integrate (pi&0) sin t/(2-cost) dt

I noticed the following ...

A denominator, whose derivative shows up in the numberator...
Ahhh, logs!

∫ sin t/(2-cost) dt
= ln (2-cost)

take it from there.

My calculation:

u=2-cos t
du/dt=sin t
du=sin t dt
When t=pi,
u=2-cos pi=1.002
t=0, u=2-cos 0=1
integral (1.002&1) (1/u) du
=(ln|u|) (1.002&1)
stucked

and remember that log 3 is just a number, like any other. Don't worry about evaluating it as a decimal. Who cares what the decimal approximation is? Just like √7 or 3/π, it's fine to leave it as it is.

To integrate the given expression: ∫sin(t) / (2 - cos(t)) dt over the interval (0, π), we can use the method of substitution.

Let's start by making the substitution u = cos(t), which means du/dt = -sin(t) dt.

To replace dt in terms of du, we'll solve for sin(t) dt:
sin(t) dt = -du

Now our integral becomes:
∫ -du / (2 - u)

Next, we need to find the new limits of integration. When t = 0, u = cos(0) = 1, and when t = π, u = cos(π) = -1. So the new limits of integration are from u = 1 to u = -1.

The integral then becomes:
∫ -du / (2 - u) with limits (-1, 1)

The next step is to simplify the integrand:
-1 / (2 - u)

Now, we can integrate this expression:
∫ -1 / (2 - u) du

To do this, we can use a common integration technique called the logarithmic substitution. Let's substitute v = 2 - u, which means du = -dv.

Replacing du and u in the integral, we get:
∫ dv / v

This is a simple integral to solve:
ln|v|

Now, substituting back v = 2 - u, we get:
ln|2 - u|

Finally, evaluating the integral with the limits (-1, 1):
[ln|2 - u|] from -1 to 1

Substituting the limits, we have:
ln|2 - 1| - ln|2 - (-1)|

Simplifying, we get:
ln|1| - ln|3|

Since ln|1| = 0, this becomes:
-ln|3|

Therefore, the value of the given integral is -ln|3|.

Note: The absolute values are included in the natural logarithm because the logarithm function is only defined for positive values.