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Substitution and Reversing the Chain Rule · Tutorial 1152 of 1495

Successive Substitutions In A Single Integral

Follow a single integral through two substitutions, keeping each differential and transformed factor consistent from start to finish.

Intermediate 10 min read

What You'll Learn

  • Recognize when a first substitution simplifies an integral but leaves another substitution to make.
  • Track the transformed integrand and differential at each stage.
  • Apply two substitutions in sequence using the chain rule.
  • Transform definite-integral bounds through both variables.
  • Check an antiderivative by differentiating it.
  • Identify and correct a missing constant factor in a substitution.

Why Use Two Substitutions?

A substitution is useful when it replaces a complicated expression with a simpler variable and transforms the differential along with it. Sometimes that first change makes the integral easier to read, but the new integral still contains a composition that is not yet in a directly integrable form. A second substitution can then finish the calculation.

The key is to treat the changes in sequence. After the first substitution, rewrite the entire integral in the new variable: the integrand, the differential, and, for a definite integral, the bounds. Only then choose and apply the next substitution. Keeping these stages distinct makes constants and factors easier to track.

Central idea: Successive substitution means applying one valid change of variable, rewriting the whole integral in that variable, and then applying another valid change to the resulting integral.

The Rule Behind Successive Substitutions

For an indefinite integral, the substitution rule comes from the chain rule. If \(F'(u)=f(u)\) and \(u=g(x)\), then

$$ \frac{d}{dx}F(g(x))=f(g(x))g'(x). $$

Thus, wherever the functions involved are differentiable, an integrand of the form \(f(g(x))g'(x)\) has antiderivative \(F(g(x))\). In differential notation, the substitution \(u=g(x)\) uses \(du=g'(x)\,dx\). This notation records how the differential factor changes; it does not mean that other factors in the integrand can be left untransformed.

For successive substitutions, suppose \(u=g(x)\), then \(v=h(u)\), and the final integral in \(v\) has integrand \(H(v)\). If \(K'(v)=H(v)\), the chain rule gives

$$ \frac{d}{dx}K(h(g(x))) = H(h(g(x)))h'(g(x))g'(x). $$

Therefore, the two substitutions work when the original integrand, including all its factors, becomes exactly \(H(v)\,dv\) after both changes. The proof is the chain rule applied twice. In practice, the safest method is to show each transformed integral explicitly rather than trying to combine both changes in one step.

Condition to check: Each substitution must be differentiable on the interval in question, and its differential relation must account for the corresponding factor in the integrand. For a definite integral, also change the limits to the values of the new variable at the original endpoints.

A Reliable Sequence of Steps

1
Make the first substitution.
Choose a variable that simplifies part of the integrand, and write its differential relation.
2
Rewrite every factor.
Express the full integrand and differential in the new variable. Do not carry an old-variable factor into the next line without converting it.
3
Make the second substitution.
Use a variable that simplifies what remains, and again transform the full integrand and differential.
4
Integrate and return to the original variable.
For an indefinite integral, express the answer in the original variable and check it by differentiation.

Worked Example: Two Substitutions in a Definite Integral

Worked Example: Two Substitutions in a Definite Integral

Evaluate

$$ \int_0^1 \frac{x}{\sqrt{1+x^2}\left(1+\sqrt{1+x^2}\right)} \,dx. $$

First substitution. Set \(u=1+x^2\). Then \(du=2x\,dx\), so \(x\,dx=\frac12\,du\). When \(x=0\), \(u=1\); when \(x=1\), \(u=2\). Also, \(\sqrt{1+x^2}=\sqrt{u}\). Therefore the integral becomes

$$ \int_0^1 \frac{x}{\sqrt{1+x^2}\left(1+\sqrt{1+x^2}\right)} \,dx = \frac12\int_1^2 \frac{1}{\sqrt{u}(1+\sqrt{u})}\,du. $$

The first substitution has removed \(x\), but a square root of \(u\) remains. Use a second substitution, \(v=\sqrt{u}\). Since \(u=v^2\), \(du=2v\,dv\). The new limits are \(v=1\) and \(v=\sqrt{2}\). Thus

$$ \frac12\int_1^2 \frac{1}{\sqrt{u}(1+\sqrt{u})}\,du = \frac12\int_1^{\sqrt{2}} \frac{2v}{v(1+v)}\,dv = \int_1^{\sqrt{2}}\frac{1}{1+v}\,dv. $$

An antiderivative of \(1/(1+v)\) is \(\ln(1+v)\), since its derivative is \(1/(1+v)\). Hence

$$ \int_1^{\sqrt{2}}\frac{1}{1+v}\,dv = \ln(1+\sqrt{2})-\ln(2). $$

Check. An antiderivative of the original integrand is \(F(x)=\ln(1+\sqrt{1+x^2})\). Differentiating gives

$$ F'(x) = \frac{1}{1+\sqrt{1+x^2}}\, \frac{x}{\sqrt{1+x^2}} = \frac{x}{\sqrt{1+x^2}\left(1+\sqrt{1+x^2}\right)}. $$

Thus the integrand is correct, and \(F(1)-F(0)=\ln(1+\sqrt{2})-\ln(2)\), matching the value obtained by the two substitutions.

