Distinguish cos(x²) from cos²x

(a) Find ∫xcos⁡x2dx\int x\cos x^2 dx. mark

(b) Use integration by parts to find ∫xcos⁡2xdx\int x\cos2x dx. marks

Hence or otherwise find ∫π/4π/2xcos⁡2xdx\int_{\pi/4}^{\pi/2}x\cos^2x dx. marks

[2020/RI/Prelim/I/5]

Total marks: 7

Solution

(a)

Reverse chain rule

Find ∫xcos⁡x2 dx\int x\cos x^2\,dx.

Explore this idea

Hint

  1. The inner function x2x^2 has derivative 2x2x, a multiple of the other factor.

Answer · (a)

12sin⁡(x2)+C\frac12\sin(x^2)+C.

Explore the idea

Oscillation speeds up with |x|

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 = 1.25
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
0.5
Below-axis contribution
0
Signed integral
0.5

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

Signed integral from 0.00 to x
0.5
½ sin(x²) − F(0)
0.5
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 signed running integral from 0 is 12sin⁡(x2)\frac12\sin(x^2), so it stays between −1/2-1/2 and 1/21/2. This bounded signed value does not bound the total geometric area across successive oscillations.

Ideas to try
Why does the signed running integral oscillate between ±12\pm\frac12?
(b) indefinite

Integration by parts

Use integration by parts to find ∫xcos⁡2x dx\int x\cos2x\,dx.

Explore this idea

Hint

  1. u=xu=x simplifies on differentiation.

  2. dvdx=cos⁡2x\frac{dv}{dx}=\cos2x.

Answer · (b) indefinite

x2sin⁡2x+14cos⁡2x+C\frac x2\sin2x+\frac14\cos2x+C.

Explore the idea

An odd integrand and its antiderivative

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 = 1.50
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
0.143
Below-axis contribution
-0.534
Signed integral
-0.392

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

Signed integral from 0.00 to x
-0.392
antiderivative − F(0)
-0.392
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 integrand is odd, so the signed running integral from 0 is even: it has the same value at xx and −x-x.

Ideas to try
Is the area to the left of 0 positive or negative?
(b) definite

Power reduction

Hence, or otherwise, find ∫π/4π/2xcos⁡2x dx\int_{\pi/4}^{\pi/2}x\cos^2x\,dx.

Explore this idea

Hint

  1. Use cos⁡2x=12(1+cos⁡2x)\cos^2x=\frac12(1+\cos2x) to reach ∫xcos⁡2x\int x\cos2x.

  2. Evaluate between the limits.

Answer · (b) definite

3π2/64−1/8−π/163\pi^2/64-1/8-\pi/16.

Explore the idea

The area from π/4 to π/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 = 1.57
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
0.141
Below-axis contribution
0
Signed integral
0.141

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

Signed integral from 0.79 to x
0.141
antiderivative − F(π/4)
0.141
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 is 3π264−18−π16≈0.141\frac{3\pi^2}{64}-\frac18-\frac{\pi}{16}\approx0.141.

Ideas to try
Drag xx to π2≈1.571\frac\pi2\approx1.571 and read the area.