Integrating f(ax+b)
\(\DeclareMathOperator{cosec}{cosec}\)
\(\int sec^{2} 3x \, dx\)
- Consider \(y=\tan 3x\)
- \(\frac{dy}{dx}=3sec^{2}3x\)
- \(\therefore =\frac{1}{3}\tan 3x\)
Info on how to integrate \(f(ax+b)\)
- Guess as to what you could differentiate to get the equation you are given
- E.G. If the equation was \(\frac{1}{3x+2}\), you would guess \(y=\ln |3x+2|\)
- If you get a result as the thing you want but multiplied by a constant, just divide by that constant.
- \(\frac{dy}{dx}=\frac{1}{3x+2}\times 3\)
- and don't forget to plus \(c\)
- So our answer would be \(\frac{1}{3}\ln |3x+2|+c\)
\(\int \cos(2x+1) \, dx\)
- Let \(y=\sin(2x+1)\)
- \(\frac{dy}{dx}=\sin(2x+1)\times 2\)
- So \(\int \cos(2x+1) \, dx=\frac{1}{2}\sin(2x+1)+c\)
\(\int \sin(2x+3) \, dx\)
- Let \(y=-\cos(2x+3)\)
- \(\frac{dy}{dx}=\sin(2x+3)\times 2\)
- So \(\int \sin(2x+3) \, dx=-\frac{1}{2}\cos(2x+3)+c\)
\(\int 3e^{2x} \, dx\)
- Let \(y=e^{2x}\)
- \(\frac{dy}{dx}=2e^{2x}\)
- So \(\int 3e^{2x} \, dx=\frac{3}{2}e^{2x}+c\)
\(\int 4e^{x+5} \, dx\)
- Let \(y=e^{x+5}\)
- \(\frac{dy}{dx}=e^{x+5}\)
- So \(\int 4e^{x+5} \, dx=4e^{x+5}+c\)
\(\int \cos(1-2x) \, dx\)
- Let \(y=\sin(1-2x)\)
- \(\frac{dy}{dx}=-2\cos(1-2x)\)
- So \(\int \cos(1-2x) \, dx=-\frac{1}{2}\sin(1-2x)+c\)
\(\int \sec4x \tan4x \, dx\)
\(\frac{dy}{dx}\sec 4x=16\sec4x \tan4x\)
So \(\int \sec4x \tan4x \, dx=\frac{1}{4}\sec4x+c\)
\(\int \cosec 2x \cot 2x \, dx\)
- \(\frac{dy}{dx}(-\cosec 2x)=4\cosec 2x \cot 2x\)