P3.7 Integration of standard functions
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A2 content for Pure Mathematics 3 (specification sections 5.1 and 5.2). Integration by substitution, by parts, partial fractions, volumes of revolution and differential equations follow in P4.
What this page covers
7 points
- Integrating \(e^x\), \(\frac{1}{x}\), \(\sin x\), \(\cos x\) and \(\sec^2 x\), and sums and multiples of them
- Why the integral of \(\frac{1}{x}\) is \(\ln|x|\), with a modulus sign
- Integrating functions of \(ax+b\): \((ax+b)^n\), \(\frac{1}{ax+b}\), \(e^{ax+b}\), \(\sin(ax+b)\) and \(\cos(ax+b)\)
- Integrating by recognition: \(\frac{f'(x)}{f(x)}\) gives \(\ln|f(x)|\), including \(\int \tan x\,dx\)
- Integrating by recognition: \(f'(x)[f(x)]^n\) gives a power of \(f(x)\)
- Using identities to integrate \(\sin^2 x\), \(\cos^2 x\) and \(\tan^2 x\)
- Exact definite integrals and areas, including regions below the \(x\)-axis and between two curves
In one minute
6 points
- Every standard integral in P3 is a P3 derivative read backwards, so \(\int e^x\,dx = e^x + c\), \(\int \frac{1}{x}\,dx = \ln|x| + c\), \(\int \sin x\,dx = -\cos x + c\) and \(\int \cos x\,dx = \sin x + c\), and you can check any answer by differentiating it.
- Linear inside? Divide by \(a\): \(\int f(ax+b)\,dx = \frac{1}{a}F(ax+b) + c\). This only works when the inside is linear.
- Top is (a multiple of) the derivative of the bottom? \(\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c\); adjust any constant factor.
- A power of a function times the derivative of its inside? \(\int f'(x)[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n+1} + c\).
- Squares of trigonometric functions must be rewritten first: \(\sin^2 x = \tfrac12(1-\cos 2x)\), \(\cos^2 x = \tfrac12(1+\cos 2x)\), \(\tan^2 x = \sec^2 x - 1\).
- A region below the \(x\)-axis gives a negative integral, so for an area you split at the roots and add the sizes.
Contents
12 sections
- Which functions can you integrate on sight?
- How do you integrate a function of ax + b?
- When does an integral give a logarithm?
- How do you spot f′(x)[f(x)]ⁿ?
- How do identities help with sin²x, cos²x and tan²x?
- How do you find areas with these functions?
- Exam tips for P3.7
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
Which functions can you integrate on sight?
Integration is the reverse of differentiation. In P3 you met the derivatives of \(e^x\), \(\ln x\), \(\sin x\), \(\cos x\) and \(\tan x\). Read each of those results from right to left and you have a standard integral. That is the whole idea of this page: before you integrate anything, ask which derivative it came from.
The result for \(\frac{1}{x}\) fills the one gap in the P1 power rule. The rule \(\int x^n\,dx = \frac{x^{n+1}}{n+1} + c\) breaks down when \(n = -1\), because you would divide by zero. The function \(x^{-1}\) is the only power that does not integrate to a power; it integrates to a logarithm instead.
Why is there a modulus in ln|x|?
The function \(\ln x\) only exists for \(x > 0\), but \(y = \frac{1}{x}\) has a second branch for \(x < 0\). On that branch, differentiate \(\ln(-x)\) with the chain rule: \(\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}\). So \(\ln(-x)\) is also an antiderivative of \(\frac1x\), valid for negative \(x\). Writing \(\ln|x|\) covers both branches in one expression.
Each trigonometric result needs radians. If \(x\) were measured in degrees, the derivative of \(\sin x\) would carry a factor \(\frac{\pi}{180}\), and so would every integral. Keep your calculator in radian mode for the whole topic.
(a) Find \(\displaystyle\int \left(4e^x - \frac{3}{x} + 5\sin x - 2\cos x\right)dx\), \(x > 0\). (b) Evaluate \(\displaystyle\int_0^{\pi/2} (5\sin x - 2\cos x)\,dx\).
