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'.