Integrate sqrt(x^2 + 1) dx over [0,2*pi]
I can substitute u=arctan x to get:
Integrate (sec u)^3 du over [0,arctan(2*pi)]
From there, I'm stuck.
(thanks Count Iblis for your last help)
nevermind, I got it. The trick was using trigonometric reduction formulas.
To continue solving the integral, let's start by simplifying the expression (sec u)^3.
Recall that sec^2(u) = 1 + tan^2(u). We can rewrite this as:
(sec u)^2 = 1 + tan^2(u).
Multiply both sides of this equation by sec u:
(sec u)^3 = sec u + sec u * tan^2(u).
Now we have the expression (sec u)^3 in terms of sec u and tan u.
Next, let's find the limits of integration for the new variable u.
We have the original limits for x as [0, 2π]. Using the substitution u = arctan x, we need to find the values of u that correspond to x = 0 and x = 2π.
When x = 0, we have u = arctan(0) = 0.
When x = 2π, we have u = arctan(2π).
Now, let's focus on the integral:
∫(sec u)^3 du over [0, arctan(2π)].
To evaluate this integral, we can use integration by parts. The integration by parts formula is:
∫u dv = uv - ∫v du.
In this case, we can choose to write the integral as:
∫(sec u)^2 * sec u du.
Let's assign:
u = sec u,
dv = sec u du.
Then, we have:
du = sec u * tan u du,
v = ∫sec u du.
Using the integral of sec u, we have:
v = ln|sec u + tan u|.
Applying the integration by parts formula, we have:
∫sec^3(u) du = sec u * ln|sec u + tan u| - ∫(sec u + tan u) * (sec u * tan u) du.
Now, we can simplify the remaining integral:
∫(sec u + tan u) * (sec u * tan u) du.
Distribute sec u and tan u:
∫(sec^2 u * tan u + sec u * tan^2 u) du.
Using the identity sec^2(u) = 1 + tan^2(u), we can replace sec^2 u with 1 + tan^2 u:
∫(tan u + sec u - 1) * tan u du.
Now we have:
∫(2tan u - 1) * tan u du.
To integrate this, we can use the substitution v = tan u.
Differentiating both sides, we have:
dv = sec^2 u du.
Substituting back into the integral, we get:
∫(2v - 1) dv.
Integrating, we have:
[v^2 - v] + C,
where C is the constant of integration.
Now, substitute back for v:
(tan^2 u - tan u) + C.
Finally, we can evaluate this expression from u = 0 to u = arctan(2π) to find the definite integral.
Hope this helps!