integral (sqrt(100x^2-324)/x)dx
To find the integral of √(100x^2 - 324) / x, we can simplify the expression inside the square root and then evaluate the integral.
First, let's simplify the expression inside the square root:
100x^2 - 324 = 100(x^2 - 9) = 100(x - 3)(x + 3)
Now we can rewrite the integral as:
∫ (√(100(x - 3)(x + 3))) / x dx
Next, we can separate the expression into two fractions:
∫ (√(100(x - 3)(x + 3))) / x dx = ∫ (√(100(x - 3)(x + 3))) / x dx
= ∫ (√(100(x - 3)(x + 3))) / x dx
= ∫ (√(100(x - 3)(x + 3))) / x dx
= ∫ (√(100(x - 3)(x + 3))) / x dx
Now, we can use U-substitution to solve this integral. Let's substitute U = x - 3, which means dU = dx.
When x = 3, U = 0, and when x = -3, U = -6.
Now, the integral becomes:
∫ (√(100U(x + 3)) / (U + 3)) dU
= ∫ (√(100U(U + 6))) / (U + 3)) dU
= 10 ∫ (√(U(U + 6))) / (U + 3)) dU
To simplify further, we can expand the square root term using the identity (a + b)(a - b) = a^2 - b^2:
= 10 ∫ (√(U^2 + 6U)) / (U + 3)) dU
Now, we can make another substitution, V = U + 3, which means dV = dU.
When U = 0, V = 3, and when U = -6, V = -3.
The integral becomes:
10 ∫ (√((V - 3)^2 - 9)) / V dV
Simplifying the square root term:
= 10 ∫ (√(V^2 - 6V)) / V dV
We can further simplify this expression by separating the fraction:
= 10 ∫ (√(V^2 - 6V) / V) dV
The integral of √(V^2 - 6V) / V can be calculated using a trigonometric substitution. Let's substitute V = 3secθ.
Differentiating both sides with respect to θ, we get dV = 3secθ tanθ dθ.
When V = 3, θ = 0, and when V = -3, θ = π.
Substituting this into the integral, we get:
= 30 ∫ √(9sec^2θ - 6secθ) / (3secθ) (3secθ tanθ) dθ
= 30 ∫ √(9sec^2θ - 6secθ) / secθ tanθ dθ
= 30 ∫ √(9tan^2θ - 6tanθ) dθ
= 30 ∫ √(3tan^2θ - 2tanθ) dθ
Now, we can solve this integral using a substitution method. Let's substitute W = tanθ.
Differentiating both sides with respect to θ, we get dW = sec^2θ dθ.
When W = 0, θ = 0, and when W = √3/2, θ = π/3.
Substituting this into the integral, we get:
= 30 ∫ √(3W^2 - 2W) dW
This integral can be solved by combining the terms inside the square root into a perfect square:
= 30 ∫ √(3(W - 1/3)^2 - 1/3) dW
Next, we can perform a u-substitution, where U = W - 1/3. This means dU = dW.
When W = 0, U = -1/3, and when W = √3/2, U = √3/2 - 1/3.
Substituting this into the integral, we get:
= 30 ∫ √(3U^2 - 1/3) dU
= 30 ∫ √(9U^2 - 1) / 3 dU
Now, we can simplify the square root term by factoring out 9:
= 30 ∫ (3√(U^2 - 1/9)) / 3 dU
= 10 ∫ √(U^2 - 1/9) dU
Integrating this expression gives us:
= 10 ∫ U dU
= 5U^2 + C
Substituting back the value of U, we have:
= 5(W - 1/3)^2 + C
Remembering our substitutions, W = tanθ and V = 3secθ, we can substitute back to get the final result:
= 5(tanθ - 1/3)^2 + C