Match a denominator derivative and remove a radical

(a) Given that A(sin⁡θ+cos⁡θ)+B(cos⁡θ−sin⁡θ)≡4sin⁡θA(\sin\theta+\cos\theta)+B(\cos\theta-\sin\theta)\equiv4\sin\theta, find the constants A,BA,B. marks

Hence find the exact value of ∫0π/44sin⁡θsin⁡θ+cos⁡θdθ\int_0^{\pi/4}\frac{4\sin\theta}{\sin\theta+\cos\theta}d\theta. marks

(b)(i) Express 2t(t+1)2\frac{2t}{(t+1)^2} in partial fractions. marks

(ii) Hence find the exact value of ∫151x+2x−1dx\int_1^5\frac1{x+\sqrt{2x-1}}dx, using t=2x−1t=\sqrt{2x-1}. marks

[2010/SAJC/Prelim/I/11]

Total marks: 13

Solution

(a) constants

Comparing coefficients

Given A(sin⁡θ+cos⁡θ)+B(cos⁡θ−sin⁡θ)≡4sin⁡θA(\sin\theta+\cos\theta)+B(\cos\theta-\sin\theta)\equiv4\sin\theta, find AA and BB.

Hint

  1. Compare the coefficients of sin⁡θ\sin\theta and of cos⁡θ\cos\theta.

Answer · (a) constants

A=2,B=−2A=2,B=-2.

(a) integral

Logarithmic derivative

Hence find the exact value of ∫0π/44sin⁡θsin⁡θ+cos⁡θ dθ\int_0^{\pi/4}\dfrac{4\sin\theta}{\sin\theta+\cos\theta}\,d\theta.

Explore this idea

Hint

  1. Replace the numerator with 2(sin⁡θ+cos⁡θ)−2(cos⁡θ−sin⁡θ)2(\sin\theta+\cos\theta)-2(\cos\theta-\sin\theta).

  2. Divide through: a constant plus −2f′f-2\frac{f'}f with f=sin⁡θ+cos⁡θf=\sin\theta+\cos\theta.

Answer · (a) integral

π/2−ln⁡2\pi/2-\ln2.

Explore the idea

The area from 0 to π/4

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 = 0.79
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
0.878
Below-axis contribution
0
Signed integral
0.878

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

Signed integral from 0.00 to x
0.878
antiderivative − F(0)
0.878
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 exact value π2−ln⁡2\frac\pi2-\ln2 is the area of the shaded region up to θ=π4\theta=\frac\pi4.

Ideas to try
Drag θ\theta to π4\frac\pi4: the readout should be π2−ln⁡2≈0.878\frac\pi2-\ln2\approx0.878.
(b)(i)

Partial fractions

Express 2t(t+1)2\dfrac{2t}{(t+1)^2} in partial fractions.

Explore this idea

Hint

  1. A repeated linear factor needs two terms: At+1+B(t+1)2\frac A{t+1}+\frac B{(t+1)^2}.

Answer · (b)(i)

2t/(t+1)2=2/(t+1)−2/(t+1)22t/(t+1)^2=2/(t+1)-2/(t+1)^2.

Explore the idea

The two partial fractions add to the original

Switch on each piece.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Original fraction
  • 2/(t+1)
  • −2/(t+1)²
Ideas to try
At t=1t=1 add the two dashed values and compare with the teal curve.
(b)(ii)

Radical substitution

Hence find the exact value of ∫151x+2x−1 dx\int_1^5\dfrac1{x+\sqrt{2x-1}}\,dx, using t=2x−1t=\sqrt{2x-1}.

Explore this idea

Hint

  1. Write xx and dxdx in terms of tt and change the limits.

  2. The integrand should become 2t(t+1)2\frac{2t}{(t+1)^2}.

  3. Use part (i).

Answer · (b)(ii)

2ln⁡2−1/22\ln2-1/2.

Explore the idea

The area from 1 to 5

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 = 5.00
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
0.886
Below-axis contribution
0
Signed integral
0.886

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

Signed integral from 1.00 to x
0.886
antiderivative − F(1)
0.886
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

Substituting t=2x−1t=\sqrt{2x-1} turned the radical into a rational function, and the area is 2ln⁡2−122\ln2-\frac12.

Ideas to try
Drag xx to 5: the readout should be 2ln⁡2−12≈0.8862\ln2-\frac12\approx0.886.