/u

This simplifies the integral to ∫ du/u.

The integral of 1/u is ln|u| (natural logarithm of the absolute value of u).

So, the result is ln|u| + C, where C is the constant of integration.

Substituting back u = 2 + sin(x), the final answer is ln|2 + sin(x)| + C.

Now let's move on to the second integral:

∫ (sin(ln(x)))/(x) dx

In this case, we can use a substitution where u = ln(x). Differentiating both sides, we get du = (1/x) dx.

Now, substitute u and du into the integral:

∫ (sin(u))/(x) dx = ∫ (sin(u)) du (since du = (1/x) dx)

We know that the integral of sin(u) du is -cos(u) + C', where C' is the constant of integration.

Substituting u = ln(x), the final answer is -cos(ln(x)) + C'.