Another trig sub integral question?

Question

Given integral sqrt (1-4x^2 )dx, how would you answer the question?

substitute 2x = sin theta
x = 1/2 sin theta
dx = 1/2 cos theta
By formula, I get integral sqrt. (1-sin^2 theta) dx

then which is integral cos theta dx ? Do I sub in dx = 1/2 cos theta ?

If so, 1/2 integral cos^2 theta is what I get...yet I'm pretty certain that I'm not going about this in the right way.

Thank you for answering.

Answer 1

You are on the right way.

Just solve the integral ∫ cos²θ dθ and perform back
substitution:

Integrate by parts
∫ cos²θ dθ = sinθ·cosθ - ∫-sin²θ dθ
<=>
∫ cos²θ dθ = sinθ·cosθ + ∫ 1-cos²θ dθ
<=>
∫ cos²θ dθ = sinθ·cosθ + θ - ∫ cos²θ dθ
<=>
∫ cos²θ dθ = (1/2) · (sinθ·cosθ + θ)

perform back substitution
substitution with
sinθ = 2x
cosθ = sqrt( 1 - sin² θ) = sqrt(1 - 4x²)
θ = arcsin(2x)

Hence:
∫ sqrt(1-4x²) dx
= (1/2) · (2x·sqrt(1 - 4x²) + arcsin(2x))
= x·sqrt(1 - 4x²) + (1/2)·arcsin(2x)

Answer 2

that's much less complicated in case you talk approximately looking the antiderivative first. as quickly as you have completed this, you need to use the fundamental Theorem of Calculus to evaluate the crucial. the simplest thank you to proceed is to apply the substitution u = 9 - x^2, du = -2x dx. INT x/sqrt(9 - x^2) dx = (-a million/2)INT( a million/sqrt(9 - x^2 ) * -2x )dx = (-a million/2)INT a million/sqrt(u) du = (-a million/2) INT u^(-a million/2) du = (-a million/2)(2/a million)u^(a million/2) = -sqrt(u) = -sqrt(9 - x^2) yet otherwise to reach at it is to apply the trig substitution t = Arcsin(x/3), x = 3sin t , dx = 3cos t dt , sqrt(9 - x^2) = 3 cos t. INT x/sqrt(9 - x^2) dx = INT( 3sin t/3cos t * 3cos t )dt = INT 3 sin t dt = -3cos t = - sqrt(9 - x^2) call the respond g( x ). The given particular crucial = g(a million) - g(-3) = -sqrt(8) - 0 = - 2sqrt(2)

Answer 3

∫sqrt (1 - 4x^2 )dx
Let sinθ = 2x
cosθdθ = 2dx
∫sqrt (1 - sin^2θ )dx =
(1/2)∫cosθsqrt (cos^2θ )dθ =
(1/2)∫cos^2θdθ =
(1/4)(arcsin(2x) + 2x√(1 - 4x^2)) + C