(a) Integrate term by term, watching the signs on the trigonometric terms: \(4e^x\) stays \(4e^x\); \(-\frac{3}{x}\) gives \(-3\ln x\); \(5\sin x\) gives \(-5\cos x\); \(-2\cos x\) gives \(-2\sin x\). Because \(x > 0\), \(\ln|x|\) can be written \(\ln x\). \(4e^x - 3\ln x - 5\cos x - 2\sin x + c\)
(b) Use the antiderivative \(-5\cos x - 2\sin x\) and substitute the limits: \((-5\times 0 - 2\times 1) - (-5\times 1 - 2\times 0) = -2 + 5\). \(3\)
Getting the sign of \(\int \sin x\,dx\) wrong. Differentiating \(\cos x\) gives \(-\sin x\), so going back, \(\int \sin x\,dx = -\cos x + c\). Check by differentiating: \(\frac{d}{dx}(-\cos x) = \sin x\).
How do you integrate a function of ax + b?
By the chain rule, \(\frac{d}{dx}\big[F(ax+b)\big] = a\,F'(ax+b)\). The derivative picks up an extra factor \(a\), the coefficient of \(x\). Reversing this, you integrate as if the bracket were a single \(x\), then divide by \(a\) to cancel the factor the chain rule would put back.
- \((ax+b)^n \to \dfrac{(ax+b)^{n+1}}{a(n+1)}\), \(n \ne -1\); \(\dfrac{1}{ax+b} \to \dfrac{1}{a}\ln|ax+b|\)
- \(e^{ax+b} \to \dfrac{1}{a}e^{ax+b}\); \(\sin(ax+b) \to -\dfrac{1}{a}\cos(ax+b)\); \(\cos(ax+b) \to \dfrac{1}{a}\sin(ax+b)\)
There is a neat picture behind the factor \(\frac{1}{a}\). The graph of \(y = \cos 2x\) is the graph of \(y = \cos x\) squashed horizontally by a factor of 2. Every vertical strip becomes half as wide but keeps its height, so every area is halved. That is exactly the \(\frac12\) in \(\int \cos 2x\,dx = \frac12\sin 2x + c\).
The rule only works when the inside is linear. For \(\int (x^2+1)^3\,dx\) the derivative of the inside is \(2x\), which is not a constant, so you cannot simply divide by it. Either expand the bracket or, if a factor of \(x\) is present outside, use the pattern in the section on \(f'(x)[f(x)]^n\).
Find (a) \(\int (3x-2)^4\,dx\) (b) \(\int \frac{8}{2x+5}\,dx\) (c) \(\int e^{1-4x}\,dx\) (d) \(\int \sin\!\left(2x + \frac{\pi}{6}\right)dx\).
(a) Raise the power to 5, divide by the new power and by \(a = 3\): \(\frac{(3x-2)^5}{5\times 3}\). \(\frac{1}{15}(3x-2)^5 + c\)
(b) \(8 \times \frac{1}{2}\ln|2x+5|\). \(4\ln|2x+5| + c\)
(c) Here \(a = -4\), so divide by \(-4\). \(-\frac14 e^{1-4x} + c\)
(d) \(\int \sin u\,du = -\cos u\), then divide by 2. \(-\frac12\cos\!\left(2x + \frac{\pi}{6}\right) + c\)
Find the exact value of (a) \(\displaystyle\int_1^4 \frac{6}{2x+1}\,dx\) (b) \(\displaystyle\int_0^{\ln 2} e^{3x}\,dx\).
(a) Integrate, dividing by 2: \(\big[3\ln|2x+1|\big]_1^4 = 3\ln 9 - 3\ln 3\). Combine with the laws of logarithms: \(3\ln\frac{9}{3} = 3\ln 3\) (≈ 3.30). \(3\ln 3\)
(b) \(\big[\tfrac13 e^{3x}\big]_0^{\ln 2}\). Use \(e^{3\ln 2} = e^{\ln 8} = 8\): \(\frac13(8 - 1)\). \(\frac{7}{3}\)
"Exact value" means leave ln, e and π in your answer and simplify with the laws of logarithms: \(3\ln 9 - 3\ln 3 = 3\ln 3\). A calculator decimal is not an exact answer; use it only as a check.
When does an integral give a logarithm?
By the chain rule, \(\frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)}\). So whenever the numerator of a fraction is the derivative of its denominator, or a constant multiple of it, the integral is a logarithm. This is the most useful recognition pattern on the paper, and the test is quick: differentiate the bottom and compare it with the top.
