# Integration by Parts Integration by parts is used to undo [[The Derivative Product Rule]]. It boils down to the following formula $\int u \,dv = uv - \int v \, du$ (where $u$ and $v$ are functions of $x$) ^aftqxa By freely defining what parts of an expression correspond to $u$ and $dv$, we can simplify the integration process of the whole expression by picking $dv$ that can be easily integrated. The devil is in the detail is a mnemonic which can be used to memorize the tactic used to pick $dv$. - When picking $dv$ use the following in decreasing order of preference: - $e$xponential functions - trigonometric functions - algebraic functions (polynomials) - inverse trigonometric functions - logarithms You can simplify the process by writing a 2x2 table whose first column contains $u$ and $v$ and the second column contains $du$ and $dv$. You then fill it in clockwise order. ## Example Integrate $\int x^2 \sin x\, dx$. Define $u = x^2$ and $dv = \sin x\, dx$ | $u=x^2$ | $du = 2x\, dx$ | | ------------- | ----------------- | | $v = -\cos x$ | $dv = \sin x\, dx$ | ![[Integration by Parts#^aftqxa]] $\int x^2 \sin x \, dx= -x^2\cos x + 2\int x\cos x\, dx$ Define $u=x$ and $dv=\cos x \, dx$: | $u=x$ | $du=dx$ | | ---------- | ------------------- | | $v=\sin x$ | $dv = \cos x \, dx$ | $\int x \cos x \, dx = x\sin x - \int \sin x \, dx = x\sin x +\cos x + C$ $\int x^2 \sin x\, dx = -x^2 \cos x + 2 (x\sin x + \cos x) + C$ ## The LIDI Method Instead of writing out the $u$ and $v$ explicitly like in the example above, a slightly faster approach is advocated by Neil Does Maths called the LIDI method. Split the integral into first and second parts (the first corresponding to $u$ and second to $v$) and write out four brackets like: $I =(L)(I)-\int (D)(I)$ LIDI corresponds to: - L - leave the first part - I - integrate the second part - D - differentiate the first part - I - integrate the second part So applying it to the example above we simply write: $ \begin{align} \int x^2\sin x \, dx &= (x^2) (-\cos x) - \int (2x) (-\cos x \, dx) \\ &= -x^2 \cos x +2\int x \cos x \, dx \\ &= -x^2 \cos x +2\left[ (x)(\sin x)-\int(1)(\sin x \, dx) \right] \\ &= -x^2 \cos x +2(x\sin x+\cos x) + C \end{align} $ ## Proof Start with the product rule: ![[The Derivative Product Rule#^a1ucyf]] but express it a bit differently ($u$ and $v$ will replace $f$ and $g$ for common naming consistency): $ \frac d {dx} \bigg[uv \bigg] = \frac {du} {dx}v + u\frac{dv}{dx} $ Integrate both sides with respect to $x$: $ \int \frac d {dx} \bigg[uv \bigg] dx = \int \frac {du} {dx}v \, dx + \int u\frac{dv}{dx} \, dx $ The left hand side is just $uv$ because we're reversing the derivative. On the right side $dx$ cancel out. After rewriting we get: $ \begin{gather} uv = \int v \, du + \int u \,dv \\ \int u \, dv = uv - \int v \,du \end{gather} $