Reduction formulas

What you will be able to do

• derive a reduction formula by integration by parts or by an identity;
• apply a reduction formula again and again until an easy integral is left;
• use the Wallis integrals $\displaystyle\int_0^{\pi/2}\sin^n x\,dx$ and their products of fractions.

An integral with a whole number in it

Suppose you need $\displaystyle\int x^5 e^x\,dx$. Integration by parts works, but it has to be used five times. A better plan is to find one rule that connects the integral with $x^n$ to the integral with $x^{n-1}$. Then you apply the rule five times, and you never repeat the derivation.

Call the integral $I_n$. A reduction formula is a rule that gives $I_n$ from $I_{n-1}$, or from $I_{n-2}$. It reduces the number $n$. You stop at a case that you can do directly, such as $I_0$ or $I_1$.

The first example: $x^n e^x$

Let $I_n = \displaystyle\int x^n e^x\,dx$. Take $u = x^n$ and $dv = e^x\,dx$. Then $du = n x^{n-1}\,dx$ and $v = e^x$. By parts,

For $n = 3$ the chain is $I_3 = x^3 e^x - 3I_2$, $I_2 = x^2 e^x - 2I_1$ and $I_1 = x\,e^x - I_0$. Putting them together gives

Powers of sine, cosine and tangent

Take $I_n = \displaystyle\int \sin^n x\,dx$ with $n \ge 2$. Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and use parts with $u = \sin^{n-1}x$ and $dv = \sin x\,dx$. Then $du = (n - 1)\sin^{n-2}x\cos x\,dx$ and $v = -\cos x$. After replacing $\cos^2 x$ by $1 - \sin^2 x$, we get

For cosine the same steps give $\displaystyle\int \cos^n x\,dx = \dfrac{\cos^{n-1}x\sin x}{n} + \dfrac{n - 1}{n}\int \cos^{n-2}x\,dx$.

For tangent we use an identity instead. Write $\tan^n x = \tan^{n-2}x\left(\sec^2 x - 1\right)$. Then

For the logarithm, take $u = (\ln x)^n$ and $dv = dx$. Then $\displaystyle\int (\ln x)^n\,dx = x(\ln x)^n - n\int (\ln x)^{n-1}\,dx$.

The Wallis integrals

Over $\left[0, \dfrac{\pi}{2}\right]$ the boundary term of the sine formula is zero, because $\cos\dfrac{\pi}{2} = 0$ and $\sin 0 = 0$. Let $W_n = \displaystyle\int_0^{\pi/2}\sin^n x\,dx$. Then

The same numbers come from $\displaystyle\int_0^{\pi/2}\cos^n x\,dx$, because the reflection $x \to \dfrac{\pi}{2} - x$ swaps sine and cosine. Here are the first few values.

• $W_2 = \dfrac{\pi}{4}$, $W_3 = \dfrac{2}{3}$, $W_4 = \dfrac{3\pi}{16}$, $W_5 = \dfrac{8}{15}$, $W_6 = \dfrac{5\pi}{32}$.

For an even $n$ the chain ends at $W_0$, so $\pi$ appears. For an odd $n$ it ends at $W_1$, and the answer is a fraction.

Over $[0, \pi]$ the area is twice as large, because of the symmetry of the sine curve. So $\displaystyle\int_0^{\pi}\sin^n x\,dx = 2W_n$ for every $n$. For an even $n$, the integral over $[0, 2\pi]$ is four times $W_n$.

Worked examples

Example 1. Find $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx$.

1. Use $W_n = \dfrac{n - 1}{n}W_{n-2}$ three times: $W_6 = \dfrac{5}{6}W_4 = \dfrac{5}{6} \times \dfrac{3}{4}W_2 = \dfrac{5}{6} \times \dfrac{3}{4} \times \dfrac{1}{2}W_0$.
2. With $W_0 = \dfrac{\pi}{2}$ this is $\dfrac{15}{48} \times \dfrac{\pi}{2} = \dfrac{5\pi}{32}$.

Example 2. Find $\displaystyle\int_0^{\pi}\sin^5 x\,dx$.

1. $\displaystyle\int_0^\pi\sin^5 x\,dx = 2W_5 = 2 \times \dfrac{4}{5} \times \dfrac{2}{3} \times 1 = \dfrac{16}{15}$.

Example 3. Find $\displaystyle\int_0^1 x^3 e^x\,dx$ with the reduction formula.

1. For $[0, 1]$ the formula is $J_n = e - n\,J_{n-1}$, because $x^n e^x$ equals $e$ at $x = 1$ and $0$ at $x = 0$ for $n \ge 1$. Also $J_0 = e - 1$.
2. $J_1 = e - (e - 1) = 1$. $J_2 = e - 2$. $J_3 = e - 3(e - 2) = 6 - 2e$.

Practise with new numbers

Try it in Python

The program builds $W_n$ from $W_0$ and $W_1$ with the reduction formula, and checks every value by adding thin strips.

Common mistakes

• Stopping the chain too early. Keep going down by $2$ until you reach $W_0$ or $W_1$.
• Starting from the wrong end. An even $n$ ends at $W_0 = \dfrac{\pi}{2}$, and an odd $n$ ends at $W_1 = 1$.
• Using the Wallis formula without the limits. The boundary term vanishes only over $\left[0, \dfrac{\pi}{2}\right]$.
• A sign slip in the tangent formula. It is $\dfrac{\tan^{n-1}x}{n - 1} - I_{n-2}$, with a minus.
• Forgetting the factor $n$ in $x^n e^x$. The formula is $I_n = x^n e^x - n\,I_{n-1}$.

Remember these

• A reduction formula connects $I_n$ with $I_{n-1}$ or $I_{n-2}$. Use it repeatedly.
• $\int x^n e^x\,dx = x^n e^x - n\int x^{n-1}e^x\,dx$.
• $W_n = \dfrac{n - 1}{n}W_{n-2}$ for $\int_0^{\pi/2}\sin^n x\,dx$, with $W_0 = \dfrac{\pi}{2}$ and $W_1 = 1$.
• $\int \tan^n x\,dx = \dfrac{\tan^{n-1}x}{n - 1} - \int \tan^{n-2}x\,dx$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\int_0^{\pi/2}\cos^7 x\,dx$.

2. Use the reduction formula to find $\displaystyle\int x^2 e^x\,dx$.

3. Show that $\displaystyle\int_0^{\pi}\sin^2 x\,dx = \dfrac{\pi}{2}$.

4. Let $T_n = \displaystyle\int_0^{\pi/4}\tan^n x\,dx$. Show that $T_n + T_{n-2} = \dfrac{1}{n - 1}$.

5. Does $W_n$ grow or fall as $n$ grows? Why?

References

• OpenStax, Calculus Volume 2, Chapter 3, sections 3.1 Integration by Parts and 3.2 Trigonometric Integrals.