The most famous case is \(\tan x\). Write it as \(\frac{\sin x}{\cos x}\). The derivative of \(\cos x\) is \(-\sin x\), so the numerator is \(-1\) times the derivative of the denominator:
\[\int \tan x\,dx = -\int \frac{-\sin x}{\cos x}\,dx = -\ln|\cos x| + c = \ln|\sec x| + c.\]The last step uses \(-\ln|\cos x| = \ln|\cos x|^{-1} = \ln|\sec x|\). Similarly \(\int \cot x\,dx = \int \frac{\cos x}{\sin x}\,dx = \ln|\sin x| + c\).
The numerator must be a constant multiple of the derivative of the denominator. \(\int \frac{x}{x^2+1}\,dx = \frac12\ln(x^2+1) + c\) works, because \(x\) is half of \(2x\). But \(\int \frac{1}{x^2+1}\,dx\) does not fit the pattern at all: the top would need to contain \(x\).
(a) Find \(\displaystyle\int \frac{4x-2}{x^2-x+3}\,dx\). (b) Show that \(\displaystyle\int_0^{\pi/2} \frac{\cos x}{3+\sin x}\,dx = \ln\frac{4}{3}\).
(a) Differentiate the denominator: \(\frac{d}{dx}(x^2-x+3) = 2x-1\), and \(4x-2 = 2(2x-1)\). The integrand is twice \(\frac{f'}{f}\), so it integrates to twice a logarithm. No modulus is needed: \(x^2-x+3\) has discriminant \(1 - 12 < 0\), so it is always positive. \(2\ln(x^2-x+3) + c\)
(b) The numerator \(\cos x\) is the derivative of \(3 + \sin x\), which is always positive. So \(\big[\ln(3+\sin x)\big]_0^{\pi/2} = \ln 4 - \ln 3\). \(\ln\frac{4}{3}\) as required.
Writing \(\int \frac{1}{x^2+1}\,dx = \ln(x^2+1)\). Differentiate it: you get \(\frac{2x}{x^2+1}\), not the integrand. The \(\frac{f'}{f}\) pattern needs the derivative of the denominator on top.
How do you spot f′(x)[f(x)]ⁿ?
The chain rule gives \(\frac{d}{dx}\frac{[f(x)]^{n+1}}{n+1} = f'(x)[f(x)]^n\). So a power of a function, multiplied by the derivative of that function, integrates by raising the power, exactly as if \(f(x)\) were a single variable.
- Look for a bracket (or a function such as \(\cos x\) or \(e^x + 1\)) raised to a power, multiplied by a constant multiple of its derivative.
- Guess \([f(x)]^{n+1}\), differentiate it, compare with the integrand, then fix the constant.
- \(n = -1\) is the \(\frac{f'}{f}\) case of the previous section.
The method "guess, differentiate, adjust" is more reliable than trying to remember a formula. You only ever have to fix a constant factor. If the factor you need is a function of \(x\), the pattern does not apply and you need a different method (substitution, in P4).
(a) Find \(\int x^2(x^3+2)^4\,dx\). (b) Evaluate \(\int_0^{\pi/2} \sin x\cos^3 x\,dx\).
(a) Guess \((x^3+2)^5\) and differentiate: \(\frac{d}{dx}(x^3+2)^5 = 5(x^3+2)^4 \times 3x^2 = 15x^2(x^3+2)^4\). That is 15 times too big, so divide by 15. \(\frac{1}{15}(x^3+2)^5 + c\)
(b) \(\cos^3 x = (\cos x)^3\), and its inside, \(\cos x\), has derivative \(-\sin x\). Guess \(\cos^4 x\): \(\frac{d}{dx}\cos^4 x = -4\sin x\cos^3 x\), so \(\int \sin x\cos^3 x\,dx = -\frac14\cos^4 x + c\). With limits: \(\big[-\tfrac14\cos^4 x\big]_0^{\pi/2} = 0 - \left(-\tfrac14\right)\). \(\frac14\)
Questions often say "by differentiating \((x^3+2)^5\), or otherwise". That is a hint to use this pattern: write down the derivative, then divide by the constant that appeared.
How do identities help with sin²x, cos²x and tan²x?
There is no standard integral for \(\sin^2 x\), \(\cos^2 x\) or \(\tan^2 x\). The trick is to rearrange an identity so that the square disappears, and then use the earlier sections. The double-angle formula \(\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1\) gives the first two, and \(1 + \tan^2 x = \sec^2 x\) gives the third.
