Integration by reduction formulae

How to compute the integral

To compute the integral, we set n to its value and use the reduction formula to express it in terms of the (n – 1) or (n – 2) integral. The lower index integral can be used to calculate the higher index ones; the process is continued repeatedly until we reach a point where the function to be integrated can be computed, usually when its index is 0 or 1. Then we back-substitute the previous results until we have computed I_n.

Examples

Below are examples of the procedure.

Cosine integral

Typically, integrals like

${\displaystyle \int {\cos }^{n}x\text{d}x,}$

can be evaluated by a reduction formula.

Start by setting:

${\displaystyle I_n=\int {\cos }^{n}x\text{d}x.}$

Now re-write as:

${\displaystyle I_n=\int {\cos }^{n-1}x\cos x\text{d}x,}$

Integrating by this substitution:

${\displaystyle \cos x\text{d}x=\text{d}(\sin x),}$

${\displaystyle I_n=\int {\cos }^{n-1}x\text{d}(\sin x).}$

Now integrating by parts:

${\displaystyle \begin{array}{cc}\int {\cos }^{n}x\text{d}x& =\int {\cos }^{n-1}x\text{d}(\sin x)={\cos }^{n-1}x\sin x-\int \sin x\text{d}({\cos }^{n-1}x)\\ & ={\cos }^{n-1}x\sin x+(n-1)\int \sin x{\cos }^{n-2}x\sin x\text{d}x\\ & ={\cos }^{n-1}x\sin x+(n-1)\int {\cos }^{n-2}x{\sin }^{2}x\text{d}x\\ & ={\cos }^{n-1}x\sin x+(n-1)\int {\cos }^{n-2}x(1-{\cos }^{2}x)\text{d}x\\ & ={\cos }^{n-1}x\sin x+(n-1)\int {\cos }^{n-2}x\text{d}x-(n-1)\int {\cos }^{n}x\text{d}x\\ & ={\cos }^{n-1}x\sin x+(n-1)I_{n-2}-(n-1)I_n,\end{array}}$

solving for I_n:

${\displaystyle I_n+(n-1)I_n={\cos }^{n-1}x\sin x+(n-1)I_{n-2},}$

${\displaystyle n{I}_n={\cos }^{n-1}(x)\sin x+(n-1)I_{n-2},}$

${\displaystyle I_n=\frac{1}{n}{\cos }^{n-1}x\sin x+\frac{n-1}{n}{I}_{n-2},}$

so the reduction formula is:

${\displaystyle \int {\cos }^{n}x\text{d}x=\frac{1}{n}{\cos }^{n-1}x\sin x+\frac{n-1}{n}\int {\cos }^{n-2}x\text{d}x.}$

To supplement the example, the above can be used to evaluate the integral for (say) n = 5;

${\displaystyle I_5=\int {\cos }^{5}x\text{d}x.}$

Calculating lower indices:

${\displaystyle n=5,I_5=\frac{1}{5}{\cos }^{4}x\sin x+\frac{4}{5}{I}_3,}$

${\displaystyle n=3,I_3=\frac{1}{3}{\cos }^{2}x\sin x+\frac{2}{3}{I}_1,}$

back-substituting:

${\displaystyle \because I_1=\int \cos x\text{d}x=\sin x+C_1,}$

${\displaystyle \therefore I_3=\frac{1}{3}{\cos }^{2}x\sin x+\frac{2}{3}\sin x+C_2,C_2=\frac{2}{3}{C}_1,}$

${\displaystyle I_5=\frac{1}{5}{\cos }^{4}x\sin x+\frac{4}{5}\left[\frac{1}{3}{\cos }^{2}x\sin x+\frac{2}{3}\sin x\right]+C,}$

where C is a constant.

Exponential integral

Another typical example is:

${\displaystyle \int x^n{e}^{ax}\text{d}x.}$

Start by setting:

${\displaystyle I_n=\int x^n{e}^{ax}\text{d}x.}$

Integrating by substitution:

${\displaystyle x^n\text{d}x=\frac{\text{d}(x^{n+1})}{n+1},}$

${\displaystyle I_n=\frac{1}{n+1}\int e^{ax}\text{d}(x^{n+1}),}$

Now integrating by parts:

${\displaystyle \begin{array}{cc}\int e^{ax}\text{d}(x^{n+1})& =x^{n+1}{e}^{ax}-\int x^{n+1}\text{d}(e^{ax})\\ & =x^{n+1}{e}^{ax}-a\int x^{n+1}{e}^{ax}\text{d}x,\end{array}}$

