Integration by parts
\(\DeclareMathOperator{cosec}{cosec}\)
Equation
\(\(\int u\frac{dv}{dx} \, dx =uv-\int v\frac{du}{dx} \, dx\)\)
- The \(x\) part (the part which can disappear) is the one which we use for \(u\)
- If part of the integral loops round (like a trig identity), then we use that part for the integration bit (\(v\)) , not the differentiation bit
\(\int x\cos x \, dx\)
- \(u=x\)
- \(\frac{du}{dx}=1\)
- \(\frac{dv}{dx}=\cos x\)
- \(v=\sin x\)
- \(\int x\cos x \, dx=x\sin x--\cos x+c\)
- \(=x\sin x+\cos x+c\)
\(\int x^2\ln x \, dx\)
- \(u=\ln x\)
- \(\frac{du}{dx}=\frac{1}{x}\)
- \(\frac{dv}{dx}=x^2\)
- \(v=\frac{x^3}{3}\)
- \(\int x^2\ln x \, dx=\frac{x^3}{3}\ln x-\int x^{\cancel{ 3 }2}\times\frac{1}{\cancel{ x }} \, dx\)
- \(=x^3\ln x-\int \frac{x^2}{3} \, dx\)
- \(=\frac{x^3}{3}\ln x-\frac{x^3}{9}+c\)
\(\int x^2e^x \, dx\)
- \(u=x^2\)
- \(\frac{du}{dx}=2x\)
- \(\frac{dv}{dx}=e^x\)
- \(v=e^x\)
- \(=x^2e^x-\int e^x2x \, dx\)