The graphs show why these identities are natural. The curve \(y = \cos^2 x\) never goes below zero; it oscillates between 0 and 1 about the mean level \(\frac12\), with period \(\pi\). That is exactly a cosine wave \(\frac12\cos 2x\) lifted up by \(\frac12\).
(a) Show that \(\displaystyle\int_0^{\pi/4} \cos^2 x\,dx = \frac{\pi+2}{8}\). (b) Find \(\int \tan^2 3x\,dx\).
(a) Use the identity: \(\int_0^{\pi/4} \tfrac12(1+\cos 2x)\,dx\). Integrate, dividing the \(\cos 2x\) term by 2: \(\left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_0^{\pi/4}\). With \(\sin\frac{\pi}{2} = 1\): \(\left(\frac{\pi}{8} + \frac14\right) - 0\). \(\frac{\pi+2}{8}\) as required.
(b) \(\tan^2 3x = \sec^2 3x - 1\), and \(\int \sec^2 kx\,dx = \frac1k\tan kx\). \(\frac13\tan 3x - x + c\)
A useful sanity check: over a full period the average of \(\sin^2 x\) is \(\frac12\), so \(\int_0^{\pi} \sin^2 x\,dx = \frac{\pi}{2}\). The formula agrees: \(\left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^{\pi} = \frac{\pi}{2}\).
Writing \(\int \sin^2 x\,dx = \frac{\sin^3 x}{3}\). That treats \(\sin x\) as if it were \(x\). Differentiating \(\frac{\sin^3 x}{3}\) gives \(\sin^2 x\cos x\), not \(\sin^2 x\). Use the identity.
How do you find areas with these functions?
The rules for definite integrals are the same as in P2; only the functions are new.
- \(\int_a^b f(x)\,dx = F(b) - F(a)\). The constant cancels, so leave it out.
- A region below the \(x\)-axis gives a negative integral. Find where the curve crosses the axis, integrate each part separately and add the sizes.
- The area between two curves is \(\int(\text{upper} - \text{lower})\,dx\) between their intersections; it does not matter whether the region is above or below the axis.
- For trigonometric functions use radians and exact values such as \(\cos\frac{\pi}{3} = \frac12\).
The curves \(y = \sin 2x\) and \(y = \sin x\) meet at the origin. (a) Show that, for \(0 < x < \pi\), the curves meet only where \(x = \frac{\pi}{3}\). (b) Find the exact area of the finite region enclosed by the two curves for \(0 \le x \le \frac{\pi}{3}\).
(a) Use \(\sin 2x = 2\sin x\cos x\): \(2\sin x\cos x = \sin x \Rightarrow \sin x(2\cos x - 1) = 0\). Since \(\sin x \ne 0\) for \(0 < x < \pi\), \(\cos x = \frac12\), and \(x = \frac{\pi}{3}\) is the only solution in \((0, \pi)\). \(x = \frac{\pi}{3}\) only
(b) \(\sin 2x\) is the upper curve on \(\left(0, \frac{\pi}{3}\right)\) (at \(x = \frac{\pi}{6}\): \(\frac{\sqrt3}{2} > \frac12\)). Area \(= \int_0^{\pi/3}(\sin 2x - \sin x)\,dx = \left[-\tfrac12\cos 2x + \cos x\right]_0^{\pi/3}\). With \(\cos\frac{2\pi}{3} = -\frac12\) and \(\cos\frac{\pi}{3} = \frac12\): \(\left(\frac14 + \frac12\right) - \left(-\frac12 + 1\right) = \frac34 - \frac12\). \(\frac14\)
The curve \(C\) has equation \(y = x - 5 + \frac{4}{x}\), \(x > 0\). (a) Find the \(x\)-coordinates of the points where \(C\) crosses the \(x\)-axis. (b) Show that the area of the finite region bounded by \(C\) and the \(x\)-axis is \(\frac{15}{2} - 8\ln 2\). (c) Find, in the form \(p + q\ln 2 + r\ln 3\), the total area of the regions between \(C\), the \(x\)-axis and the lines \(x = 1\) and \(x = 6\).
