5.16 Integration by substitution and by parts

Differentiation follows fixed rules: any function built from standard functions can be differentiated by applying the chain, product and quotient rules in turn. Integration has no such complete set of rules. Many integrals can only be found by turning them into simpler integrals that are already known. The two most important ways of doing this reverse the two main rules of differentiation: integration by substitution reverses the chain rule, and integration by parts reverses the product rule.

These methods extend the range of functions that can be integrated exactly to products such as $x e^x$ and $x^2 \sin x$, to single functions such as $\ln x$ and $\arctan x$

Integration by substitution

Reversing the chain rule

The chain rule (5.6) differentiates a composite function. If $F$ is an antiderivative of $f$, so that $F'(u) = f(u)$, then

Reading this backwards gives an integral:

Writing $u = g(x)$, the right-hand side is $F(u) + C = \int f(u)\,du$. So an integral of the form $\int f\big(g(x)\big)\,g'(x)\,dx$

In 5.10 integrals of this form were found by inspection: guessing the answer and checking it by differentiation. Substitution does the same job systematically, and it also works when the integrand is not exactly in the form $f\big(g(x)\big)\,g'(x)$.

The method step by step

1. Choose the substitution $u = g(x)$ (or use the one given). A good choice is usually the "inside" of a composite function, such as the expression inside a bracket, a root or an exponent.

2. Differentiate to find $\frac{du}{dx} = g'(x)$

Differentiating the answer is a reliable check. By the chain rule, $\frac{d}{dx}\left(\frac{(x^3 + 2)^6}{18}\right) = \frac{6(x^3 + 2)^5 \cdot 3x^2}{18} = x^2(x^3 + 2)^5$

The inside function does not have to be a polynomial. When the integrand contains a power of $\sin x$ multiplied by $\cos x$, the substitution $u = \sin x$ turns it into a power of $u$; likewise $u = e^x$

Substitutions that leave x in the integrand

Sometimes the $x$ terms do not all cancel after replacing $dx$. The substitution still works if $x$ itself is written in terms of $u$ by rearranging the substitution. A typical case is a polynomial multiplied by a power or a root of a linear expression, such as $\frac{x}{\sqrt{x + 1}}$

Different substitutions can lead to answers that look different. Two correct answers to the same indefinite integral always differ by a constant, which is absorbed into $C$. Expanding or factorising both forms, or differentiating them, shows whether they agree.

Definite integrals by substitution

For a definite integral, the limits $x = a$ and $x = b$ are values of $x$. After substituting, the integral is in terms of $u$, so the limits must be changed to the matching values of $u$: $u = g(a)$

With the new limits there is no need to return to $x$ at the end. The integral in $u$ gives the same number as the original integral in $x$.

The two integrals describe regions of equal area under different curves. For example, with $u = x^2 + 1$, the integral $\int_0^1 2x\left(x^2 + 1\right)^3 dx$

The limits in $u$ are not always in increasing order. If the substitution is a decreasing function, the lower limit in $x$ gives the larger value of $u$. The new limits are kept in the order they come from the substitution: the $u$-value from the lower $x$-limit goes at the bottom, whichever is larger. The property $\int_p^q f(u)\,du = -\int_q^p f(u)\,du$

Choosing a substitution

When no substitution is given, the form of the integrand suggests one. The aim is always the same: after substituting, only $u$ should remain, and the new integral should be one that is already known.

Here it was easier to differentiate $x = u^2$ with respect to $u$ than to differentiate $u = \sqrt{x}$

A substitution can also turn an integral into one of the standard forms of 5.15. With $u = e^x$, the integral $\int \frac{e^x}{e^{2x} + 1}\,dx$

Trigonometric substitutions

Expressions such as $\sqrt{a^2 - x^2}$ and $a^2 + x^2$

The substitution is written as $x$ in terms of $\theta$, so differentiate it with respect to $\theta$: for $x = a\sin\theta$, $\frac{dx}{d\theta} = a\cos\theta$

For an indefinite integral, the answer must be written back in terms of $x$. A right-angled triangle is the quickest way to do this. With $x = \tan\theta$, draw a triangle with opposite side $x$ and adjacent side 1, so that $\tan\theta = \frac{x}{1}$

Integration by parts

Reversing the product rule

Substitution cannot integrate a product of two unrelated functions, such as $x e^x$, because neither factor is the derivative of a function inside the other. For these, the product rule (5.6) is reversed. For two differentiable functions $u$ and $v$ of $x$,

Integrating both sides with respect to $x$ gives

and rearranging gives the formula for integration by parts.

