integral from 5 to 0
1/(4+x^2)^1/2
this is what i have so far
1/(sq.root(4+x^2)) as (1/2) (1+y)^-(1/2) with y=(x/2)^2
The binomial series of (1+y)^a = sum_k (a,k) y^k
where (a,k) is the binomial symbol.
In this case a=-1/2.
for some reason when i plug in the values i get the wrong answer, help please.
To evaluate the integral ∫(1/(√(4+x^2))) dx from 5 to 0, you seem to be on the right track using the substitution y = (x/2)^2. However, there are a couple of mistakes in your approach.
First, when you substitute y = (x/2)^2, you should also express dx in terms of dy. In this case, dx = 2√y dy.
Second, the binomial series expansion you mentioned is not applicable in this case since the exponent (-1/2) is not a positive integer.
To correctly evaluate the integral, we can use a trigonometric substitution. Since we have the expression (√(4+x^2)), it suggests that we use x = 2tanθ or x = -2tanθ as the substitution.
If we let x = 2tanθ, then dx = 2sec^2θ dθ. Also, we need to find the new limits of integration when x goes from 5 to 0.
When x = 5, we have 5 = 2tanθ, so θ = arctan(5/2). Similarly, when x = 0, we have 0 = 2tanθ, so θ = 0.
Now, let's rewrite the integral using the new variable θ:
∫(1/(√(4+x^2))) dx = ∫(1/(√(4+(2tanθ)^2))) (2sec^2θ) dθ.
Simplifying, this becomes:
2∫(1/(√(4+4tan^2θ))) sec^2θ dθ.
Using the trigonometric identity sec^2θ = 1 + tan^2θ, we can rewrite the integral as:
2∫(1/(√(4(1+tan^2θ)))) (1 + tan^2θ) dθ.
Simplifying further, we have:
2∫(1/√(4sec^2θ)) sec^2θ dθ.
The term inside the square root can be simplified as follows:
√(4sec^2θ) = 2√sec^2θ = 2secθ.
Therefore, the integral becomes:
2∫(1/(2secθ)) sec^2θ dθ = ∫secθ dθ.
The integral of secθ is ln|secθ + tanθ| + C, where C is the constant of integration.
Now, we can evaluate the integral from the new limits of integration:
∫secθ dθ from arctan(5/2) to 0.
Using the integral result, we have:
ln|sec(0) + tan(0)| - ln|sec(arctan(5/2)) + tan(arctan(5/2))|.
simplifying further, sec(0) = 1, and since tan(arctan(x)) = x, we have:
ln(1) - ln(√(1+(5/2)^2)).
ln(1) is 0, and simplifying the denominator inside the ln, we have:
-ln(√(1+25/4)) = -ln(√(29/4)) = -ln(√29/2) = -ln(√29) - ln(√2).
Taking the logarithm of a square root is the same as dividing the exponent by 2, so we have:
-1/2 ln(29) - 1/2 ln(2).
Combining the terms, we get:
-1/2 (ln(29) + ln(2)).
Therefore, the final result of the integral from 5 to 0 is:
-1/2 (ln(29) + ln(2)).