(a) Multiply by \(x\) (allowed, since \(x > 0\)): \(x^2 - 5x + 4 = 0 \Rightarrow (x-1)(x-4) = 0\). \(x = 1\) and \(x = 4\)
(b) \(\int_1^4\left(x - 5 + \frac4x\right)dx = \left[\frac{x^2}{2} - 5x + 4\ln x\right]_1^4\). With \(\ln 4 = 2\ln 2\): \((8 - 20 + 8\ln 2) - \left(\frac12 - 5\right) = -\frac{15}{2} + 8\ln 2\), which is negative because the region is below the axis. The area is its size. \(\frac{15}{2} - 8\ln 2\) (≈ 1.955)
(c) Between 4 and 6 the curve is above the axis: \(\left[\frac{x^2}{2} - 5x + 4\ln x\right]_4^6 = (18 - 30 + 4\ln 6) - (8 - 20 + 4\ln 4) = 4\ln\frac64 = 4\ln 3 - 4\ln 2\) (≈ 1.622). Add the two sizes: \(\frac{15}{2} - 8\ln 2 + 4\ln 3 - 4\ln 2\). \(\frac{15}{2} - 12\ln 2 + 4\ln 3\) (≈ 3.58)
Integrating straight from 1 to 6 gives \(4\ln 6 - \frac{15}{2} \approx -0.333\): the parts below and above the axis have partly cancelled. That is not an area.
Using your calculator
- Many scientific calculators can evaluate a definite integral numerically. Use it to check an exact answer: \(3\ln 3 \approx 3.30\) should match \(\int_1^4 \frac{6}{2x+1}\,dx\).
- Set radian mode before any trigonometric integral; in degree mode the check will disagree.
- A numerical value never replaces the working when a question asks for an exact answer or says "show that".
Integrating straight across a root. If the curve crosses the axis between the limits, a single definite integral gives the area above minus the area below. Sketch or test the sign first, then split the integral at each root.
Exam tips for P3.7
Every result on this page can be checked in a few seconds by differentiating your answer. The factor \(\frac1a\), the sign of a cosine and the constant in an \(f'f^n\) integral are the usual slips, and differentiation catches all three.
Before integrating, ask in order: is it a standard function? A function of \(ax+b\)? Is the top the derivative of the bottom? Is a power multiplied by the derivative of its inside? Is it a trigonometric square that needs an identity? Write the rewritten form down before you integrate.
Calculus with trigonometric functions only works in radians. Keep \(\pi\) in limits, use exact values such as \(\sin\frac{\pi}{3} = \frac{\sqrt3}{2}\), and combine logarithms with the laws of logs for exact answers.
Write \(\ln|f(x)|\) unless you know \(f(x) > 0\) on the interval. With limits, both must lie where the denominator has the same sign: \(\int \frac{1}{2x-1}\,dx\) cannot be evaluated across \(x = \frac12\).
For an area, find the roots in the interval first. A negative definite integral is a perfectly good integral but never a final area: report its size, and add separate pieces.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\int \sin x\,dx = \cos x + c\). | \(\frac{d}{dx}(\cos x) = -\sin x\), so the sign is wrong. | \(\int \sin x\,dx = -\cos x + c\). |
| \(\int e^{3x}\,dx = 3e^{3x} + c\). | That is the derivative: it multiplies by \(a\). | Integrating divides by \(a\): \(\frac13 e^{3x} + c\). |
| \(\int \frac{1}{5x+2}\,dx = \ln|5x+2| + c\). | Differentiating gives \(\frac{5}{5x+2}\). | \(\frac15\ln|5x+2| + c\). |
| Any fraction integrates to a logarithm. | Only if the top is a multiple of the derivative of the bottom. | Differentiate the denominator and compare before writing ln. |
| \(\int \cos^2 x\,dx = \frac{\cos^3 x}{3} + c\). | \(\cos x\) is not the variable; the chain rule factor is missing. | Use \(\cos^2 x = \frac12(1+\cos 2x)\) first. |
| A definite integral always gives the area. | Parts below the axis count as negative. | Split at the roots and add the sizes. |
How this topic connects
P3 Differentiation
Every result here is a P3 derivative reversed: the derivatives of \(e^x\), \(\ln x\), \(\sin x\) and \(\cos x\), the chain rule (for \(f(ax+b)\), \(\frac{f'}{f}\) and \(f'f^n\)), and the double-angle and Pythagorean identities from P3 trigonometry.
P1 and P2 Integration
The power rule and the constant of integration come from P1; definite integrals, areas and the trapezium rule from P2. This page widens the set of functions you can apply them to.
