Computing integrals (3): Integration by parts

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 Sometimes, we may simplify integration by using the product rule of differentiation. This technique is called integration by parts.



Theorem (Integration by parts)

Let \(f(x)\) and \(g(x)\) be differentiable functions on an open interval \(I\). Then,

  1.  \(\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx\);
  2. For any \(a, b \in I\), \[\int_a^bf(x)g'(x)dx = \left[f(x)g(x)\right]_a^b - \int_a^bf'(x)g(x)dx.\]
Proof. By the product rule,
\[[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)\]
so
\[f(x)g'(x) = [f(x)g(x)]' - f'(x)g(x).\]
By integrating both sides, we have the desired results. ■

Example. Let us find \(\int x\cosh x dx\).
\[ \begin{eqnarray*} \int x\cosh x dx &=& \int x(\sinh x)'dx \\ &=& x \sinh x - \int 1 \cdot \sinh x dx\\ &=& x \sinh x - \cosh x + C. \end{eqnarray*} \]

Example (eg:recur). Let us study how we can compute
\[I_n = \int \frac{dx}{(x^2 + 1)^n}\]
for \(n\in \mathbb{N}\). Note
\[I_{n} = \int \frac{x^2+1}{(x^2 + 1)^{n+1}}dx = \int \frac{x^2}{(x^2 + 1)^{n+1}}dx + I_{n+1}.\]
Since
\[\left(-\frac{1}{2n(x^2+1)^n}\right)' = \frac{x}{(x^2+1)^{n+1}},\]
we have
\[ \begin{eqnarray*} \int \frac{x^2}{(x^2 + 1)^{n+1}}dx &=& \int x\cdot\frac{x}{(x^2 + 1)^{n+1}}dx\\ &=& -\frac{x}{2n(x^2+1)^n} + \frac{1}{2n}\int\frac{dx}{(x^2+1)^{n}}\\ &=&\frac{1}{2n}I_{n} -\frac{x}{2n(x^2+1)^n}. \end{eqnarray*} \]
Therefore, we have the following recurrence equation
\[ \begin{eqnarray} I_{n+1} &=& I_n - \int\frac{x^2}{(x^2+1)^{n+1}}dx \\ &=& \left(1 - \frac{1}{2n}\right)I_n + \frac{1}{2n}\cdot\frac{x}{(x^2+1)^n}.\tag{eq:Irec} \end{eqnarray}\]
Starting from
\[I_1 = \int \frac{dx}{x^2 + 1} = \arctan x + C,\]
we can compute \(I_2, I_3, \cdots\) recursively by using Eq. (eq:Irec).

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