Editing Integration by substitution (section)

== Substitution for a single variable ==

=== Introduction (indefinite integrals) ===
Before stating the result [[mathematical rigor|rigorously]], consider a simple case using [[indefinite integral|indefinite integrals]].

Compute <math display="inline">\int(2x^3+1)^7(x^2)\,dx.</math><ref>{{harvnb|Swokowski|1983|p=258}}</ref>

Set <math>u=2x^3+1.</math> This means <math display="inline">\frac{du}{dx}=6x^2,</math> or as a [[differential form]], <math display="inline">du=6x^2\,dx.</math> Now: 
<math display="block">\begin{aligned}
    \int(2x^3 +1)^7(x^2)\,dx
    &= \frac{1}{6}\int\underbrace{(2x^3+1)^{7}}_{u^{7}}\underbrace{(6x^2)\,dx}_{du} \\
    &= \frac{1}{6}\int u^{7}\,du \\
    &= \frac{1}{6}\left(\frac{1}{8}u^{8}\right)+C \\
    &= \frac{1}{48}(2x^3+1)^{8}+C,
\end{aligned}</math>
where <math>C</math> is an arbitrary [[constant of integration]].

This procedure is frequently used, but not all integrals are of a form that permits its use. In any event, the result should be verified by differentiating and comparing to the original integrand.
<math display="block">\frac{d}{dx}\left[\frac{1}{48}(2x^3+1)^{8}+C\right] = \frac{1}{6}(2x^3+1)^{7}(6x^2) = (2x^3+1)^7(x^2).</math>
For definite integrals, the limits of integration must also be adjusted, but the procedure is mostly the same.

=== Statement for definite integrals ===
Let <math>g:[a,b]\to I</math> be a [[differentiable function]] with a [[continuous function|continuous]] derivative, where <math>I \subset \mathbb{R}</math> is an [[interval (mathematics)|interval]]. Suppose that <math>f:I\to\mathbb{R}</math> is a [[continuous function]]. Then:<ref>{{harvnb|Briggs|Cochran|2011|p=361}}</ref>
<math display="block">\int_a^b f(g(x))\cdot g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\ du. </math>

In Leibniz notation, the substitution <math>u=g(x)</math> yields: 
<math display="block">\frac{du}{dx} = g'(x).</math>
Working heuristically with [[infinitesimal]]s yields the equation
<math display="block">du = g'(x)\,dx,</math>
which suggests the substitution formula above.  (This equation may be put on a rigorous foundation by interpreting it as a statement about [[differential form]]s.)  One may view the method of integration by substitution as a partial justification of [[Leibniz's notation]] for integrals and derivatives.

The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be read from left to right or from right to left in order to simplify a given integral. When used in the former manner, it is sometimes known as '''''u''-substitution''' or '''''w''-substitution''' in which a new variable is defined to be a function of the original variable found inside the [[function composition|composite]] function multiplied by the derivative of the inner function. The latter manner is commonly used in [[trigonometric substitution]], replacing the original variable with a [[trigonometric function]] of a new variable and the original [[differential of a function|differential]] with the differential of the trigonometric function.

=== Proof ===

Integration by substitution can be derived from the [[fundamental theorem of calculus]] as follows.  Let <math>f</math> and <math>g</math> be two functions satisfying the above hypothesis that <math>f</math> is continuous on <math>I</math> and <math>g'</math> is integrable on the closed interval <math>[a,b]</math>.  Then the function <math>f(g(x))\cdot g'(x)</math> is also integrable on <math>[a,b]</math>.  Hence the integrals
<math display="block">\int_a^b f(g(x))\cdot g'(x)\ dx</math>
and
<math display="block">\int_{g(a)}^{g(b)} f(u)\ du</math>
in fact exist, and it remains to show that they are equal.

Since <math>f</math> is continuous, it has an [[antiderivative]] <math>F</math>. The [[function composition|composite function]] <math>F \circ g</math> is then defined. Since <math>g</math> is differentiable, combining the [[chain rule]] and the definition of an antiderivative gives:
<math display="block">(F \circ g)'(x) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x).</math>

Applying the [[fundamental theorem of calculus]] twice gives:
<math display="block">\begin{align}
\int_a^b f(g(x)) \cdot g'(x)\ dx
&= \int_a^b (F \circ g)'(x)\ dx \\
&= (F \circ g)(b) - (F \circ g)(a) \\
&= F(g(b)) - F(g(a)) \\
&= \int_{g(a)}^{g(b)} f(u)\, du,
\end{align}</math>
which is the substitution rule.

=== Examples: Antiderivatives (indefinite integrals) ===

Substitution can be used to determine [[antiderivative]]s. One chooses a relation between <math>x</math> and <math>u,</math> determines the corresponding relation between <math>dx</math> and <math>du</math> by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between <math>x</math> and <math>u</math> is then undone.

