quality:corruption, information:extraction
Tabular integration by parts

The essential process of the above formula can be summarized in a table; the resulting method is called "tabular integration" and was featured in the film Stand and Deliver (1988).

For example, consider the integral

∫ x 3 cos x d x {\displaystyle \int x^{3}\cos x\,dx\quad } and take u (0) = x 3, v (n) = cos x. {\displaystyle \quad u^{(0)}=x^{3},\quad v^{(n)}=\cos x.}

Begin to list in column A the function u (0) = x 3 {\displaystyle u^{(0)}=x^{3}} and its subsequent derivatives u (i) {\displaystyle u^{(i)}} until zero is reached. Then list in column B the function v (n) = cos x {\displaystyle v^{(n)}=\cos x} and its subsequent integrals v (n − i) {\displaystyle v^{(n-i)}} until the size of column B is the same as that of column A. The result is as follows:

The product of the entries in row i of columns A and B together with the respective sign give the relevant integrals in step i in the course of repeated integration by parts. Step i = 0 yields the original integral. For the complete result in step i > 0 the i th integral must be added to all the previous products (0 ≤ j < i) of the j th entry of column A and the (j + 1) st entry of column B (i.e., multiply the 1st entry of column A with the 2nd entry of column B, the 2nd entry of column A with the 3rd entry of column B, etc....) with the given j th sign. This process comes to a natural halt, when the product, which yields the integral, is zero (i = 4 in the example). The complete result is the following (with the alternating signs in each term):

(+ 1) (x 3) (sin x) ⏟ j = 0 + (− 1) (3 x 2) (− cos x) ⏟ j = 1 + (+ 1) (6 x) (− sin x) ⏟ j = 2 + (− 1) (6) (cos x) ⏟ j = 3 + ∫ (+ 1) (0) (cos x) d x ⏟ i = 4: → C. {\displaystyle \underbrace {(+1)(x^{3})(\sin x)} _{j=0}+\underbrace {(-1)(3x^{2})(-\cos x)} _{j=1}+\underbrace {(+1)(6x)(-\sin x)} _{j=2}+\underbrace {(-1)(6)(\cos x)} _{j=3}+\underbrace {\int (+1)(0)(\cos x)\,dx} _{i=4:\;\to \;C}.}

This yields

∫ x 3 cos x d x ⏟ step 0 = x 3 sin x + 3 x 2 cos x − 6 x sin x − 6 cos x + C. {\displaystyle \underbrace {\int x^{3}\cos x\,dx} _{\text{step 0}}=x^{3}\sin x+3x^{2}\cos x-6x\sin x-6\cos x+C.}

The repeated partial integration also turns out useful, when in the course of respectively differentiating and integrating the functions u (i) {\displaystyle u^{(i)}} and v (n − i) {\displaystyle v^{(n-i)}} their product results in a multiple of the original integrand. In this case the repetition may also be terminated with this index i. This can happen, expectably, with exponentials and trigonometric functions. As an example consider

∫ e x cos x d x. {\displaystyle \int e^{x}\cos x\,dx.}

In this case the product of the terms in columns A and B with the appropriate sign for index i = 2 yields the negative of the original integrand (compare rows i = 0 and i = 2).

∫ e x cos x d x ⏟ step 0 = (+ 1) (e x) (sin x) ⏟ j = 0 + (− 1) (e x) (− cos x) ⏟ j = 1 + ∫ (+ 1) (e x) (− cos x) d x ⏟ i = 2. {\displaystyle \underbrace {\int e^{x}\cos x\,dx} _{\text{step 0}}=\underbrace {(+1)(e^{x})(\sin x)} _{j=0}+\underbrace {(-1)(e^{x})(-\cos x)} _{j=1}+\underbrace {\int (+1)(e^{x})(-\cos x)\,dx} _{i=2}.}

Observing that the integral on the RHS can have its own constant of integration C ′ {\displaystyle C'}, and bringing the abstract integral to the other side, gives:

2 ∫ e x cos x d x = e x sin x + e x cos x + C ′, {\displaystyle 2\int e^{x}\cos x\,dx=e^{x}\sin x+e^{x}\cos x+C',}

and finally:

∫ e x cos x d x = 1 2 (e x (sin x + cos x)) + C, {\displaystyle \int e^{x}\cos x\,dx={\frac {1}{2}}\left(e^{x}(\sin x+\cos x)\right)+C,}

where C = C ′ 2 {\displaystyle C={\frac {C'}{2}}}.