${\displaystyle (n+1)I_n=x^{n+1}{e}^{ax}-a{I}_{n+1},}$

shifting indices back by 1 (so n + 1 → n, n → n – 1):

${\displaystyle n{I}_{n-1}=x^n{e}^{ax}-a{I}_n,}$

solving for I_n:

${\displaystyle I_n=\frac{1}{a}\left(x^n{e}^{ax}-n{I}_{n-1}\right),}$

so the reduction formula is:

${\displaystyle \int x^n{e}^{ax}\text{d}x=\frac{1}{a}\left(x^n{e}^{ax}-n\int x^{n-1}{e}^{ax}\text{d}x\right).}$

An alternative way in which the derivation could be done starts by substituting ${\displaystyle e^{ax}}$.

Integration by substitution:

${\displaystyle e^{ax}\text{d}x=\frac{\text{d}(e^{ax})}{a},}$

${\displaystyle I_n=\frac{1}{a}\int x^n\text{d}(e^{ax}),}$

Now integrating by parts:

${\displaystyle \begin{array}{cc}\int x^n\text{d}(e^{ax})& =x^n{e}^{ax}-\int e^{ax}\text{d}(x^n)\\ & =x^n{e}^{ax}-n\int e^{ax}{x}^{n-1}\text{d}x,\end{array}}$

which gives the reduction formula when substituting back:

${\displaystyle I_n=\frac{1}{a}\left(x^n{e}^{ax}-n{I}_{n-1}\right),}$

which is equivalent to:

${\displaystyle \int x^n{e}^{ax}\text{d}x=\frac{1}{a}\left(x^n{e}^{ax}-n\int x^{n-1}{e}^{ax}\text{d}x\right).}$

Another alternative way in which the derivation could be done by integrating by parts:

${\displaystyle I_n=\int x^{n}x{e}^{ax}\text{d}x,}$

${\displaystyle u=x^n\text{ , }dv=e^{ax},}$

${\displaystyle \frac{du}{dx}=n{x}^{n-1}\text{ , }v=\frac{e^{ax}}{a}}$

${\displaystyle I_n=\frac{x^n{e}^{ax}}{a}-\int n{x}^{n-1}\frac{e^{ax}}{a}\text{d}x}$

${\displaystyle I_n=\frac{x^n{e}^{ax}}{a}-\frac{n}{a}\int x^{n-1}{e}^{ax}\text{d}x}$

Remember:

${\displaystyle I_{n-1}=\int x^{n-1}{e}^{ax}\text{d}x}$

${\displaystyle \therefore I_n=\frac{x^n{e}^{ax}}{a}-\frac{n}{a}{I}_{n-1}}$

which gives the reduction formula when substituting back:

${\displaystyle I_n=\frac{1}{a}\left(x^n{e}^{ax}-n{I}_{n-1}\right),}$

which is equivalent to:

${\displaystyle \int x^n{e}^{ax}\text{d}x=\frac{1}{a}\left(x^n{e}^{ax}-n\int x^{n-1}{e}^{ax}\text{d}x\right).}$

Tables of integral reduction formulas

Rational functions

The following integrals contain:

• Factors of the linear radical ${\displaystyle \sqrt{ax+b}}$
• Linear factors ${\displaystyle px+q}$ and the linear radical ${\displaystyle \sqrt{ax+b}}$
• Quadratic factors ${\displaystyle x^2+a^2}$
• Quadratic factors ${\displaystyle x^2-a^2}$, for ${\displaystyle x>a}$
• Quadratic factors ${\displaystyle a^2-x^2}$, for ${\displaystyle x<a}$
• (Irreducible) quadratic factors ${\displaystyle a{x}^2+bx+c}$
• Radicals of irreducible quadratic factors ${\displaystyle \sqrt{a{x}^2+bx+c}}$

note that by the laws of indices:

${\displaystyle I_{n+\frac{1}{2}}=I_{\frac{2n+1}{2}}=\int \frac{1}{(a{x}^2+bx+c{)}^{\frac{2n+1}{2}}}\text{d}x=\int \frac{1}{\sqrt{(a{x}^2+bx+c{)}^{2n+1}}}\text{d}x}$

Transcendental functions

The following integrals contain:

• Factors of sine
• Factors of cosine
• Factors of sine and cosine products and quotients
• Products/quotients of exponential factors and powers of x
• Products of exponential and sine/cosine factors