#math/calculus #math $ \huge\int u\d v = uv - \int v\d u $ ## Proof Give the [[Derivative Product Rule]] for [[Derivative|Derivatives]], we know that: $\huge\begin{align*} y &=u v\\ \frac{\mathrm{d} y}{\mathrm{~d} x}&=\frac{\mathrm{d}(u v)}{\mathrm{d} x}=u \frac{\mathrm{d} v}{\mathrm{~d} x}+v \frac{\mathrm{d} u}{\mathrm{~d} x} \end{align*}$ Rearranging, we get: $ $ [[Integration|Integrating]] both sides: $ \int u \frac{\mathrm{d} v}{\mathrm{~d} x} \mathrm{~d} x=\int \frac{\mathrm{d}(u v)}{\mathrm{d} x} \mathrm{~d} x-\int v \frac{\mathrm{d} u}{\mathrm{~d} x} \mathrm{~d} x $ Simplyfing: $ \int u \frac{\mathrm{d} v}{\mathrm{~d} x} \mathrm{~d} x = uv -\int v \mathrm{d}u $ ___ ## Examples ### $xe^x$ $\huge \int xe^xdx $ $\huge\begin{align*} \int \left(u\right)dv &= uv - \int(v)dv \\ u &= x \\ du &= 1dx \\ v &= e^x \\ dv &= e^xdx \\ \\ \int xe^xdx&=xe^x-\int e^x \cdot 1dx \\ &= xe^x - (e^x + C)\\ &= e^x(x-1) + C \end{align*}$ ___ ### $xsin(x)$ $ \int x\sin (x)dx $ $\begin{align*} \int \left(u\right)dv &= uv - \int(v)du \\ u&=x\\ du &= 1dx \\ dv &= sin(x) \\ v &= -\cos x \\ \int x\sin (x)dx &= -x\cos(x) - \int-\cos(x)dx \\ &= -x\cos (x) - (-\sin (x) + C) \\ &= \sin(x) - x\cos(x) + C \\ \end{align*}$ ___ ### x^2ln(x) $ \int x^2\ln(x) dx $ $\begin{align*} \int \left(u\right)dv &= uv - \int(v)du \\ u &= \ln (x) \\ du &= \frac{dx}{x} \\ dv &= x^2 \\ v &= \frac{x^3}{3} \\ \int x^2\ln(x) dx &= \ln (x)x^2 -\int\frac{x^3dx}{3x}\\ &=\frac{\ln(x)}x^3{3} - \int \frac{x^2dx}{3} \\ &=\frac{\ln(x)x^3}{3} - \frac{x^3}{9} + C \\ \end{align*}$ ___ ### bruh moment $\int e^{2z} \cos\left(\frac{z}{4}\right) \mathrm{d}z$ $\begin{align*} \int \left(u\right)dv &= uv - \int(v)du \\ u &= \cos \left(\frac{z}{4}\right)\\ \mathrm{d}u &= -\frac{1}{4}\sin\left(\frac{z}{4}\right)\mathrm{d}z \\ \mathrm{d}v&= e^{2z} \\ v &= \int e^{2z}\mathrm{d}x=\frac{1}{2}e^{2z} \\ \int e^{2z} \cos\left(\frac{z}{4}\right) \mathrm{d}z &= \frac{1}{2}e^{2x}\cos \left(\frac{z}{4}\right) + \int \frac{1}{2}e^{2z}\cdot \frac{1}{4}\sin\left(\frac{z}{4}\right)\mathrm{d}z \\ &= \frac{1}{2}e^{2x}\cos \left(\frac{z}{4}\right) + \frac{1}{8}\int e^{2z}\sin\left(\frac{z}{4}\right)\mathrm{d}z \\ \\ \\ \int e^{2z}\sin\left(\frac{z}{4}\right)\mathrm{d}z &= f\rm{d}g -\int g \mathrm{d}f \\ f &= \sin \left(\frac{z}{4} \right) \\ \mathrm{d}f &= \frac{1}{4} \cos\left(\frac{z}{4} \right)\mathrm{d}z \\ \mathrm{d}g &= e^{2z} \\ g &= \int e^{2z} = \frac{1}{2}e^{2z} \\ \int e^{2z}\sin\left(\frac{z}{4}\right)\mathrm{d}z &= \sin\left(\frac{z}{4} \right)e^{2z} - \int \frac{1}{2}e^{2z}\cdot \frac{1}{4}\cos\left(\frac{z}{4}\right) \\ &= \sin\left(\frac{z}{4} \right)e^{2z} - \frac{1}{8}\int e^{2z}\cdot \cos\left(\frac{z}{4}\right) \\ \\ \end{align*}$ $\begin{align*} \int e^{2z} \cos\left(\frac{z}{4}\right) \mathrm{d}z &= \frac{1}{2}e^{2x}\cos \left(\frac{z}{4}\right) + \frac{1}{8}\int e^{2z}\sin\left(\frac{z}{4}\right)\mathrm{d}z \\ &=\frac{1}{2}e^{2x}\cos \left(\frac{z}{4}\right) + \frac{1}{8} \left( \sin\left(\frac{z}{4} \right)e^{2z} - \frac{1}{8}\int e^{2z}\cdot \cos\left(\frac{z}{4}\right) \right) \\ \text{let } h &= e^{2z}\cos \left(\frac{z}{4} \right) \\ \int h \mathrm{d}z &= \frac{1}{2}h+ \frac{1}{8}\sin\left( \frac{z}{4} \right)e^{2z} - \frac{1}{64}\int h \mathrm{d}z \\ 8\int h \mathrm{d}z &= 4h+ \sin\left( \frac{z}{4} \right)e^{2z} - \frac{1}{8}\int h \mathrm{d}z \\ \frac{65}{8} \int h \mathrm{d}z &= 4h+ \sin\left( \frac{z}{4} \right)e^{2z} \\ \int h \mathrm{d}z &= \frac{8}{65} \left( 4h+ \sin\left( \frac{z}{4} \right)e^{2z} \right) \\ \int e^{2z}\cos\left(\frac{z}{4}\right) \mathrm{d}z &= \frac{8}{65} \left( 4e^{2z}\cos\left(\frac{z}{4}\right) + \sin\left( \frac{z}{4} \right)e^{2z} \right) \\ &= \frac{8}{65}e^{2z} \left( 4\cos\left(\frac{z}{4}\right) + \sin\left( \frac{z}{4} \right) \right) \\ \end{align*}$ >[!example] > #example using integration by parts along with [[Integral U Substitution]]. >$\huge \int \cos(\sqrt x)\d x $ >$\huge \begin{align} \let a &= \sqrt{x} \\ \let \d a &= \frac{1}{2\sqrt x} \d x \\ \\ \int \cos(\sqrt x)\d x&= \int2\sqrt{x} \cos(a)\d a \\ &= 2\int\sqrt{x} \cos(a)\d a \\ &= 2\int a \cos(a)\d a \\ \\ \let u &= a\\ \let \d u &= \d a \\ \let \d v &=\cos(a)\d a\\ \let v &= \sin(a) \\ \\ 2\int a \cos(a)\d a &= \sin(a)a - \int \sin(a)\d a \\ &=2\pa{\sin(a)a + \cos(a)} \\ &=2 \sqrt x \sin \pa{\sqrt{x}} + 2\cos \pa{\sqrt x} \end{align}$