Reverse Chain Rule
\(\DeclareMathOperator{cosec}{cosec}\)
What is Reverse Chain Rule: Example
- A function in the form: \(\frac{f^{'}(x)}{f(x)}\)
- \(\int \frac{2x}{x^2+1} \, dx\)
- Consider \(y=\ln |x^2+1|\)
- \(\frac{dy}{dx}=\frac{1}{x^2+1}\times 2x\) (using chain rule)
- = \(\frac{2x}{x^2+1}\)
- \(\int \frac{2x}{x^2+1} \, dx=\ln |x^2+1|+c\)
Using fractions
Example 1
- E.G. \(\int x(x^2+1)^6\, dx\)
- Try \((x^2+1)^7\)
- \(\frac{d}{dx}=7(x^2+1)^6\)
- \(= 14x(x^2+1)^6\)
- \(= \frac{1}{14}(x^2+1)^7\)
Example 2
- E.G. \(\int \sin^4x\cos x\, dx\)
- Try \(\sin^5x\)
- \(\frac{d}{dx}=5\sin^4x\cos x\)
- \(\int =\frac{1}{5}\sin^5x\)
Using \(\ln\)
Example 4
- E.G. \(\int \frac{\cos x}{3+2\sin x}\, dx\)
- Try \(\ln|3+2\sin x|\)
- \(\frac{1}{3+2\sin x}\times 2\cos x\)
- \(= \frac{2\cos x}{3+2\sin x}\)
- This result is double of the equation we want so we put a \(\frac{1}{2}\) in front of our \(\ln\) equation we tried earlier.
- \(\frac{1}{2}\ln|3+2\sin x|+c\)
Ex4 Q1a
- E.G. \(\frac{e^{2x}}{e^{2x}+1}\)
- Try: \(\ln|e^{2x}+1|\)
- \(= \frac{1}{e^{2x}+1}\times2e^{2x}\)
- \(= \frac{2x}{x^2+4}\)
- So our final answer is:\(\frac{1}{2}\ln|x^2+4|+c\)
Ex4 Q1b
- E.G. \(\frac{e^{2x}}{e^{2x}+1}\)
- Try: \(\ln|e^{2x}+1|\)
- \(= \frac{1}{e^{2x}}\times2e^{2x}\)
- \(= \frac{2e^{2x}}{2e^{2x}+1}\)
- So our final answer is:\(\frac{1}{2}\ln|e^{2x}+1|+c\)
Using \(\int kf^{'}(x)(f(x))^n \, dx\)
Given that \(\int ^\theta_{0}5\tan x \sec^4x \, dx=\frac{15}{4}\) find the exact value of \(\theta\)
- Let \(I=\int ^\theta_{0}5\tan x \sec^4x \, dx\)
- Let \(y=\sec^4x\)
- \(\frac{dy}{dx}=4\sec^3x\times\sec x\tan x=4\sec^4x\times \tan x\)
- So \(I=\left[ \frac{5}{4}\sec^4x \right]^\theta_{0}=\frac{15}{4}\)
- \(\left( \frac{5}{4}\sec^4\theta \right)-\left( \frac{5}{4}\sec^40 \right)=\frac{15}{4}\)
- \(\frac{5}{4}\sec^4\theta-\frac{5}{4}=\frac{15}{4}\)
- \(\frac{5}{4}\sec^4\theta=\frac{20}{4}\)
- \(\sec^4\theta=4\)
- \(\sec\theta=^+_{-}\sqrt{ 2 }\)
- \(\theta=\frac{\pi}{4}\)
Practice Questions
\(\frac{x}{x^2+4}\)
- Try \(\ln |x^2+4|\)
- \(\frac{1}{x^2+4}\times 2x\)
- \(\frac{2x}{x^2+4}\)
- Add constant to adjust:
\(\frac{1}{2}\ln |x^2+4|\)
\(\frac{e^{2x}}{e^{2x}+1}\)
- Try \(\ln |e^{2x}+1|\)
- \(\frac{dy}{dx}=\frac{1}{e^{2x}+1}\times e^{2x}\)
- \(=\frac{2e^{2x}}{e^{2x}+1}\)
- Adjust with constant
- \(\frac{1}{2}\ln |e^{2x}+1|\)
\(\frac{x}{(x^2+4)^3}\)
- Try \((x^2+4)^{-2}\)
- \(\frac{dy}{dx}=(x^2+4)^{-3}\times 2x\)
- \(2x(x^2+4)^{-3}\)
- \(\frac{2x}{(x^2+4)^3}\)
- Adjust with constant
- \(\frac{1}{2}(x^2+4)^{-2}\)
\(\frac{e^{2x}}{(e^{2x}+1)^3}\)
- Let \(y=(e^{2x}+1)^{-2}\)
- \(\frac{dy}{dx}=-2(e^{2x}+1)^{-3}\times2e^{2x}\)
- \(=-4e^{2x}(e^{2x+1})^{-3}\)
- \(=\frac{-4e^{2x}}{(e^{2x}+1)^{-3}}\)
- Add constant to adjust
- \(-4(e^{2x}+1)^{-2}\)