The formula does not give the answer directly. It exchanges one integral, $\int u\frac{dv}{dx}\,dx$, for another, $\int v\frac{du}{dx}\,dx$

The formula also has a picture. If $u$ is plotted against $v$, the integral $\int u\,dv$ is the area between the curve and the $v$-axis, and $\int v\,du$

Using the formula

The working is clearest when the four quantities $u$, $\frac{du}{dx}$, $\frac{dv}{dx}$

The same choice, $u = x$, works whenever a power of $x$ multiplies $\sin kx$, $\cos kx$ or $e^{kx}$

Choosing u: the LIATE rule

The choice of $u$ decides whether the method works. Choosing the wrong factor makes the new integral harder than the original. For $\int x e^x\,dx$ with $u = e^x$ and $\frac{dv}{dx} = x$

and the power of $x$ has gone up instead of down.

A useful guide is the LIATE rule: choose as $u$ the factor whose type comes first in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The functions early in the list become simpler when differentiated; those late in the list are easy to integrate and stay much the same.

The rule is a guide, not a law. When both factors are of types T and E, either choice works (see cyclic integrals). For $\int x\ln x\,dx$, the rule puts $\ln x$ first, so $u = \ln x$ and $\frac{dv}{dx} = x$

Integrals with a hidden factor of 1

The functions $\ln x$, $\arcsin x$ and $\arctan x$ have simple derivatives but no integral in the standard list. Writing each as a product with 1, as in $\ln x = 1 \times \ln x$, allows integration by parts with $u = \ln x$

The result $\int \ln x\,dx = x\ln x - x + C$ is used often. It is worth remembering, and deriving it takes only one line.

Definite integrals by parts

For a definite integral, the formula becomes

The term $uv$ is evaluated between the limits straight away, and the second integral keeps the same limits. The limits are values of $x$ throughout, because integration by parts does not change the variable.

Repeated integration by parts

When $u$ is $x^2$ or a higher power, one application of the formula lowers the power by one but leaves another product. The method is then applied again to the new integral, keeping the same type of choice ($u$ = the power of $x$) each time, until no power of $x$ remains. A power $x^n$

The table for $\int x^2\sin x\,dx$ shows the method. Each row's derivative is multiplied by the next row's integral.

Reading the diagonals: $+x^2(-\cos x) - 2x(-\sin x) + 2\cos x$.

Cyclic integrals

For a product of an exponential and a sine or cosine, such as $\int e^x\sin x\,dx$, neither factor becomes simpler when differentiated: the derivatives of $e^x$ and $\sin x$ are again exponential and trigonometric. After two applications of integration by parts, the original integral appears again. Calling it $I$

Cyclic integrals with coefficients, such as $\int e^{2x}\cos x\,dx$, work the same way. The fractions from integrating $e^{2x}$ make the algebra longer, so the final step is to collect the multiples of $I$ carefully. The result is

Combining substitution and integration by parts

Some integrals need both methods. Usually a substitution comes first, to simplify the inside of a composite function, and integration by parts finishes the job.

Choosing an integration method

The hardest part of many integrals is deciding how to start. The table summarises the methods available in the course and the features of the integrand that point to each. Check the earlier rows first: a direct or inspection method is always quicker than substitution or parts when it applies.

Integration by parts could also find (c), with $u = x$ and $\frac{dv}{dx} = (x + 3)^{\frac{1}{2}}$

Exam-style questions on substitution and integration by parts

Cyclic integrals appear in non-calculator questions because the answer is exact and the method is long enough to test careful algebra.

Self-check

• To do:

  I can explain why integration by substitution reverses the chain rule, and why integration by parts reverses the product rule.

• To do:

  I can use a given or chosen substitution, replacing every $x$ and the $dx$, and write $x$ in terms of $u$ when it does not cancel.

• To do:

  I can change the limits of a definite integral to values of the new variable, including when they come out in decreasing order.

• To do:

  I can use a given trigonometric substitution such as $x = a\sin\theta$