∫_0^(1/2)xdx/√(1-x²)
∫_0^(1/2)xdx/√(1-x²)
I do not know what
0^(1/2)
is. It is probably a symbol that I do not have.
Integral of x dx/sqrt(1-x^2)
let x = sin t
then dx = cos t dt
1-x^2 = 1 -sin^2 t = cos^2 t
so
integral of
sin t cos t dt / cos t
= integral of
sin t dt
=-cos t
but
-cos t = -sqrt(1-x^2)
To evaluate the definite integral ∫(0 to 1/2) xdx/√(1-x²), we can use a trigonometric substitution.
Let's substitute x = sinθ, which implies dx = cosθ dθ.
When x = 0, θ = 0, and when x = 1/2, θ = π/6.
Now, we can rewrite the integral as:
∫(0 to π/6) (sinθ)(cosθ dθ)/√(1-sin²θ)
Simplifying the expression under the square root, we have:
√(1-sin²θ) = √(cos²θ) = cosθ
So the integral becomes:
∫(0 to π/6) (sinθ)(cosθ dθ)/cosθ
Simplifying further, we cancel out the common factor of cosθ:
∫(0 to π/6) sinθ dθ
The antiderivative of sinθ is -cosθ, so the integral is:
[-cosθ] evaluated from 0 to π/6
Plugging in the limits:
[-cos(π/6)] - [-cos(0)]
Simplifying, we get:
[-√3/2] - [-1]
= -(-1) + √3/2
= 1 + √3/2
So, the value of the definite integral ∫(0 to 1/2) xdx/√(1-x²) is 1 + √3/2.