Use the indicated substitution to evaluate the integral
integral 18sec(2x) tan(2x)dx, u=2x
recall that integral secu tanu du = secu
Let u=2x and you have du=2dx, so
integral 9 secu tanu du = 9secu = 9sec2x
To evaluate the integral ∫18sec(2x)tan(2x)dx using the substitution u = 2x, we need to express dx in terms of du.
Taking the derivative of both sides of u = 2x with respect to x, we get du/dx = 2, which implies dx = du/2.
Now, let's substitute these expressions into the integral:
∫18sec(2x)tan(2x)dx = ∫18sec(u)tan(u)(du/2)
Next, we can simplify the integrand using trigonometric identities.
Recall that sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x).
∫18sec(u)tan(u)(du/2) = 9∫(1/cos(u))(sin(u)/cos(u))(du/2)
Next, we can simplify further by combining the two fractions:
9∫(sin(u))/(cos^2(u))(du/2)
Now, let's simplify the expression inside the integral even further.
cos^2(u) = 1 - sin^2(u)
Therefore, we have:
9∫(sin(u))/(1 - sin^2(u))(du/2)
Now, we can make another substitution to simplify the integral further. Let v = sin(u), then dv = cos(u)du.
Rearranging, we get du = dv/cos(u). Substituting this into the integral:
9/2 ∫(v)/(1 - v^2) dv
This is a standard integral that can be solved using partial fractions or the natural logarithm method.
Once you solve this new integral, you can substitute back for u = 2x to obtain the final value of the integral.