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 🙂
Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume
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!