Infinite Integrals by Parts

Techniques of  Integration

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

Let $u = \cos 2x$, $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}x} = e^{2x}$, $v = \frac{1}{2}e^{2x}$

$\displaystyle \displaystyle\int e^{2x} \cos 2x \, \mathrm{d}x = \frac{1}{2}e^{2x} \cos 2x - \displaystyle\int (-2\sin 2x) \cdot \frac{1}{2}e^{2x} \, \mathrm{d}x$

$\displaystyle = \frac{1}{2}e^{2x} \cos 2x + \displaystyle\int e^{2x} \sin 2x \, \mathrm{d}x$

Second round: $u_1 = \sin 2x$, $\displaystyle \frac{dv_1}{\mathrm{d}x} = e^{2x}$, $v_1 = \frac{1}{2}e^{2x}$

$\displaystyle = \frac{1}{2}e^{2x} \cos 2x + \left[\frac{1}{2}e^{2x} \sin 2x - \displaystyle\int 2\cos 2x \cdot \frac{1}{2}e^{2x} \, \mathrm{d}x\right]$

$\displaystyle = \frac{1}{2}e^{2x} \cos 2x + \frac{1}{2}e^{2x} \sin 2x - \displaystyle\int e^{2x} \cos 2x \, \mathrm{d}x$

Let $I = \displaystyle\int e^{2x} \cos 2x \, \mathrm{d}x$

$\displaystyle I = \frac{1}{2}e^{2x} \cos 2x + \frac{1}{2}e^{2x} \sin 2x - I$

$\displaystyle 2I = \frac{1}{2}e^{2x}(\cos 2x + \sin 2x)$

$\displaystyle \displaystyle\int e^{2x} \cos 2x \, \mathrm{d}x = \frac{1}{4}e^{2x}(\cos 2x + \sin 2x) + C$

Techniques of Integration - Infinite Integrals by Parts

Question b1a6acca

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