5.7 Integration by substitution

This is the integral equivalent of the chain rule. If z = f(x) and x = g(t) then the chain rule says, {dz\over dt} = {dz\over dx} {dx\over dt} . We can rearrange this “by multiplying by” dt to get,

. (This can be proven from the rule for finite steps,

which can be rearranged as

In the limit that δx goes to zero, as it must in the integral, we find the required result). This is the basic formula we need to convert an integral with respect to a new variable z. It is true as a substitution rule inside the integral, not as a general equality.

5.7.1 Type 1

Replace some function of x by z.

Example 5.14: 

Evaluate I ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2}x\mathop{sin}\nolimits ({x}^{2})\kern 1.66702pt dx.

Solution: 

Substitute z = {x}^{2} (try this), then dz = (dz∕dx)dx = 2xdx. We can only use this substitution if we can identify 2x\kern 1.66702pt dx as part of I. To that end write I = (1∕2){\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2}\mathop{ sin}\nolimits {x}^{2}\kern 1.66702pt 2x\kern 1.66702pt dx. We can now substitute for {x}^{2} = z and for 2xdx = dz, and thus I = {1\over 2}\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits \mathop{sin}\nolimits z\kern 1.66702pt dz, where the limits still need to be filled in. Since I is now an integral w.r.t. z, the limits must be starting and finishing values of z. At the start, where x = 0, z = {x}^{2} = 0. At the finish x = 2, z = 4, so

\eqalignno{ I & = {1\over 2}{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{4}\mathop{ sin}\nolimits z\kern 1.66702pt dz & & \cr & = {1\over 2}{[−\mathop{cos}\nolimits z]}_{0}^{4} & & \cr & = {1\over 2}(−\mathop{cos}\nolimits 4 + 1) & & \cr & ≈ 0.8268\quad . & & }

Note: The integrand, (i.e., the object being integrated) changes from x\mathop{sin}\nolimits ({x}^{2}) to (1∕2)\mathop{sin}\nolimits z. Part of this change is due to the change from dx to dz.

Note: The integration limits change (for definite integrals only).

Example 5.15: 

Calculate the indefinite integral I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { a}^{2} + {x}^{2}}.

Solution: 

Use substitution, and take z = (x∕a), dz = (1∕a)dx, x = az.

\begin{eqnarray*} I& =& \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { a}^{2} + {x}^{2}}\kern 1.66702pt a\kern 1.66702pt {1\over a}dx%& \\ & =& \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over { a}^{2} + {a}^{2}{z}^{2}}adz%& \\ & =& \mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {a\over { a}^{2}(1 + {z}^{2})}dz %& \\ & =& {1\over a}\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits {dz\over (1 + {z}^{2})} %& \\ & =& {1\over a}{\mathop{tan}\nolimits }^{−1}z + c. %& \\ \end{eqnarray*}

Finally we must substitute back using z = x∕a,

\class{boxed}{I = {1\over a}{\mathop{tan}\nolimits }^{−1}\left ({x\over a}\right ) + c. }

Several standard integrals can be generalised using this substitution (left as exercise).

Example 5.16: 

Evaluate I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits \kern 1.66702pt {1\over \sqrt{{a}^{2 } − {x}^{2}}}dx\quad .

Solution: 

Using the substitution x = az we find

I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dz\over \sqrt{1 − {z}^{2}}} ={\mathop{ sin}\nolimits }^{−1}z + c ={\mathop{ sin}\nolimits }^{−1}(x∕a + c)

Thus

\class{boxed}{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits \kern 1.66702pt {1\over \sqrt{{a}^{2 } − {x}^{2}}}dx ={\mathop{ sin}\nolimits }^{−1}(x∕a + c)\quad . }

5.7.2 Type 2

Replace x by a function of z. Sometimes, instead of putting

e.g., z = {x}^{2}, we replace x directly by putting

This is really same as using (5.1) since we can rearrange this equation, (i.e., solve for x) to get (5.2). However, we can work directly from (5.2) by calculating dx∕dz. We then use the formula

(Remember that we also must change limits on a definite integral!)

Example 5.17: 

Evaluate I ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{4} {1\over 1 + \sqrt{x}}\kern 1.66702pt dx.

Solution: 

Put x = {z}^{2}, dx∕dz = 2z, dx = 2zdz. The limits change, x = 0 ⇒ z = 0, x = 4 ⇒ z = 2. We obtain

\begin{eqnarray*} I& =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2} {1\over 1 + \sqrt{{z}^{2}}}2zdz ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2} {2z\over (1 + z)}dz %& \\ & =& {\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2}{2(z + 1) − 2\over (z + 1)} dz ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{2}\left (2 − {2\over (z + 1)}\right )dz %& \\ & =& {[2z − 2\mathop{ln}\nolimits (z + 1)]}_{0}^{2} = (4 − 2\mathop{ln}\nolimits 3) − 0 = 4 − 2\mathop{ln}\nolimits 3 = 1.8028%& \\ \end{eqnarray*}