evaluate the integral of
dx/square root of 9-8x-x^2
9-8x-x^2 = 25 - (16 + 8x + x^2) = 3^2 - (x-4)^2
Integral sqrt(a^2 - u^2) du = u/2 * sqrt(a^2 - u^2) + a^2/2 arcsin(u/a)
just plug in a=5 and u=x-4 and plug away.
This is a standard trig substitution and integration by parts.
Go on: wolframalpha dot com
When page be open in rectangle type:
integrate dx/sqrt(9-8x-x^2)
and click option =
After few seconds when you see result click option: Show steps
typo: 5^2 not 3^2
To evaluate the given integral, we need to use a technique called completing the square. Here's how you can do it:
1. First, let's rewrite the denominator of the integrand: 9 - 8x - x^2. We want to express it in the form of a perfect square. To do this, we need to manipulate the terms within the square root.
2. Rearrange the terms in the denominator by grouping the x terms: -x^2 - 8x + 9.
3. Next, we need to complete the square for the quadratic expression -x^2 - 8x. To do this, take half of the coefficient of x (-8) and square it: (-8/2)^2 = 16.
4. Add and subtract 16 within the square root to maintain the value of the expression: -x^2 - 8x + 16 - 16 + 9.
5. Now, we can split the expression: (-(x^2 + 8x + 16) + 25).
6. Simplify the square: (-(x + 4)^2 + 25).
7. Finally, rewriting the integral: ∫(dx/√(-(x + 4)^2 + 25)).
At this point, the integral cannot be expressed in terms of elementary functions. It can be computed using advanced techniques such as trigonometric or hyperbolic substitutions, or numerical integration methods. The exact approach depends on the context of the problem and the level of mathematical sophistication required.