Reverse Chain Rule

The chain rule allows us to differentiate in terms of something other than $x$ , and we end up with a product of two derivatives.

We can do this in reverse to integrate complicated functions where a function and its derivative both appear in that which is to be integrated.

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The Reverse Chain Rule

Recall: The chain rule.

$dfrac{dy}{dx}=dfrac{dy}{dt}dfrac{dt}{dx}$

Now the reverse chain rule:

${LARGE int}dfrac{du}{dx}f'(u)dx=f(u)+c$

The easiest way to spot when to use this is to look for a function and its derivative.

Look for a Function and Its Derivative

Some general results from the reverse chain rule appear so frequently that they are worth remembering.

${LARGE int}(n+1)f'(x)(f(x))^{n}dx=(f(x))^{n+1}+c$

The multiple in the integral is $n+1$ , not $n$ , which is very important to note.

Example 1: Reverse Chain Rule

Find the integral of $2cos(2x)e^{sin(2x)}$

[2 marks]

$dfrac{d}{dx}(sin(2x))=2cos(2x)$ so our integral is of the form ${LARGE int}dfrac{du}{dx}f'(u)dx$ where $f'(u)=e^{u}$ . Hence:

$int2cos(2x)e^{sin(2x)}dx=e^{sin(2x)}+c$

Example 2: A Function and its Derivative

Find $intdfrac{4}{x}(ln(x))^{3}dx$

[2 marks]

Since $dfrac{d}{dx}(ln(x))=dfrac{1}{x}$ , this is of the form $int(n+1)f'(x)(f(x))^{n}dx$ where $n=3$ and $f(x)=ln(x)$ . Hence:

${LARGE int}dfrac{4}{x}(ln(x))^{3}dx=(ln(x))^{4}+c$

Reverse Chain Rule Example Questions

Question 1: Integrate:

i) $3x^{2}e^{x^{3}}$

ii) $4x^{7}left(dfrac{1}{2}x^{8}+4right)^{2}$

iii) $x^{2}cosleft(dfrac{1}{3}x^{3}right)$

[9 marks]

i) $dfrac{d}{dx}(x^{3})=3x^{2}$ so this is in reverse chain rule form. Hence:

$begin{aligned}int3x^{2}e^{x^{3}}dx=e^{x^{3}}+cend{aligned}$

ii) $dfrac{d}{dx}left(dfrac{1}{2}x^{8}+4right)=4x^{7}$ so this is in reverse chain rule form. Hence:

$begin{aligned}int4x^{7}left(dfrac{1}{2}x^{8}+4right)^{2}dx=dfrac{1}{3}left(dfrac{1}{2}x^{8}+4right)^{3}+cend{aligned}$

iii) $dfrac{d}{dx}left(dfrac{1}{3}x^{3}right)=x^{2}$ so this is in reverse chain rule form. Hence:

$begin{aligned}int x^{2}cosleft(dfrac{1}{3}x^{3}right)dx=sinleft(dfrac{1}{3}x^{3}right)+cend{aligned}$

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Question 2: Integrate:

i) $12x^{3}e^{x^{4}}$

ii) $dfrac{2}{x}(log(x))^{5}$

iii) $4sec^{2}(x)e^{tan(x)}$

[9 marks]

i) $dfrac{d}{dx}(x^{4})=4x^{3}$ so this is in reverse chain rule form. Hence:

$begin{aligned}int12x^{3}e^{x^{4}}dx&=3int4x^{3}e^{x^{4}}dx[1.2em]&=3e^{x^{4}}+cend{aligned}$

ii) $dfrac{d}{dx}(log(x))=dfrac{1}{x}$ so this is in reverse chain rule form. Hence:

$begin{aligned}intdfrac{2}{x}(log(x))^{5}dx&=2timesdfrac{1}{6}(log(x))^{6}+c[1.2em]&=dfrac{1}{3}(log(x))^{6}+cend{aligned}$

iii) $dfrac{d}{dx}(tan(x))=sec^{2}(x)$ so this is in reverse chain rule form. Hence:

$begin{aligned}int4sec^{2}(x)e^{tan(x)}dx=4e^{tan(x)}+cend{aligned}$

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Question 3: Integrate:

i) $(6x^{2}+12x+2)e^{x^{3}+3x^{2}+x+5}$

ii) $(12x+4)(3x^{2}+2x+9)^{6}$

iii) $19cosec^{2}(3x)cot^{12}(3x)$

[9 marks]

i) $dfrac{d}{dx}(x^{3}+3x^{2}+x+5)=3x^{2}+6x+1$ so this is in reverse chain rule form. Hence:

$begin{aligned}&int(6x^{2}+12x+2)e^{x^{3}+3x^{2}+x+5}dx[1.2em]&=2int(3x^{2}+6x+1)e^{x^{3}+3x^{2}+x+5}dx[1.2em]&=e^{x^{3}+3x^{2}+x+5}+cend{aligned}$

ii) $dfrac{d}{dx}(3x^{2}+2x+9)=6x+2$ so this is in reverse chain rule form. Hence:

$begin{aligned}&int(12x+4)(3x^{2}+2x+9)^{6}dx[1.2em]&=2int(6x+2)(3x^{2}+2x+9)^{6}dx[1.2em]&=2timesdfrac{1}{7}(3x^{2}+2x+9)^{7}+c[1.2em]&=dfrac{2}{7}(3x^{2}+2x+9)^{7}+cend{aligned}$

iii) $dfrac{d}{dx}(cot(x))=-cosec^{2}(x)$ so this is in reverse chain rule form. Hence:

$begin{aligned}&int19cosec^{2}(3x)cot^{12}(3x)dx[1.2em]&=-19int-cosec^{2}(3x)cot^{12}(3x)dx[1.2em]&=-19timesdfrac{1}{3}timesdfrac{1}{13}cot^{13}(3x)+c[1.2em]&=-dfrac{19}{39}cot^{13}(3x)+cend{aligned}$