Mathwizurd.com is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

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