Worked Example: A Logarithm After the First Change

Worked Example: A Logarithm After the First Change

Find an antiderivative of

$$ \int \frac{e^x}{(1+e^x)\sqrt{\ln(1+e^x)}}\,dx. $$

First substitution. Let \(u=1+e^x\). Then \(du=e^x\,dx\). The factor \(1+e^x\) becomes \(u\), while the logarithm becomes \(\ln u\). Therefore

$$ \int \frac{e^x}{(1+e^x)\sqrt{\ln(1+e^x)}}\,dx = \int\frac{1}{u\sqrt{\ln u}}\,du. $$

The first substitution has removed the exponential expression, but the logarithm and its square root remain. Let \(v=\ln u\). Then \(dv=\frac{1}{u}\,du\), so the remaining integral is

$$ \int\frac{1}{u\sqrt{\ln u}}\,du = \int v^{-1/2}\,dv = 2\sqrt{v}+C. $$

Returning to \(x\), an antiderivative is

$$ 2\sqrt{\ln(1+e^x)}+C. $$

All expressions are defined for real \(x\): \(1+e^x>1\), so \(\ln(1+e^x)>0\).

Check. Differentiating the result gives

$$ \frac{d}{dx}\left(2\sqrt{\ln(1+e^x)}\right) = \frac{1}{\sqrt{\ln(1+e^x)}}\, \frac{e^x}{1+e^x} = \frac{e^x}{(1+e^x)\sqrt{\ln(1+e^x)}}. $$

This is the original integrand, so the antiderivative is verified.

Worked Example: An Exponential Appearing After the First Change

Worked Example: An Exponential Appearing After the First Change

Find an antiderivative of

$$ \int\frac{x e^{x^2}}{1+e^{x^2}}\,dx. $$

First substitution. Set \(u=x^2\), so \(du=2x\,dx\) and \(x\,dx=\frac12\,du\). The integral becomes

$$ \int\frac{x e^{x^2}}{1+e^{x^2}}\,dx = \frac12\int\frac{e^u}{1+e^u}\,du. $$

Now set \(v=1+e^u\). Then \(dv=e^u\,du\), and the integral becomes

$$ \frac12\int\frac{e^u}{1+e^u}\,du = \frac12\int\frac{1}{v}\,dv = \frac12\ln(v)+C. $$

Since \(v=1+e^{x^2}>0\), the answer can be written as

$$ \frac12\ln(1+e^{x^2})+C. $$

Check. Differentiate:

$$ \frac{d}{dx}\left(\frac12\ln(1+e^{x^2})\right) = \frac12\frac{e^{x^2}}{1+e^{x^2}}(2x) = \frac{x e^{x^2}}{1+e^{x^2}}. $$

The derivative matches the original integrand.

Why the Intermediate Lines Matter

Each substitution changes more than the name of a variable. It changes the differential and may change several appearances of the inner expression. For example, in the last worked example the first change gave \(x\,dx=\frac12\,du\), not \(du\). That factor of \(\frac12\) must remain when the second substitution is made.

A common mistake is to omit that factor:

$$ \int\frac{x e^{x^2}}{1+e^{x^2}}\,dx \overset{\text{wrong}}{=} \int\frac{e^u}{1+e^u}\,du = \ln(1+e^{x^2})+C. $$

The error is that \(du=2x\,dx\), so \(x\,dx=\frac12\,du\). The correct calculation retains the one-half through both substitutions:

$$ \int\frac{x e^{x^2}}{1+e^{x^2}}\,dx = \frac12\int\frac{e^u}{1+e^u}\,du = \frac12\ln(1+e^{x^2})+C. $$

The wrong answer is easy to detect by differentiating: its derivative is \(2x e^{x^2}/(1+e^{x^2})\), twice the required integrand. A derivative check is especially useful after successive substitutions because small factors can be lost at either stage.

Using Successive Substitutions Carefully

When performing two changes of variable, keep a clear record of what each variable means. If the first is \(u=g(x)\) and the second is \(v=h(u)\), then the final expression for the answer must ultimately be written in terms of \(x\), unless the integral is definite and its transformed limits have already been used.

For a definite integral, track the endpoints in order. If the original limits are \(a\) and \(b\), the first substitution changes them to \(g(a)\) and \(g(b)\); the second changes those to \(h(g(a))\) and \(h(g(b))\). In the first worked example, the bounds moved from \(0,1\) to \(1,2\), then to \(1,\sqrt{2}\). The order of the limits follows the endpoint values; it should not be changed just because the variable has changed.

The purpose of successive substitution is not to make an integral look more complicated by introducing extra variables. It is to turn a remaining composition into an integral that can be evaluated cleanly. Showing every transformed factor makes that purpose visible and gives a direct route to checking the result.

Check Your Understanding

For each question, keep track of the differential as carefully as the transformed expression.

  1. In the first worked example, why does the first substitution produce a factor of \(\frac12\)?
  2. For \(u=1+e^x\), what are \(du\) and the resulting integrand in the second worked example?
  3. In the third worked example, what is the relation between \(du\) and \(dv\) when \(v=1+e^u\)?
  4. Why is differentiating a proposed antiderivative a useful check after two substitutions?
  5. For a definite integral, how are the limits changed when a second substitution follows the first?

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