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Parts applied to its own output: a recurrence $I_n = c + kI_{n-1}$ with a base case, or an equation $I = c - \lambda I$ to solve.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to derive a reduction formula by applying parts to a general power, identify the base case a recurrence needs and notice when a step of two forces two separate ladders, climb a ladder collecting its factors, recognise a loop when two passes return the original integral, solve for the integral algebraically instead of integrating a third time, and check the direction of a recurrence against the size the integrand says its values should have.
Integration by parts trades one integral for another. In the examples of the last lesson the new integral was elementary and the calculation ended. It very often is not.
Two things can happen instead. The new integral can be the same shape as the old one with a smaller exponent — then parts has not solved the problem, it has produced a recurrence, and a recurrence with a base case is as good as an answer. Or the new integral can be the same shape with the same exponent — then parts appears to have achieved nothing, and in fact it has handed you an equation.
A reduction formula expresses $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$: a recurrence in the index. The base case is the smallest index, evaluated directly; a recurrence without one names a number without ever producing it.
A loop is what happens when parts returns the original integral multiplied by a constant. The integral is then the unknown in a linear equation, and the last step of the calculation is algebra rather than integration.
Note which way a recurrence steps. A formula that drops the index by two splits the problem into two independent ladders, one for odd indices and one for even, each with its own base case.
Reduction. Suppose parts turns $I_n$ into boundary terms plus a multiple of $I_{n-1}$. Then $$I_n = c_n + k_n I_{n-1},$$ a recurrence. To use it: evaluate the base case $I_0$ (or $I_1$) directly, then apply the formula upward, collecting factors. The classic examples are $\int x^n e^x dx$, $\int \sin^n x\,dx$, $\int \sec^n x\,dx$ and $\int (x^2 + a^2)^{-n}dx$.
The loop. Suppose two passes of parts return the original integral: $$I = c - \lambda I \quad\text{with } \lambda \ne -1.$$ Then $(1 + \lambda)I = c$ and $I = \dfrac{c}{1 + \lambda}$. This is how $\int e^{ax}\sin bx\,dx$ and $\int e^{ax}\cos bx\,dx$ are done, and the denominator that appears is always $a^2 + b^2$.
Why the loop closes. Differentiating $e^{ax}\sin bx$ twice returns a combination of $e^{ax}\sin bx$ and $e^{ax}\cos bx$ — the pair is closed under differentiation. Parts inherits that closure, which is why exactly two passes are needed and a third is wasted.
Stopping. A reduction stops at its base case. A loop stops the moment the original integral reappears. Continuing past either point is the commonest way a correct calculation is undone.
Another way: picture
A reduction formula is a ladder with the base at the bottom. You cannot start at the top and climb down forever; you go to the bottom rung, which is a plain integral, and climb up collecting one multiplier per rung. A loop is not a ladder at all — it is a circle, and the way out of a circle is to give the thing a name and solve for it.
Another way: steps
For $I_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$, write $\sin^n = \sin^{n-1}x \cdot \sin x$ and take $u = \sin^{n-1}x$, $dv = \sin x\,dx$. Then $v = -\cos x$ and $$I_n = \left[-\sin^{n-1}x\cos x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx.$$ The boundary term vanishes at both ends. Replacing $\cos^2 = 1 - \sin^2$ gives $I_n = (n-1)I_{n-2} - (n-1)I_n$, and solving for $I_n$ — a loop inside a reduction — gives $$I_n = \frac{n-1}{n}I_{n-2}.$$
The two base cases are $I_0 = \dfrac{\pi}{2}$ and $I_1 = 1$. So even exponents produce a multiple of $\pi$ and odd exponents produce a rational number, and that difference is a consequence of which base case the ladder lands on rather than anything about the integrand.
No base case. Applying $I_n = \frac{n-1}{n}I_{n-2}$ without ever evaluating $I_0$ or $I_1$ produces an expression in terms of an integral nobody has computed. A recurrence is not an answer until it reaches the ground.
The wrong base case. The step of two means even and odd are separate ladders. Reducing $I_5$ down to $I_0$ is impossible; it lands on $I_1$.
Running the recurrence upside down. $I_{n-2} = \frac{n}{n-1}I_n$ is the same formula read the other way and it is almost always the wrong direction to work in. The check is that each rung should be smaller than the one below when the integrand is shrinking.
A third pass on a loop. It returns you exactly to where the second pass started. If the algebra looks as though it is cancelling to $0 = 0$, that is what has happened.
Losing the boundary terms along the ladder. In the $\sin^n$ derivation they vanish; in $\int x^n e^x dx$ they do not, and every rung contributes one.
When parts hands back an integral of the same shape, the natural reading is that the technique did not work and a different one is needed. That reading costs learners the whole of this lesson, because in both cases the returning integral is the point.
