integral of square root of (9u^4 + 4u^2 + 1) = ?
To find the integral of √(9u^4 + 4u^2 + 1), we can start by using the substitution method.
Let's make the substitution: let v = u^2.
Taking the derivative of both sides with respect to u, we have dv/du = 2u.
Now, let's solve for du:
dv/du = 2u
du = (1/2u) dv
Next, let's substitute u^2 = v into our original equation:
√(9u^4 + 4u^2 + 1) du = √(9v^2 + 4v + 1) (1/2u) dv.
We can simplify this expression further by substituting v back in for u^2:
√(9v^2 + 4v + 1) (1/2u) dv = √(9v^2 + 4v + 1) (1/2√v) dv.
Now our integral becomes:
∫ √(9v^2 + 4v + 1) (1/2√v) dv.
At this point, we can simplify further by factoring out the 1/2 term:
(1/2) ∫ √(9v^2 + 4v + 1) / √v dv.
Now we have a rational function, so we can use partial fraction decomposition or the method of completing the square to integrate it further. I'll use the method of completing the square.
Let's complete the square inside the square root by adding and subtracting 1/4:
(1/2) ∫ √(9v^2 + 4v + 1) / √v dv = (1/2) ∫ √((3v + 1)^2 - 1/4) / √v dv.
Now, let's make another substitution:
Let w = 3v + 1.
Taking the derivative of both sides with respect to v, we have dw/dv = 3.
Simplifying, we have dv = (1/3) dw.
Substituting v back in for w, we get:
(1/2) ∫ √w^2 - 1/4 / √((w - 1)/3) (1/3) dw.
We can simplify this further:
(1/6) ∫ √w^2 - 1/4 / √((w - 1)/3) dw.
Now, we have a standard form of a square root integral. We can use a trigonometric substitution.
Let's make the substitution: w - 1 = (3/2) cos θ.
Taking the derivative of both sides with respect to w, we have dw = (3/2) (-sin θ) dθ.
Substituting back in, we get:
(1/6) ∫ √((3/2)^2 cos^2 θ - 1/4) / √((3/2) cos θ) (-sin θ) dθ.
Now, we can simplify further:
(1/6) ∫ √(9/4 cos^2 θ - 1/4) / √((3/2) cos θ) (-sin θ) dθ.
After simplifying the square root expression:
(1/6) ∫ √(9cos^2 θ - 1) / (3/2√cos θ) (-sin θ) dθ.
Now, we notice that the square root expression matches the Pythagorean identity sin^2 θ + cos^2 θ = 1.
Using this identity, we can rewrite the integral:
(1/6) ∫ √(sin^2 θ) / (3/2√cos θ) (-sin θ) dθ.
Simplifying further, we have:
(1/6) ∫ (sin θ / (3/2√cos θ)) (-sin θ) dθ.
Now, we can cancel out the negative sign:
(1/6) ∫ (sin θ / (3/2√cos θ)) sin θ dθ.
After canceling out the 6 and the 3/2, our integral becomes:
∫ (2/9) (sin θ / √cos θ) sin θ dθ.
Now, we have a standard integral that we can solve. Let's simplify further:
(2/9) ∫ (sin^2 θ / √cos θ) dθ.
We know that sin^2 θ = 1 - cos^2 θ.
Let's substitute that into our equation:
(2/9) ∫ ((1 - cos^2 θ) / √cos θ) dθ.
Expanding out the integral, we get:
(2/9) ∫ (√cos θ - cos^2 θ / √cos θ) dθ.
Now, we can split the integral into two parts:
(2/9) ∫ (√cos θ dθ - cos^2 θ / √cos θ dθ).
Let's solve each integral separately:
1. ∫ √cos θ dθ: This can be solved using a trigonometric substitution.
Let's make the substitution: cos θ = t.
Taking the derivative of both sides with respect to θ, we have -sin θ dθ = dt.
Substituting back in, we get:
- ∫ √t dt.
Using the power rule, the integral becomes:
- (2/3) t^(3/2) + C.
Substituting back in for t, we have:
- (2/3) cos^(3/2) θ + C1.
2. ∫ (cos^2 θ / √cos θ) dθ: We can simplify this expression first.
Using the power rule for integration, the integral becomes:
∫ (cos^(3/2) θ / √1) dθ.
Simplifying further, we have:
∫ cos^(3/2) θ dθ.
At this point, we have reached the stage where we need to use special functions like the Fresnel integral to evaluate the integral exactly. However, it is worth noting that there is no elementary closed-form solution for this integral.
Therefore, the integral of √(9u^4 + 4u^2 + 1) with respect to u cannot be expressed in terms of elementary functions. It is a non-elementary, or so-called "special" function, and would require the use of numerical approximation methods or specialized software to obtain an approximate value.