Integration by substitution

Let $Q$ be a composition of two functions $P$ and $g$, say $Q(x)=P[g(x)]$ for all $x$ in some interval $I.$ If we know the derivative of $P$, say MATH the chain rule tells us that the derivative of $Q$ is given by the formula MATH Since $P^{\prime }=f$, this states that MATH. In other words,

MATH implies MATH

In Leibniz notation, this statement can be written as follows: If we have the integration formula
MATH
then we have also have the more general formula
MATH

For example, if $f(x)=\cos x,$ then
MATH
holds with $P(x)=\sin x,$ so
MATH
becomes
MATH
In particular, if $g(x)=x^3,$ this gives us
MATH
a result that is easily verified directly since the derivative of $\sin x^3$ is $3x^2\cos x^3$. We notice that the general formula in
MATH
is related to
MATH
by a simple mechanical process. Suppose we replace $g(x)$ everywhere in
MATH
by a new symbol $u$ and replace $g^{\prime }(x)$ by $du/dx$, the Leibniz notation for derivatives. Then it becomes
MATH
At this stage the temptation is strong to replace the combination $\frac{du}{dx}dx$ by $du$. If we do the last formula becomes
MATH
Notice that this has exactly the same form as
MATH
except that the symbol $u$ appears everywhere instead of $x$. In other words, every integration formula such as
MATH
can be yield a more general integration formula if we simply substitute symbols. We replace in
MATH
by a new symbol $u$ to obtain
MATH
and then we think of $u$ as representing a new function of $x$, say $u=g(x)$. Then we replace the symbol $du$ by the combination $g^{\prime }(x)dx,$ and equation
MATH
reduces to the general formula in
MATH

For example, if we replace $x$ by $u$ in the formula MATH we obtain
MATH
In this latter formula, $u$ may be replaced by $g(x)$ and $du$ by $g^{\prime }(x)dx$, and a correct integration formula,
MATH
results.

When this mechanical process is used in reverse, it becomes the method of integration by substitution. The object of the method is to transform an integral with a complicated integrand, such as MATH, into a more familiar integral, such as $\int \cos udu$. The method is applicable whenever the original integral can be written in the form
MATH
since the substitution
MATH
transforms this to $\int f(u)du.$ If we succeed in carrying out the integration indicated by $\int f(u)du$, we obtain a primitive, say $P(u)$, and then the original integral may be evaluated by replacing $u$ by $g(x)$ in the formula for $P(u).$

EXAMPLE 1. Integrate MATH

Solution. Let us keep in mind that we are trying to write $x^3\cos x^4$ in the form MATH with a suitable choice of $f$ and $g$. Since $\cos x^4$ is a composition, this suggests that we take $f(x)=\cos x$ and $g(x)=x^4$ so $\cos x^4$ becomes $f[g(x)].$ This choice of $g$ gives MATH, and hence MATH. The extra factor $4$ is easily taken care of by multiplying and dividing the integrand by $4$. Thus we have
MATH
Now, we make the substitution $u=g(x)=x^4$, MATH, and obtain
MATH
Replacing $u$ by $x^4$ in the end result, we obtain the formula
MATH
which can be verified directly by differentiation. After a little practice one can perform some of the above steps mentally, and the entire calculation can be given more briefly as follows: Let $u=x^4$. Then $du=4x^3dx$, and we obtain
MATH
Notice that the method works in this example because the factor $x^3$ has an exponent one less than the power of $x$ which appears in $\cos x^4$.

EXAMPLE 2. Integrate MATH.

Solution. Let $u=\cos x.$ Then $du=-\sin xdx,$ and we get
MATH
Again, the final result is easily verified by differentiation.

EXAMPLE 3. Integrate MATH.

Solution. Let MATH Then MATH or $dx/\sqrt{x}=2du,$ Hence
MATH

EXAMPLE 4. Integrate MATH

Solution. Let $u=1+x^2.$ Then $du=2xdx$ so $xdx=\frac 12du,$and we obtain
MATH

The method of substitution is, of course, also applicable to definite integrals.

EXAMPLE 5. Evaluate MATH

Solution. Let $u=x^2+2x+3.$ Then $du=(2x+2)dx$, so that
MATH
Now we obtain new limits of integration by noting that $u=11$ when $x=2,$ and that $u=18$ when $x=3.$Then we write
MATH

The same result is arrived at by expressing everything in terms of $x$. Thus we have
MATH

Now we prove a general theorem which justifies the process used in Example 5.

THEOREM SUBSTITUTION THEOREM FOR INTEGRALS. Assume $g$ has a continuous derivative $g^{\prime }$ on an open interval $I$. Let $J$ be the set of values taken by $g$ on $I$ and assume that $f$ is continuous on $J$. Then for each $x$ and $c$ in $I$, we have
MATH

Proof. Let $a=g(c)$ and define two new functions $P$ and $Q$ as follows:
MATH
Since $P$ and $Q$ are indefinite integrals of continuous functions they have derivatives given by the formulas
MATH
Now, let $R$ denote the composite function, $R(x)=P[g(x)].$ Using the chain rule, we find
MATH

Applying the second fundamental theorem twice, we obtain
MATH
and
MATH
This shows that the two integrals are equal.

Integration by Parts


MATH

In order to find MATH, the integration by parts formula follows the steps
MATH

MATH

MATH

MATH

MATH

MATH
to obtain


MATH

A moment's reflection shows that the result was obtained by taking successive derivatives of $x^5$ and taking successive integrals of $\cos \ x$ and combine the result as indicated below:
MATH
The procedure can be justified from
MATH

MATH
where MATH

MATH

In view of the above formula we see that if $p$ is a polynomial of degree $n$ then
MATH

Table of Integrals for $\int x^n\exp (x)dx$

The integration by parts formula yields
MATH
Thus if MATH, then the polynomial $p_n(x)$ is obtained from the previous polynomial $p_{n-1}(x)$ by multiplying by $n$ and adding the term $x^n.$

Exercise: Build up the following table of coefficients for the $p_n^{\prime}s.$

Exercise: Compute $\int \ln (x)^ndx$ for MATH
MATH

It is clear now that if $p(x)$ is a polynomial of degree $n$ then
MATH

MATH

Table of Integrals for MATH, MATH

a. Compute MATH for $n=1,2,3,4,5.$

b. Write down the reduction formula for MATH and MATH

c. Compute the polynomials appeared in MATH and in MATH for MATH

Chebyshev Polynomials

a. Establish the identity
MATH

b. Set $\cos \ \theta =x$, MATH. Show that
MATH

c. List the coefficients of $T_n$ for $n=0,1,2,...,12.$

d. Draw the graphs of $T_1$, $T_2$, $T_3,...,T_6$ on the square MATH

e. Calculate the coefficients $c_k$ in the expansion MATH for $n=1,2,\cdots ,12.$

f. Find the coefficients $a_k$ in the integral
MATH
for MATH

MATH

a. Establish the identities


MATH


MATH

b. Deduce from the identities
MATH

MATH

MATH

MATH

MATH

c. Compute the coefficients $a_k$ and $b_k$ in the expansions
MATH

MATH
for MATH

d. Compute MATH for MATH

MATH

The reduction formula follows from
MATH

MATH
Hence
MATH

MATH

MATH

For even $n$, write MATH to reduce it to MATH and expand the binomial expression. For odd $n$, set MATH, MATH and when $\tan ^2\theta $ appears, use the identity MATH

Example. Find MATH.

Write $u = \sec \ \theta $, MATH

MATH, $v = \tan \ \theta $

MATH

MATH

Therefore
MATH
and so
MATH

This document created by Scientific WorkPlace 4.0.