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Integration by Parts

Integration by Parts

Integration by parts is basically the derivative product rule for integrals.

MathJax TeX Test Page Derivation: $$\dfrac{d}{dx}(uv) = uv' + u'v$$ If you integrate both sides, you get $$uv = \int{uv'}\, \mathrm{d}x + \int{vu'}\, \mathrm{d}x$$ $$uv = \int{u}\, \mathrm{d}v + \int{v}\, \mathrm{d}u$$ You can rewrite this as: $$\boxed{\int{\mathrm{u}}\, \mathrm{d}v = uv - \int{\mathrm{v}}\, \mathrm{d}u}$$

In order to solve problems involving integration by parts, you must choose a term that you can differentiate for "u", and a term that you can integrate for "dv".

Example 1- Regular

MathJax TeX Test Page $$\int xe^x\, \mathrm{d}x$$ $u = x, dv = e^x\, \mathrm{d}x$
$\, \mathrm{d}u = \,\mathrm{d}x, v = e^x$

$$xe^x - \int e^x\, \mathrm{d}x =xe^x - e^x + C = \boxed{e^x(x-1) + C}$$

Example 2- One Term

MathJax TeX Test Page $$\int{\mathrm{\arcsin(x)}}\, \mathrm{d}x$$ $u = \arcsin(x), dv = dx$

$du = \dfrac{1}{\sqrt{1 - x^2}}dx, v = x$ $$=x\arcsin(x) - \int{\dfrac{x}{\sqrt{1 - x^2}}}\,\mathrm{d}x$$ You can solve the second integral by using u-sub. You let $u = 1 - x^2$, so $du = -2xdx$, so $-\frac{1}{2}du = xdx$ $$=x\arcsin(x) + \frac{1}{2}\int{\dfrac{1}{\sqrt{u}}}\, \mathrm{d}u$$ $$ = x\arcsin(x) + \sqrt{u} + C$$ $$ = \boxed{x\arcsin(x) + \sqrt{1 - x^2} + C}$$ $$\int{\mathrm{\ln(x)}}\, \mathrm{d}x$$ $u = \ln(x), dv = dx$

$du = \frac{1}{x}, v = x$ $$= x\ln(x) - \int{\frac{x}{x}}\, \mathrm{d}x$$ $$= x\ln(x) - x$$ $$ = \boxed{x(\ln(x) - 1) + C}$$

Example 3- Swingy-Swingy

(That's what my teacher called those problems)

MathJax TeX Test Page $$\int \sec^3(x)\, \mathrm{d}x$$ $u = \sec(x), dv = \sec^2(x)\, \mathrm{d}x$
$du = \sec(x)\tan(x)dx, v = \tan(x)$

$$= \sec(x)\tan(x) - \int\tan^2(x)\sec(x)\,\mathrm{d}x$$ Now, we have to evaluate the second integral. $$\tan^2(x) = \sec^2(x) - 1 \to \int\tan^2(x)\sec(x)\,\mathrm{d}x = \int\left(\sec^2(x) - 1\right)\sec(x)\,\mathrm{d}x$$ $$\int\sec^3(x)\, \mathrm{d}x = \sec(x)\tan(x) - \int\left(\sec^2(x) - 1\right)\sec(x)\,\mathrm{d}x$$ $$\int\sec^3(x)\, \mathrm{d}x = \sec(x)\tan(x) - \int\sec^3(x)\, \mathrm{d}x + \int\sec(x)\,\mathrm{d}x$$ Remember, our goal is to figure out what $\int\sec^3(x)\, \mathrm{d}x$ is, so we put those terms on one side. $$2\int\sec^3(x)\, \mathrm{d}x = \sec(x)\tan(x) + \int\sec(x)\,\mathrm{d}x$$ Recall that the integral of sec(x) = $\ln|\sec(x) + \tan(x)| + C$

$$\int\sec^3(x)\, \mathrm{d}x = \boxed{\dfrac{1}{2}\sec(x)\tan(x) + \dfrac{1}{2}\ln|\sec(x) + \tan(x)| + C}$$ So, swingy-swingy means that you repeat the original integral on both sides of the equation, so you can just combine the terms on one side.

 

 

David Witten

L'Hopital's Rule

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