Build an inverse-sine integral from an earlier result

(i) Use the substitution u=2x−1u=2x-1 to find ∫x1−(2x−1)2dx\int\frac x{\sqrt{1-(2x-1)^2}}dx. marks

(ii) Hence find ∫sin⁡−1(2x−1)dx\int\sin^{-1}(2x-1)dx. marks

[2020/ACJC/Prelim/I/1]

Total marks: 6

Solution

(i)

Substitution

Use u=2x−1u=2x-1 to find ∫x1−(2x−1)2 dx\int\dfrac x{\sqrt{1-(2x-1)^2}}\,dx.

Explore this idea

Hint

  1. Express xx and dxdx in terms of uu.

  2. Split the numerator into a reverse-chain-rule part and a standard sin⁡−1\sin^{-1} part.

Answer · (i)

14arcsin⁡(2x−1)−141−(2x−1)2+C\frac14\arcsin(2x-1)-\frac14\sqrt{1-(2x-1)^2}+C.

Explore the idea

Between x = 0 and x = 1

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 0.74
Reset the controls

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.156
Below-axis contribution
0
Signed integral
0.156

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.50 to x
0.156
antiderivative − F(½)
0.156
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

At x=0x=0 the original quotient is undefined but tends to zero. It blows up only at x=1x=1, with an integrable square-root singularity, so the endpoint area is finite.

Ideas to try
The integrand blows up at both ends. Does the area stay finite?
(ii)

Integration by parts

Hence find ∫sin⁡−1(2x−1) dx\int\sin^{-1}(2x-1)\,dx.

Explore this idea

Hint

  1. Use parts with u=sin⁡−1(2x−1)u=\sin^{-1}(2x-1) and dvdx=1\frac{dv}{dx}=1.

  2. The remaining integral is a multiple of part (i).

Answer · (ii)

(x−1/2)arcsin⁡(2x−1)+121−(2x−1)2+C(x-1/2)\arcsin(2x-1)+\frac12\sqrt{1-(2x-1)^2}+C.

Differentiable working interval 0<x<1.

Explore the idea

An S-shaped curve and its area

Drag x. The shading shows the selected interval; the lower curves compare the numerical signed integral with the change in the antiderivative.

Upper limit xx = 0.75
Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.064
Below-axis contribution
0
Signed integral
0.064

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 0.50 to x
0.064
antiderivative − F(½)
0.064
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

The area from 12\frac12 is symmetric: the integrand is odd about x=12x=\frac12, so the area is even.

Ideas to try
The curve is odd about x=12x=\frac12: what does that say about the area?