Integration by Reverse Chain Rule
By recalling the chain rule, the Integration Reverse Chain Rule comes from the usual chain rule of differentiation. This skill is to be used to integrate composite functions such as
\( e^{x^2+5x}, \cos{(x^3+x)}, \log_{e}{(4x^2+2x)} \).
Let’s take a close look at the following example of applying the chain rule to differentiate and then reverse its order to obtain the result of its integration.
\( \begin{aligned} \displaystyle
\frac{d}{dx} \sin{x^2} &= \sin{x^2} \times \frac{d}{dx} x^2 \\
&= \sin{x^2} \times 2x \\
&= 2x \sin{x^2} \\
2x \sin{x^2} &= \frac{d}{dx} \sin{x^2} \\
\therefore \int{2x \sin{x^2}} dx &= \sin{x^2} +C
\end{aligned} \)
Integration Reverse Chain Rule for Exponential Functions
(a) Differentiate \( e^{3x^2+2x-1} \).
\( \begin{aligned} \displaystyle
\frac{d}{dx} e^{3x^2+2x+1} &= e^{3x^2+2x-1} \times \frac{d}{dx} (3x^2+2x-1) \\
&= e^{3x^2+2x-1} \times (6x+2) \\
&= (6x+2)e^{3x^2+2x-1}
\end{aligned} \)
(b) Integrate \( (3x+1)e^{3x^2+2x-1} \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
(6x+2)e^{3x^2+2x-1} &= \frac{d}{dx} e^{3x^2+2x-1} &\color{red} \text{from (a)} \\
\int{(6x+2)e^{3x^2+2x-1}} dx &= e^{3x^2+2x-1} \\
\therefore \int{(3x+1)e^{3x^2+2x-1}} dx &= \frac{1}{2} e^{3x^2+2x-1} +C
\end{aligned} \)
Integration Reverse Chain Rule for Trigonometric Functions
(a) Differentiate \( \cos{3x^3} \).
\( \begin{aligned} \displaystyle
\frac{d}{dx} \cos{3x^3} &= -\sin{3x^3} \times \frac{d}{dx} (3x^3) \\
&= -\sin{3x^3} \times 9x^2 \\
&= -9x^2 \sin{3x^3}
\end{aligned} \)
(b) Integrate \( x^2 \sin{3x^3} \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
-9x^2 \sin{3x^3} &= \frac{d}{dx} \cos{3x^3} &\color{red} \text{from (a)} \\
\int{-9x^2 \sin{3x^3}} dx &= \cos{3x^3} \\
\therefore \int{x^2 \sin{3x^3}} dx &= -\frac{1}{9} \cos{3x^3} + C
\end{aligned} \)
Integration Reverse Chain Rule for Logarithmic Functions
(a) Differentiate \( \log_{e} \sin{x} \).
\( \begin{aligned} \displaystyle
\frac{d}{dx} \log_{e} \sin{x} &= \frac{1}{\sin{x}} \times \frac{d}{dx} \sin{x} \\
&= \frac{1}{\sin{x}} \times \cos{x} \\
&= \cot{x}
\end{aligned} \)
(b) Hence, integrate \( \cot{x} \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\cot{x} &= \frac{d}{dx} \log_{e} \sin{x} &\color{red} \text{from (a)} \\
\therefore \int{\cot{x}} dx &= \log_{e} \sin{x} +C
\end{aligned} \)
Quiz for your Practice
Differentiate \( \displaystyle \log_{e}{\cos{x^2}} \), hence find \( \displaystyle \int{x \tan{x^2}} dx\). Have Fun!
Feel free to let us know if you are unsure how to do this in case 🙂
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Your integral with 2x sin(x^2) should be -cos(x^2) + c.
Similarly, your integral with x^2 cos(3x^3) should be sin(3x^3)/9 + c
Good pick. Thank-you!