P4.6 Integration and beyond
Integration by substitution generalises the \(\frac{f'}{f}\) and \(f'f^n\) patterns; integration by parts, partial fractions (P4.1, which often end in ln terms), volumes of revolution and separable differential equations all rely on these standard results. In mechanics, a velocity such as \(v = 4e^{-0.5t}\) or \(3\sin 2t\) is integrated to give displacement, and in IAL Physics exponential decay and simple harmonic motion lead to the same integrals.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\int e^x\,dx = e^x + c\); \(\int \frac1x\,dx = \ln|x| + c\) | exponential and reciprocal | No — learn it |
| \(\int \sin x\,dx = -\cos x + c\); \(\int \cos x\,dx = \sin x + c\) | sine and cosine (\(x\) in radians) | No — learn it |
| \(\int \sec^2 kx\,dx = \frac1k\tan kx + c\) | secant squared | Yes |
| \(\int \tan kx\,dx = \frac1k\ln|\sec kx| + c\); \(\int \cot x\,dx = \ln|\sin x| + c\) | tangent and cotangent | Yes |
| \(\int f(ax+b)\,dx = \frac1a F(ax+b) + c\) | linear inside function, e.g. \((ax+b)^n\), \(\frac{1}{ax+b}\), \(e^{ax+b}\) | No — derive it (reverse chain rule) |
| \(\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c\) | numerator is the derivative of the denominator | No — learn it |
| \(\int f'(x)[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n+1} + c\) | power of a function times its derivative | No — derive it (reverse chain rule) |
| \(\sin^2 x = \frac12(1-\cos 2x)\); \(\cos^2 x = \frac12(1+\cos 2x)\) | integrating \(\sin^2 x\) and \(\cos^2 x\) | No — learn it (from cos 2x) |
| \(\tan^2 x = \sec^2 x - 1\) | integrating \(\tan^2 x\) | No — learn it |
| \(\int_a^b f(x)\,dx = F(b) - F(a)\) | definite integrals and areas | No — learn it |
Check your understanding
1. Find \(\int e^{-x/2}\,dx\).
Here \(a = -\frac12\), so divide by \(-\frac12\), i.e. multiply by \(-2\): \(-2e^{-x/2} + c\).
2. Find the exact value of \(\int_2^6 \frac1x\,dx\).
\(\ln 6 - \ln 2 = \ln\frac62 = \ln 3\).
3. Find \(\int \frac{3x^2}{x^3+1}\,dx\).
The numerator is the derivative of the denominator, so the integral is \(\ln|x^3+1| + c\).
4. Explain why \(\int \sin^2 x\,dx\) is not \(\frac{\sin^3 x}{3} + c\), and find it.
Differentiating \(\frac{\sin^3 x}{3}\) gives \(\sin^2 x\cos x\), not \(\sin^2 x\). Use \(\sin^2 x = \frac12(1-\cos 2x)\): \(\int \sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + c\).
5. Find \(\int \cos x\sin^4 x\,dx\).
\(\cos x\) is the derivative of \(\sin x\), so this is \(f'f^4\): \(\frac{\sin^5 x}{5} + c\).
6. Find the exact area under \(y = \tan x\) between \(x = 0\) and \(x = \frac{\pi}{3}\).
\(\tan x > 0\) on this interval, so the area is \(\big[\ln|\sec x|\big]_0^{\pi/3} = \ln 2 - \ln 1 = \ln 2\) (≈ 0.693), because \(\sec\frac{\pi}{3} = \frac{1}{\cos(\pi/3)} = 2\).
Specification checklist
- (P3 5.1) I can integrate \(e^x\), \(\frac1x\), \(\sin x\) and \(\cos x\), and sums and multiples of them, and I know why the answer for \(\frac1x\) is \(\ln|x|\).
- (P3 5.1) I can integrate functions of \(ax+b\), including \((ax+b)^n\), \(\frac{1}{ax+b}\), \(e^{ax+b}\), \(\sin(ax+b)\) and \(\cos(ax+b)\).
- (P3 5.2) I can recognise an integrand of the form \(\frac{k f'(x)}{f(x)}\) and integrate it to a logarithm, including \(\int \tan x\,dx\).
- (P3 5.2) I can recognise an integrand of the form \(k f'(x)[f(x)]^n\) and integrate it by reversing the chain rule.
- (P3 5.2) I can use trigonometric identities to integrate \(\sin^2 x\), \(\cos^2 x\) and \(\tan^2 x\), and multiples of the angle.
- (P3 5.1, 5.2) I can evaluate definite integrals of these functions exactly and find areas, splitting at roots and using upper minus lower for the region between two curves.