True or False? Any of these standard integrals can be checked by differentiating the answer.

True. Integration is the reverse of differentiation, so differentiating your answer has to return what you started with. It is the quickest way to settle a sign you are unsure of.

What does the reverse chain rule undo?

A differentiation that used the chain rule , so the integrand is a composite function multiplied by the derivative of its inside. Recognising that shape is what lets you integrate by inspection, without setting up a formal substitution.

What are the steps of the reverse chain rule?

Spot the main function, write down what would differentiate to give it, then adjust and compensate for any constant that the chain rule would have produced. Simplify at the end.

How can you check an integration answer?

Differentiate it. You should get back exactly what you set out to integrate. This is worth doing whenever the integral was reached by inspection rather than by a formal method, because inspection is where a constant is most easily dropped.

Why do you sometimes need a trigonometric identity before integrating?

Because the expression as written is not one of the standard integrals, but an identity can turn it into one that is. Most often it is a squared trigonometric term that has to be rewritten.

International A Level (IAL)MathsEdexcelPure 3FlashcardsIntegrationFurther Integration

Further Integration (Edexcel International A Level (IAL) Maths: Pure 3): Flashcards

Exam code: YMA01

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FrontIntegrating Other Functions
Complete the two standard trigonometric integrals:
$\int \sin x d x =______$
$\int \cos x d x =______$

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Cards in this collection (21)

Complete the two standard trigonometric integrals:
$\int \sin x d x =______$
$\int \cos x d x =______$
The completed integrals are:
$\int \sin x d x = - \cos x + c$
$\int \cos x d x = \sin x + c$
The minus sign attaches to the integral of $\sin$ , which is the opposite way round from differentiation, where it attaches to $\cos$ .
Why does an indefinite integral always end in $+ c$ ?
Because a constant differentiates to zero, so infinitely many functions share the same derivative.
The $+ c$ stands for all of them at once, and it is dropped only once limits are applied.
What is $\int e^{k x} d x$ ?
$\frac{1}{k} e^{k x} + c$ .
Differentiating $e^{k x}$ multiplies by $k$ , so going the other way has to divide by it.
Why can the rule for integrating $x^{n}$ not be used on $\frac{1}{x}$ ?
Because $\frac{1}{x} = x^{- 1}$ , and raising that index by $1$ would leave a denominator of zero.
The integral is $\ln |x| + c$ instead, which is why this one has to be learned separately.
True or False?
Any of these standard integrals can be checked by differentiating the answer.
True.
Integration is the reverse of differentiation, so differentiating your answer has to return what you started with.
It is the quickest way to settle a sign you are unsure of.
What is $\int \tan x d x$ ?
$\ln |\sec x| + c$ .
The modulus is needed because $\sec x$ takes negative values as well as positive ones, and a logarithm cannot accept those.
What does the reverse chain rule undo?
A differentiation that used the chain rule, so the integrand is a composite function multiplied by the derivative of its inside.
Recognising that shape is what lets you integrate by inspection, without setting up a formal substitution.
What are the steps of the reverse chain rule?
Spot the main function, write down what would differentiate to give it, then adjust and compensate for any constant that the chain rule would have produced.
Simplify at the end.
What does it mean to "adjust and compensate" when integrating?
Put in the constant your answer needs, then multiply by its reciprocal outside, so that nothing has actually been changed.
Integrating $e^{5 x}$ needs a $5$ from the chain rule, so you write $\frac{1}{5} e^{5 x}$ : the $\frac{1}{5}$ compensates for the $5$ that differentiating would bring out.
How can you check an integration answer?
Differentiate it. You should get back exactly what you set out to integrate.
This is worth doing whenever the integral was reached by inspection rather than by a formal method, because inspection is where a constant is most easily dropped.
True or False?
The reverse chain rule works whenever the integrand is a composite function.
False.
The derivative of the inside function has to be present as well, at least up to a constant multiple.
$\int 2 x e^{x^{2}} d x$ works because the $2 x$ is the derivative of $x^{2}$ , whereas $\int e^{x^{2}} d x$ cannot be done at all by elementary means.
Complete the standard result:
$\int \frac{f' (x)}{f (x)} d x =______$
The completed result is:
$\int \frac{f' (x)}{f (x)} d x = \ln |f (x)| + c$
The modulus is there for the same reason as in $\int \frac{1}{x} d x$ : the logarithm needs a positive argument.
How do you test whether a fraction is of the form $\frac{f' (x)}{f (x)}$ ?
Differentiate the denominator and compare the result with the numerator.
Ignore any coefficients while comparing: if the two match apart from a constant multiple, the form applies.
The numerator is a constant multiple of the derivative of the denominator, but not equal to it. What do you do?
Adjust for that constant, exactly as in the reverse chain rule.
In $\int \frac{x}{x^{2} + 1} d x$ the denominator differentiates to $2 x$ , so write it as $\frac{1}{2} \int \frac{2 x}{x^{2} + 1} d x$ , giving $\frac{1}{2} \ln |x^{2} + 1| + c$ .
True or False?
$\int \frac{2 x + 1}{x^{2} + x} d x = \ln |x^{2} + x| + c$
True.
Differentiating $x^{2} + x$ gives exactly $2 x + 1$ , which is the numerator, so there is no constant to adjust for.
This is the cleanest form the pattern takes.
Why does this pattern integrate to a logarithm?
Because differentiating $\ln (f (x))$ gives $\frac{f' (x)}{f (x)}$ , by the chain rule.
The integral is that result read backwards, which is why no separate rule has to be learned for it.
Why do you sometimes need a trigonometric identity before integrating?
Because the expression as written is not one of the standard integrals, but an identity can turn it into one that is.
Most often it is a squared trigonometric term that has to be rewritten.
How do you integrate $\sin^{2} x$ or $\cos^{2} x$ ?
Rewrite them with the double angle identity for $\cos 2 A$ , which turns a square into a linear expression in $\cos 2 x$ .
From $\cos 2 x = 1 - 2 \sin^{2} x$ you get $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ , which integrates term by term.
How do you integrate $\sin 3 x \cos 3 x$ ?
Use $\sin 2 A = 2 \sin A \cos A$ backwards, which gives $\sin 3 x \cos 3 x = \frac{1}{2} \sin 6 x$ .
That is a standard integral, so the answer is $- \frac{1}{12} \cos 6 x + c$ .
Complete the identities used to integrate these squared functions:
$\tan^{2} x =______- 1$
$\cot^{2} x =______- 1$
The completed identities are:
$\tan^{2} x = \sec^{2} x - 1$
$\cot^{2} x = {cosec}^{2} x - 1$
Each turns a square you cannot integrate directly into one you can, since $\sec^{2} x$ and ${cosec}^{2} x$ are both standard integrals.
True or False?
$\int \sin^{4} x \cos x d x$ needs a trigonometric identity.
False.
It looks as though it should, but $\cos x$ is the derivative of $\sin x$ , so this is a reverse chain rule integral and the answer is $\frac{1}{5} \sin^{5} x + c$ .
Anything of the form $\sin^{n} k x \cos k x$ behaves the same way and needs no identity at all.

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