Change both the variable and its differential

(a) Find ∫13−4t2dt\int\frac1{3-4t^2}dt. marks

(b) Use the substitution u=5xu=5^x to find ∫5xcos⁡2(5x)dx\int5^x\cos^2(5^x)dx. marks

[2010/HCI/Prelim/I/6]

Total marks: 8

Solution

(a)

Standard form with a modulus

Find ∫13−4t2 dt\int\dfrac1{3-4t^2}\,dt.

Explore this idea

Hint

  1. Factor out the coefficient of t2t^2 to match ∫dtt2−a2\int\frac{dt}{t^2-a^2}.

  2. Use ∫dtt2−a2=12aln⁡∣t−at+a∣+C\int\frac{dt}{t^2-a^2}=\frac1{2a}\ln\left|\frac{t-a}{t+a}\right|+C.

Answer · (a)

−312ln⁡∣2t−32t+3∣+C-\frac{\sqrt3}{12}\ln|\frac{2t-\sqrt3}{2t+\sqrt3}|+C.

t≠±√3/2, on a connected interval.

Explore the idea

Between the poles at t = ±√3/2

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 = 0.40
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
0.144
Below-axis contribution
0
Signed integral
0.144

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

Signed integral from 0.00 to x
0.144
antiderivative − F(0)
0.144
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

Inside ∣t∣<32|t|<\frac{\sqrt3}2 the integrand is positive and the area grows without bound towards the poles.

Ideas to try
Slide towards t=32≈0.866t=\frac{\sqrt3}2\approx0.866: what does the area do?
(b)

Integration by substitution

Use u=5xu=5^x to find ∫5xcos⁡2(5x) dx\int5^x\cos^2(5^x)\,dx.

Explore this idea

Hint

  1. Differentiate uu to find dxdx in terms of dudu.

  2. Rewrite the integral entirely in uu.

  3. Use cos⁡2u=12(1+cos⁡2u)\cos^2u=\frac12(1+\cos2u).

Answer · (b)

12ln⁡5[5x+12sin⁡(2⋅5x)]+C\frac1{2\ln5}[5^x+\frac12\sin(2\cdot5^x)]+C.

Explore the idea

Oscillation grows with 5ˣ

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 = 0.65
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
0.346
Below-axis contribution
0
Signed integral
0.346

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

Signed integral from 0.00 to x
0.346
antiderivative − F(0)
0.346
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 cos⁡2(5x)\cos^2(5^x) oscillation speeds up as 5x5^x grows, but the antiderivative still matches the running area.

Ideas to try
Why does the integrand oscillate faster to the right?