Evaluate the integral.
The integral from the square root of three over three to the square root of three of the function 6/(t^2+1)
This is one of your standard integrals:
∫dx/(x^2+1) = arctan(x)
now, what angle θ has tanθ = √3/3?
θ = pi/6
Now go and crank on it. If you get stuck, come on back.
To evaluate the integral, we first need to find the antiderivative of the function \(6/(t^2+1)\).
One way to find the antiderivative is by using a substitution. Let's make the substitution \(u = t^2 + 1\). Taking the derivative of \(u\) with respect to \(t\), we have \(du/dt = 2t\), which implies \(dt = du/(2t)\).
Now let's substitute for \(u\) and \(dt\) in terms of \(t\) into the integral:
\[
\int \frac{6}{t^2+1} dt = \int \frac{6}{u} \cdot \frac{du}{2t}
\]
Notice that the numerator becomes a constant 6, and the denominator becomes just \(u\). We can pull the constant out of the integral:
\[
\frac{6}{2} \int \frac{1}{u} du = 3 \ln|u| + C
\]
Now we need to substitute back \(u\) in terms of \(t\). Recall that \(u = t^2 + 1\). Substituting back, we have:
\[
3 \ln(t^2+1) + C
\]
Finally, to evaluate the definite integral from the square root of three over three to the square root of three, we need to subtract the value of the antiderivative at the lower bound from the value at the upper bound.
Substituting the upper bound (\(\sqrt{3}\)) into the antiderivative, we have:
\[
3 \ln(\sqrt{3}^2+1) + C = 3 \ln(3+1) + C = 3 \ln(4) + C
\]
Substituting the lower bound (\(\sqrt{3}/3\)) into the antiderivative, we have:
\[
3 \ln\left(\left(\frac{\sqrt{3}}{3}\right)^2+1\right) + C = 3 \ln\left(\frac{3}{9}+1\right) + C = 3 \ln\left(\frac{4}{9}\right) + C
\]
Therefore, the value of the integral is:
\[
3 \ln(4) + C - 3 \ln\left(\frac{4}{9}\right) - C = 3 \ln(4) - 3 \ln\left(\frac{4}{9}\right)
\]