Recognise the structure before integrating

Evaluate:

PartIntegral
(a)∫12x(ln⁡x)5 dx\int\frac{1}{2x}(\ln x)^5\,dx
(b)∫xe4−x2 dx\int xe^{4-x^2}\,dx
(c)∫(ln⁡x)2 dx\int(\ln x)^2\,dx
(d)∫41+cos⁡4θ dθ\int\frac4{1+\cos4\theta}\,d\theta
(e)∫1sin⁡2xecot⁡x dx\int\frac1{\sin^2x}e^{\cot x}\,dx
(f)∫1x2+2x+5 dx\int\frac1{x^2+2x+5}\,dx
(g)∫x2−xx2−16 dx\int\frac{x^2-x}{x^2-16}\,dx
(h)∫x2−6x+1(3x+1)(x2+3) dx\int\frac{x^2-6x+1}{(3x+1)(x^2+3)}\,dx
(i)∫e2x+1(e2x+1)3 dx\int\frac{e^{2x+1}}{(e^{2x}+1)^3}\,dx
(j)∫x+11−x2 dx\int\frac{x+1}{\sqrt{1-x^2}}\,dx

2025 TJC H2 Maths (9758) JC2 MSM · original exercise attribution retained on source pages

Solution

If this part uses an earlier result, revisit (c) Integration by parts, twice or open all working together.

(d)

Trigonometric identity

Evaluate ∫41+cos⁡4θ dθ\int\dfrac4{1+\cos4\theta}\,d\theta.

Explore this idea

Hint

  1. Use cos⁡2A=2cos⁡2A−1\cos2A=2\cos^2A-1 with A=2θA=2\theta to remove the 1+1+.

  2. The result is a sec⁡2\sec^2 integral.

Answer · (d)

tan⁡2θ+C\tan2\theta+C.

On intervals where cos2θ≠0.

Explore the idea

Open interactive exploration

4/(1 + cos 4θ) and tan 2θ

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 = 0.35
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
0.842
Below-axis contribution
0
Signed integral
0.842

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

Signed integral from 0.00 to x
0.842
tan 2θ − F(0)
0.842
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

tan⁡2θ\tan2\theta is the running area; both blow up at ±π4\pm\frac\pi4, where cos⁡4θ=−1\cos4\theta=-1.

Ideas to try
Slide towards θ=±π4\theta=\pm\frac\pi4: what happens to the area?