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 (d) Trigonometric identity or open all working together.

(e)

Reverse chain rule (cot)

Evaluate ∫1sin⁡2xecot⁡x dx\int\dfrac1{\sin^2x}e^{\cot x}\,dx.

Explore this idea

Hint

  1. Spot ddxcot⁡x=−csc⁡2x=−1sin⁡2x\frac d{dx}\cot x=-\csc^2x=-\frac1{\sin^2x} in the integrand.

Answer · (e)

−ecot⁡x+C-e^{\cot x}+C.

sin x≠0.

Explore the idea

Open interactive exploration

e^(cot x)/sin²x on (0, π)

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 = 2.09
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
0.432
Below-axis contribution
0
Signed integral
0.432

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

Signed integral from 1.57 to x
0.432
−e^(cot x) − F(π/2)
0.432
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 −ecot⁡x-e^{\cot x} is increasing because the integrand is positive.

Ideas to try
Slide towards x=0x=0 and towards x=πx=\pi: which way does the area go?