==== Example 1 ====
Consider the integral:
<math display="block">\int x \cos(x^2+1)\ dx.</math>
Make the substitution <math display="inline">u = x^{2} + 1</math> to obtain <math>du = 2x\ dx,</math> meaning <math display="inline">x\ dx = \frac{1}{2}\ du.</math> Therefore:
<math display="block">\begin{align}
\int x \cos(x^2+1) \,dx
&= \frac{1}{2} \int 2x \cos(x^2+1) \,dx \\[6pt]
&= \frac{1}{2} \int\cos u\,du \\[6pt]
&= \frac{1}{2}\sin u + C \\[6pt]
&= \frac{1}{2}\sin(x^2+1) + C,
\end{align}</math>
where <math>C</math> is an arbitrary [[constant of integration]].

==== Example 2: Antiderivatives of tangent and cotangent ====

The [[tangent function]] can be integrated using substitution by expressing it in terms of the sine and cosine: <math>\tan x = \tfrac{\sin x}{\cos x}</math>.

Using the substitution <math>u = \cos x</math> gives <math>du = -\sin x\,dx</math> and
<math display="block">\begin{align}
   \int \tan x \,dx &= \int \frac{\sin x}{\cos x} \,dx \\
    &= \int -\frac{du}{u} \\
    &= -\ln \left|u\right| + C \\
    &= -\ln \left|\cos x\right| + C \\
    &= \ln \left|\sec x\right| + C.
 \end{align}</math>

The [[cotangent function]] can be integrated similarly by expressing it as <math>\cot x = \tfrac{\cos x}{\sin x}</math> and using the substitution <math>u = \sin{x}, du = \cos{x}\,dx</math>:
<math display="block">\begin{align}
   \int \cot x \,dx &= \int \frac{\cos x}{\sin x} \,dx \\
    &= \int \frac{du}{u} \\
    &= \ln \left|u\right| + C \\
    &= \ln \left|\sin x\right| + C.
 \end{align}</math>

=== Examples: Definite integrals ===
When evaluating definite integrals by substitution, one may calculate the antiderivative fully first, then apply the boundary conditions. In that case, there is no need to transform the boundary terms. Alternatively, one may fully evaluate the indefinite integral ([[Integration by substitution#Examples: Antiderivatives|see above]]) first then apply the boundary conditions. This becomes especially handy when multiple substitutions are used.

==== Example 1 ====
Consider the integral:
<math display="block">\int_0^2 \frac{x}{\sqrt{x^2+1}} dx.</math>
Make the substitution <math display="inline">u = x^{2} + 1</math> to obtain <math>du = 2x\ dx,</math> meaning <math display="inline">x\ dx = \frac{1}{2}\ du.</math> Therefore:
<math display="block">\begin{align}
\int_{x=0}^{x=2} \frac{x}{\sqrt{x^2+1}} \ dx
&= \frac{1}{2} \int_{u=1}^{u=5} \frac{du}{\sqrt{u}} \\[6pt]
&= \frac{1}{2} \left(2\sqrt{5}-2\sqrt{1}\right) \\[6pt]
&= \sqrt{5}-1.
\end{align}</math>
Since the lower limit <math>x = 0</math> was replaced with <math>u = 1,</math> and the upper limit <math>x = 2</math> with <math>2^{2} + 1 = 5,</math> a transformation back into terms of <math>x</math> was unnecessary.

==== Example 2: [[Trigonometric substitution]] ====

For the integral
<math display="block">\int_0^1 \sqrt{1-x^2}\,dx,</math>
a variation of the above procedure is needed. The substitution <math>x = \sin u</math> implying <math>dx = \cos u \,du</math> is useful because <math display="inline">\sqrt{1-\sin^2 u} = \cos u.</math> We thus have:
<math display="block">\begin{align}
\int_0^1 \sqrt{1-x^2}\ dx
&= \int_0^{\pi/2} \sqrt{1-\sin^2 u} \cos u\ du \\[6pt]
&= \int_0^{\pi/2} \cos^2 u\ du \\[6pt]
&= \left[\frac{u}{2} + \frac{\sin(2u)}{4}\right]_0^{\pi/2} \\[6pt]
&= \frac{\pi}{4} + 0 \\[6pt]
&= \frac{\pi}{4}.
\end{align}</math>

The resulting integral can be computed using [[integration by parts]] or a [[List of trigonometric identities#Multiple-angle and half-angle formulae|double angle formula]], <math display="inline">2\cos^{2} u = 1 + \cos (2u),</math> followed by one more substitution. One can also note that the function being integrated is the upper right quarter of a circle with a radius of one, and hence integrating the upper right quarter from zero to one is the geometric equivalent to the area of one quarter of the unit circle, or <math>\tfrac \pi 4.</math>