integral: sqrt (16 + 6x − x^2)
dx
Note that since (x-3)^2 = x^2-6x+9, what you have is
∫√(25-(x-3)^2) dx
If u = x-3, du = dx, and it's just
∫√(25-u^2) du
which you should know. If not, just use the trig substitution u = 5sinθ
The answer can be seen at
http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%9A%2825-u^2%29+du
Or, in terms of the original integral,
http://www.wolframalpha.com/input/?i=%E2%88%AB%E2%88%9A%2816%2B6x-x^2%29+dx
thanks. i wish i could upload a screen shot of what i typed it but i did it exactly as the site gave and it was counted wrong.
nevermind! i just forgot the constant
To solve the integral ∫ sqrt (16 + 6x − x^2) dx, we can use a combination of algebraic manipulation and trigonometric substitution.
Step 1: Simplify the expression inside the square root if possible.
The expression inside the square root, 16 + 6x − x^2, can be rearranged by completing the square. We can write it as:
(16 - x^2 + 6x) = 25 - (x^2 - 6x + 9) = 25 - (x - 3)^2
Step 2: Rewrite the integral using the simplified expression.
Now we have: ∫ sqrt (25 - (x - 3)^2) dx
Step 3: Make a trigonometric substitution.
Let's substitute x - 3 with 5sinθ, where -π/2 ≤ θ ≤ π/2.
This substitution simplifies the integrand because it makes the expression inside the square root equal to a trigonometric identity.
Using the identity sin^2θ + cos^2θ = 1, we have:
25 - (x - 3)^2 = 25 - (5sinθ)^2 = 25 - 25sin^2θ = 25cos^2θ.
The derivative of x - 3 with respect to θ is dx/dθ = 5cosθ, which implies dx = 5cosθ dθ.
Substituting x - 3 = 5sinθ and dx = 5cosθ dθ in the integral, we get:
∫ sqrt (25cos^2θ) * 5cosθ dθ = ∫ 5cos^2θ dθ
Step 4: Apply the trigonometric identity to solve the integral.
Using the double-angle identity cos^2θ = (1 + cos 2θ)/2, we can rewrite the integral as:
∫ (5/2)(1 + cos 2θ) dθ
Splitting the integral, we get:
(5/2) ∫ (dθ + cos 2θ dθ)
The first term integrates to θ, and the second term integrates to (1/2)sin 2θ.
So, the final solution is:
(5/2) (θ + (1/2)sin 2θ) + C,
where C represents the constant of integration.