Integration By Parts - SS3 Mathematics Lesson Note

This method is used for functions that are in the form of products. Recall from differentiation, if \(y = uv\), where \(u\) and \(v\) are functions of \(x\) then

\[\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\]

Integrating both sides of the equation we get,

\[\int_{}^{}\frac{d(uv)}{dx}dx = \int_{}^{}{u\frac{dv}{dx}}dx + \int_{}^{}{v\frac{du}{dx}}\ dx\]

\[uv = \int_{}^{}{u\frac{dv}{dx}}dx + \int_{}^{}{v\frac{du}{dx}}\ dx\]

\[uv = \int_{}^{}udv + \int_{}^{}v\ du\]

\[\int_{}^{}u\ dv = uv - \ \int_{}^{}v\ du\]

 

Example: Evaluate \(\int_{}^{}x^{3}\sin x\ dx\) using integration by parts

Solution:

Given \(\int_{}^{}x^{3}\sin x\ dx\)

Let \(u = x^{3}\) and \(dv = \sin x\)

Then \(du = 3x^{2}\) and \(v = \ \int_{}^{}{\sin x}\ dx = \ - \cos{x\ }\)

 

By integrating by parts,

\[\int_{}^{}x^{3}\sin x\ dx = x^{3}\left( - \cos x \right) - \int_{}^{}\left( - \cos x \right)3x^{2}\ dx\]

\[= - x^{3}\cos x + \int_{}^{}{3x^{2}}\cos x\ dx\]

Now, integrating \(\int_{}^{}{3x^{2}}\cos x\ dx\) y parts,

\(\int_{}^{}{3x^{2}}\cos x\ dx = \ 3\int_{}^{}x^{2}\cos xdx = 3\left\lbrack x^{2}\sin x - \ \int_{}^{}{\sin x}\ (2x) \right\rbrack\)

\[= \ 3\lbrack x^{2}\sin x - \ \int_{}^{}{2x\ \sin x}\rbrack\]

Integrating \(\int_{}^{}{2x\ \sin x}\) by parts,

\[\int_{}^{}{2x\ \sin x} = 2\int_{}^{}x\sin x\ dx = 2\lbrack x\left( - \cos x \right) - \ \int_{}^{}{( - \cos x})1\rbrack\]

\[= 2\left\lbrack - x\cos x + \int_{}^{}{\cos x} \right\rbrack = \ 2\left\lbrack - x\cos x + \sin x + C \right\rbrack\]

\[\therefore\int_{}^{}x^{3}\sin x\ dx = \ - x^{3}\cos x + 3\lbrack x^{2}\sin x - \ 2\left\lbrack - x\cos x + \sin x + C \right\rbrack\rbrack\]

\[= - x^{3}\cos x + 3\left\lbrack x^{2}\sin x + 2x\cos x - 2\sin x + C \right\rbrack\ \]

\((note,\ multiplying\ C\ by\ a + ve\ or - ve\ number\ is\ still\ C)\)

\[= - x^{3}\cos x + 3x^{2}\sin x + 6x\cos x - 6\sin x + C\]