If it comes back smaller, you have been handed a recurrence — an infinite family of integrals solved at once, not just the one you asked about. $\int \sin^n$ for every $n$ falls out of a single application of parts, which no amount of direct integration would give you.
If it comes back the same size, you have been handed a linear equation. The unknown happens to be an integral, but nothing about solving $2I = c$ cares what $I$ is. A learner who treats the reappearance as a signal to start over will start over forever, because the second attempt does exactly what the first did.
The question to ask after each pass is not "did I finish?" but "what shape came back, and is it smaller?" Those two answers decide everything that happens next.
$I_n = \displaystyle\int_0^1 x^n e^{x}\,dx$ with $u = x^n$, $dv = e^x dx$: $I_n = [x^n e^x]_0^1 - n\displaystyle\int_0^1 x^{n-1}e^x dx = e - nI_{n-1}$.
The boundary term is $e$, not zero.
Base case: $I_0 = \displaystyle\int_0^1 e^x dx = e - 1$.
The bottom rung is a plain integral.
$I_1 = e - (e - 1) = 1$, $I_2 = e - 2 \cdot 1 = e - 2$, $I_3 = e - 3(e-2) = 6 - 2e$. The alternating signs come from the minus in the recurrence, and they are the reason these values are small although $e$ is not.
$I = \displaystyle\int e^{x}\sin x\,dx$. Parts with $u = \sin x$: $I = e^x\sin x - \displaystyle\int e^x\cos x\,dx$.
One pass changes sine into cosine.
Parts again on the new integral, again with the trigonometric factor as $u$: $\displaystyle\int e^x\cos x\,dx = e^x\cos x + I$.
The second pass brings $I$ back.
So $I = e^x\sin x - e^x\cos x - I$, hence $2I = e^x(\sin x - \cos x)$ and $I = \dfrac{e^x(\sin x - \cos x)}{2}$. Both passes had to take the trigonometric factor as $u$; swapping roles halfway round simply unwinds the first pass.
By symmetry with the sine derivation, $I_n = \dfrac{n-1}{n}I_{n-2}$.
Same recurrence, same reason.
$n = 4$ is even, so the ladder ends at $I_0 = \dfrac{\pi}{2}$.
Pick the base case the parity demands.
$I_4 = \dfrac{3}{4}I_2 = \dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{3\pi}{16}$.
Given $I_m = \displaystyle\int_0^1 x^{m}(1 - x)^{8}\,dx$ and the reduction $I_m = \dfrac{m}{8 + m}\,I_{m-1}$ taken with the exponent on $(1-x)$ rising by one each step, evaluate $\displaystyle\int_0^1 x^{2}(1 - x)^{8}\,dx$.
The integral equals answer.
Fill in the ladder for $J_m = \displaystyle\int_0^1 x^{m}(1 - x)^{4 + 2 - m}\,dx$, climbing from $m = 0$.
| Value | |
|---|---|
| $J_0$, the base case | |
| $J_1$, one application of parts | |
| $J_2$, two applications |
You have derived $I_{6} = \dfrac{5}{6}I_{4}$. Put the steps of using it into the order you do them.
Number the steps in order (write the number in the box):
Two applications of parts to $I = \displaystyle\int e^{5x}\sin(5x)\,dx$ give $I = (\text{terms}) - \dfrac{25}{25}I$. What now?
Apply parts once to each of these and say what comes back.
| The same shape with a smaller index — a recurrence | The original integral again — a loop to solve for | An elementary integral — the calculation is finished | |
|---|---|---|---|
| $\displaystyle\int_0^1 x^{7}e^{x}\,dx$ | |||
| $\displaystyle\int e^{x}\sin x\,dx$ | |||
| $\displaystyle\int x\cos x\,dx$ | |||
| $\displaystyle\int \ln x\,dx$ |
Parts gives $\displaystyle\int_0^{\pi/2}\sin^{m}x\,dx = \dfrac{m - 1}{m}\int_0^{\pi/2}\sin^{m-2}x\,dx$. Writing $\displaystyle\int_0^{\pi/2}\sin^{8}x\,dx = k\cdot\dfrac{\pi}{2}$, find $k$.
The coefficient is answer.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Given $I_m = \displaystyle\int_0^1 x^{m}(1 - x)^{3}\,dx$ and the reduction $I_m = \dfrac{m}{3 + m}\,I_{m-1}$ taken with the exponent on $(1-x)$ rising by one each step, evaluate $\displaystyle\int_0^1 x^{2}(1 - x)^{3}\,dx$.
The integral equals answer.
You can tell a ladder from a loop by looking at what came back, and finish either. Say in your own words why a recurrence without a base case is not an answer. Next: the trigonometric integrals these recurrences were invented for.
10. Your turn: $I_n = \displaystyle\int_0^{\pi/2}\cos^{n}x\,dx$ for $n = 4$, step 3
Collect the factors and multiply into the base.