2sec^2(2x+5) dx
what's the answer when dx is .3 it keeps saying i'm wrong with the one i've found
To find the value of the integral ∫ 2sec^2(2x+5) dx, we can start by using the power rule of integration.
The power rule states that ∫ x^n dx = (x^(n+1))/(n+1) + C, where C is the constant of integration.
In our case, we have 2sec^2(2x+5) dx, which can be rewritten as 2 * (sec(2x+5))^2 dx.
Now, let's make a substitution to simplify the expression. Let u = 2x+5. Taking the derivative of u with respect to x, we get du/dx = 2. Solving for dx, we have dx = du/2.
Substituting the values of u and dx, the integral becomes:
∫ 2sec^2(u) * (du/2)
Simplifying, we get:
∫ sec^2(u) du = tan(u) + C
Now, you mentioned that dx = 0.3. So, we need to evaluate the integral at a specific value of x. Let's solve for x using the given value of dx.
We have dx = du/2, so 0.3 = du/2. Multiplying both sides by 2 gives us du = 0.6.
Substituting du = 0.6 into tan(u) + C, we get:
tan(u) + C = tan(2x+5) + C
This is the final answer when dx is 0.3. However, without the specific value of x, we cannot provide a numerical value for the integral.