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 refuses to integrate directly, a trig identity reshapes it until it will.

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 resolution of the power rule's missing case.
  • Reshape sin2 x, cos2 kx and tan2 x by identity before integrating.

WHAT YOU PROBABLY THINK

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

The shelf, reversed

Each derivative fact reverses into an integral, all on the must-learn list,

∫ cos kx dx = 1ksin kx + cNOT IN THE BOOKLET — LEARN IT
∫ sin kx dx = −1kcos kx + cNOT IN THE BOOKLET — LEARN IT
∫ ekx dx = 1kekx + cNOT IN THE BOOKLET — LEARN IT
∫ 1x dx = ln|x| + cNOT IN THE BOOKLET — LEARN IT

with sec2 kx integrating to (1/k) tan kx from the booklet. Where differentiation multiplied by k, integration divides, and the sine-cosine sign dance runs in reverse. The last entry settles an old account. Trying the power rule on x−1 demands division by zero, which is why the opening lie was always doomed, and the logarithm is what actually lives there.

The curve 1 over x with the region from 1 to e shaded: the area is the natural logarithm of e, which is exactly 1, the promised resolution of the n equals minus 1 gap in the power ruley = 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, and the differentiate-back check restores each one in seconds.

Identities before integrals

sin2 x has no entry on any shelf, and no amount of staring produces one. The route is a rewrite: the double angle identity cos 2x = 1 − 2 sin2 x rearranges to sin2 x = ½ − ½ cos 2x, and both pieces integrate on sight.

The curve of sine squared x over a full period with its average height marked at one half: the double angle identity rewrites sine squared as a half minus half cos 2x, an oscillation about the level the integral seesy = 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. That division of labour is the whole topic.

YOUR TURN

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; squared trig integrands always trade through an identity first.

TRY IT UNSEEN

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.

THE EXAM BIT

  • Integration divides by k where differentiation multiplied; check each term by differentiating back.
  • ∫(1/x) dx = ln|x| + c, modulus included; the modulus is a mark on papers that set negative domains.
  • Squared trig integrands rewrite by identity first, double angle for sin² and cos², the sec² identity for tan².
  • Substitute the full angle into identities: cos² 3x involves cos 6x, and the halved coefficients follow.
  • Definite versions run the square-bracket routine unchanged; exact values of sin and cos at multiples of π finish them.

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 answers landed exact because ln e and tan (π/4) are exact-value facts, which is why these limits get chosen.

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.

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 resolution of the power rule's missing case.
  • Reshape sin2 x, cos2 kx and tan2 x by identity before integrating.

No animated video for this topic yet; these notes stand alone.