Choose the trigonometric branch before simplifying roots

(a) Use x=2tan⁡θx=2\tan\theta to find the exact value of ∫2/321xx2+4dx\int_{2/\sqrt3}^{2}\frac1{x\sqrt{x^2+4}}dx. marks

(b)(i) By writing x=A(1−x)+Bx=A(1-x)+B, where A,BA,B are constants, prove that

∫x1−xdx=23(1−x)3/2−2(1−x)1/2+C,\int\frac x{\sqrt{1-x}}dx=\frac23(1-x)^{3/2}-2(1-x)^{1/2}+C,

where CC is an arbitrary constant. marks

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

[2009/SRJC/Prelim/I/9]

Total marks: 11

Solution

(a)

Trigonometric substitution

Use x=2tan⁡θx=2\tan\theta to find the exact value of ∫2/321xx2+4 dx\int_{2/\sqrt3}^{2}\dfrac1{x\sqrt{x^2+4}}\,dx.

Explore this idea

Hint

  1. Express xx, dxdx and x2+4\sqrt{x^2+4} in θ\theta, and change the limits.

  2. Simplify to 12csc⁡θ\frac12\csc\theta.

  3. Use ∫csc⁡θ dθ=−ln⁡∣csc⁡θ+cot⁡θ∣\int\csc\theta\,d\theta=-\ln|\csc\theta+\cot\theta|.

Answer · (a)

12ln⁡[(2+3)/(1+2)]\frac12\ln[(2+\sqrt3)/(1+\sqrt2)].

Explore the idea

The area from 2/√3 to 2

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 = 2.00
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
0.218
Below-axis contribution
0
Signed integral
0.218

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

Signed integral from 1.15 to x
0.218
antiderivative − F(2/√3)
0.218
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 substitution turns a radical into a cosecant integral; the area is 12ln⁡2+31+2\frac12\ln\frac{2+\sqrt3}{1+\sqrt2}.

Ideas to try
Drag xx to 2: the readout should be about 0.218.
(b)(i) coefficients

Decomposing a numerator

Write x=A(1−x)+Bx=A(1-x)+B and find AA and BB.

Hint

  1. Compare coefficients of xx and the constant term.

Answer · (b)(i) coefficients

A=−1,B=1A=-1,B=1.

(b)(i) integral

Power rule on a radical

Prove that ∫x1−x dx=23(1−x)3/2−2(1−x)1/2+C\int\dfrac x{\sqrt{1-x}}\,dx=\dfrac23(1-x)^{3/2}-2(1-x)^{1/2}+C.

Explore this idea

Hint

  1. Use x=−(1−x)+1x=-(1-x)+1 to split the numerator.

  2. Integrate each power of (1−x)(1-x).

Answer · (b)(i) integral

23(1−x)3/2−21−x+C\frac23(1-x)^{3/2}-2\sqrt{1-x}+C.

x<1x<1.

Explore the idea

For 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 = 0.47
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
0.138
Below-axis contribution
0
Signed integral
0.138

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

Signed integral from 0.00 to x
0.138
antiderivative − F(0)
0.138
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 antiderivative matches the running area, confirming the result.

Ideas to try
Slide towards x=1x=1: the integrand blows up but the area is finite.
(b)(ii)

Integration by parts

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

Explore this idea

Hint

  1. Parts with u=sin⁡−1xu=\sin^{-1}\sqrt x and dvdx=x\frac{dv}{dx}=\sqrt x.

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

Answer · (b)(ii)

23x3/2arcsin⁡x−29(1−x)3/2+231−x+C\frac23x^{3/2}\arcsin\sqrt x-\frac29(1-x)^{3/2}+\frac23\sqrt{1-x}+C.

0<x<10<x<1.

Explore the idea

On 0 ≤ 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 = 0.50
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
0.134
Below-axis contribution
0
Signed integral
0.134

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

Signed integral from 0.00 to x
0.134
antiderivative − F(0)
0.134
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 three-term answer is exactly the area function.

Ideas to try
Why is the integrand 0 at x=0x=0 and about π2\frac\pi2 at x=1x=1?