Maths › Integration › Integrating standard functions

Integrating standard functions

Every derivative fact from the differentiation unit now runs backwards: exponentials, sines, cosines, and at long last 1 over x, whose integral is the logarithm the power rule could never produce. Where a function does not integrate directly, a trig identity can reshape it into a form that does.

Year 12-13EDEXCEL 9MA0 8.2

Builds on Definite integrals and areas and Differentiating trig, exponentials and logs.

IN THIS TOPIC

  • Integrate ekx, 1/x, sin kx, cos kx and sec2 kx, dividing by k throughout.
  • Use ∫(1/x) dx = ln|x| + c, the case the power rule could not reach.
  • Reshape sin2 x, cos2 kx and tan2 x by identity before integrating.

COMMON MISCONCEPTION

1/x follows the reversed power rule like every other power.

The shelf, reversed

Each derivative fact reverses into an integral, and all four of these are on the must-learn list.

∫coskxdx=1ksinkx+c\text{∫} \cos kx \, dx = \frac{1}{k}\sin kx + cNOT IN THE BOOKLET — LEARN IT
∫sinkxdx=−1kcoskx+c\text{∫} \sin kx \, dx = −\frac{1}{k}\cos kx + cNOT IN THE BOOKLET — LEARN IT
∫ekxdx=1kekx+c\text{∫} e^{kx} \, dx = \frac{1}{k}e^{kx} + cNOT IN THE BOOKLET — LEARN IT
∫1xdx=ln|x|+c\text{∫} \frac{1}{x} \, dx = \text{ln}|x| + cNOT IN THE BOOKLET — LEARN IT

The booklet adds sec2 kx, which integrates to (1/k) tan kx. Where differentiation multiplied by k, integration divides by it, and the sine-cosine sign dance runs in reverse. That last entry settles an old account. Try the power rule on x−1 and it demands a division by zero; the logarithm is what actually lives at n = −1.

The curve 1 over x with the region from 1 to e shaded: the area is the natural logarithm of e, exactly 1y = 1/xarea = 11e∫ of 1/x is ln x: the missing case, filled
FIG. 1The missing case, filled: the area under 1/x from 1 to e is ln e = 1 exactly. The integral of 1/x is the logarithm.

WORKED EXAMPLE

Three terms from the shelf

Find ∫(e5x + 1/(2x) + cos 3x) dx.

Termwise: e5x gives e5x/5, and 1/(2x) is ½ × 1/x, giving ½ ln|x|.

cos 3x gives (1/3) sin 3x.

The integral is e5x/5 + ½ ln|x| + ⅓ sin 3x + c.

Every k ended up dividing. Differentiate the answer back and each one reappears in seconds.

Identities before integrals

sin2 x has no entry on any shelf, and no amount of staring at it will produce one. The route in is a rewrite. The double angle identity cos 2x = 1 − 2 sin2 x rearranges to sin2 x = ½ − ½ cos 2x, and both of those pieces integrate on sight.

Sine squared x over a full period: the double angle identity rewrites it as a half minus half cos 2x, oscillating about the average height one halfy = sin² xdashed line: average ½sin² x = ½ − ½ cos 2x: integrate the right-hand side
FIG. 2Why the rewrite works: sin² x oscillates about its average ½. The constant carries the integral; the cosine wave averages itself away.

WORKED EXAMPLE

The flagship rewrite

Find ∫sin2 x dx.

Rewrite: sin2 x = ½ − ½ cos 2x.

Integrate termwise: ½x − ½ × (1/2) sin 2x.

∫sin2 x dx = x/2 − (sin 2x)/4 + c.

The identity did the mathematics. The integration that followed was two shelf lookups, and that division of labour is the whole topic in miniature.

GUIDED PRACTICE

The tangent version

Find ∫tan2 x dx, before opening the working.

Show the working

The identity sec2 x = 1 + tan2 x rearranges to tan2 x = sec2 x − 1.

Both pieces are on the shelf: ∫tan2 x dx = tan x − x + c.

The reciprocal-functions lesson built that identity for exactly this moment. A squared trig integrand always trades through an identity first.

INDEPENDENT PRACTICE

A squared cosine, with a k

Find ∫cos2 3x dx.

Show the working

cos 2A = 2 cos2 A − 1 with A = 3x gives cos2 3x = ½ + ½ cos 6x.

Integrating: x/2 + (sin 6x)/12 + c.

The doubled angle doubled again, 3x becoming 6x, and its 6 duly divided the sine. Substituting the whole angle into the identity is where this question is won or lost.

ASSESSMENT FOCUS

  • Integration divides by k where differentiation multiplied. Check each term by differentiating back.
  • ∫(1/x) dx = ln|x| + c, modulus included. On papers that set a negative domain the modulus is a mark.
  • Squared trig integrands rewrite by identity first. Double angle for sin² and cos², the sec² identity for tan².
  • Substitute the full angle into the identity. cos² 3x involves cos 6x, and the halved coefficients follow from that.

CHECK YOURSELF

Find the exact value of ∫1e (2/x) dx, and evaluate ∫0π/4 sec2 x dx.

Show a hint

Both are single shelf entries with friendly limits.

Show the answer

∫1e (2/x) dx = [2 ln|x|] from 1 to e = 2 − 0 = 2.

∫0π/4 sec2 x dx = [tan x] from 0 to π/4 = 1 − 0 = 1.

Both landed exact because ln e and tan (π/4) are exact-value facts. Examiners pick limits like these on purpose.

Reverse the shelf and divide by every k; the missing power-rule case is ln|x|.

No entry for a squared trig function exists; an identity trades it for terms that have one.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

PDF · 30 KB · screen edition · navy

Answer book

PDF · 57 KB · fully worked · navy
7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the integrating standard functions questions page.

CHECK YOUR PROGRESS

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  • Integrate ekx, 1/x, sin kx, cos kx and sec2 kx, dividing by k throughout.
  • Use ∫(1/x) dx = ln|x| + c, the case the power rule could not reach.
  • Reshape sin2 x, cos2 kx and tan2 x by identity before integrating.

Open the full revision checklist to see every objective in the course in one place.