Reduction Formulas

This is refinement of the technique of integration by parts. It is a labour saving device designed to cope with the situation where a simple-minded approach would leave us integrating by parts many times over before reaching an answer.

Example 9..1   Evaluate $\int_{0}^{1}$x⁶e^x dx.

Solution x⁶ is a function that simplifies on differentiation, e^x is a function we can integrate in our heads, and so we use integration by parts.

Let u = e^x and v = x⁶. Then $\int$u dx = e^x and v' = 6x⁵. And so

$$\int_{0}^{1} \left[\vphantom{x^6e^x}\right. \left.\vphantom{x^6e^x}\right]_{0}^{1} \int_{0}^{1}$$

$$\int_{0}^{1}$$

It is an improvement, since we have an integral involving a lower power of x, but we are still a long way from the answer. What we must do next is use parts again. This will get us down to $\int_{0}^{1}$x⁴e^x dx. Then again to get down to x³e^x. And so on.

We shall reach an answer, but it is going to take time, time spent doing a lot of repetitive processes. What a reduction formula does is take the work out of the repeats. Instead of doing lots of integrations by parts, we shall do one, with an n in place of the 6. This will give us a formula which we can then use over and over with different values for n.

Let n be a positive integer, and let

I_n = $$\int_{0}^{1}$$xⁿe^x dx

Integrate by parts with u = e^x and v = xⁿ. The result is

$$\left[\vphantom{x^ne^x}\right. \left.\vphantom{x^ne^x}\right]_{0}^{1} \int$$ I_n

$$\int_{0}^{n}$$
and so

I_n = e - nI_n-1 (*)

This is what is meant by a reduction formula. It gives I_n in terms of a simpler integral of the same type (I_n-1).

Now let us use it on our earlier problem, which was to calculate I₆.

(*) with n = 6 gives I₆ = e - 6I₅. Now use it again with n = 5 to get I₅ = e - 5I₄ and substitute to get

I₆ = e - 6I₅ = e - 6(e - 5I₄) = - 5e + 30I₄

Now use the formula again, but with n = 4. Substitute once more, and then keep going in the same sort of way.

I₆ = - 5e + 30I₄ = - 5e + 30(e - 4I₃) = 25e - 120I₃ = 25e - 120(e - 3I₂) = - 95e + 360I₂ = - 95e + 360(e - 2I₁) = 265e - 720I₁ = 265e - 720(e - I₀)

At that point we have to pause, because our formula was only valid for n > 0 and we are now down to n = 0. However, I₀ is an integral we need no help with. I₀ = $\int_{0}^{1}$e^x dx = e - 1. Therefore

I₆ = 265e - 720e + 720(e - 1) = 265e - 720

Example 9..2   Let I_n = $\int$sinⁿx dx where n is an integer $\geq$ 2. Find a reduction formula for I_n, and then use it to calculate $\int_{0}^{{\pi/2}}$sin⁶x dx.

Solution With all these questions we use integration by parts, and the aim is to recover an integral of the same shape but with a smaller n.

The largest section of the integrand that we can integrate in our heads is sin x. So set u = sin x and v = sin^n-1x (the bit that is left after we have removed u).

$$\int$$u dx = - cos x ,        $${\frac{{dv}}{{dx}}}$$ = (n - 1)sin^n-2x cos x

Therefore

$$\int$$ I_n

$$\int$$

$$\int$$

$$\int \left(\vphantom{\sin^{n-2}x-\sin^nx}\right. \left.\vphantom{\sin^{n-2}x-\sin^nx}\right)$$

$$\left[\vphantom{\int\sin^{n-2}x dx-\int\sin^nx dx}\right. \int \int \left.\vphantom{\int\sin^{n-2}x dx-\int\sin^nx dx}\right]$$

$$\left[\vphantom{I_{n-2}-I_n}\right. \left.\vphantom{I_{n-2}-I_n}\right]$$

Taking all the I_n terms on to the left we get

nI_n = - sin^n-1x cos x + (n - 1)I_n-2

To use this on a definite integral we just put in the limits.

So if we had that I_n = $\int_{0}^{{\pi/2}}$sinⁿx dx, the formula would become

$$\left[\vphantom{-\sin^{n-1}x\cos x}\right. \left.\vphantom{-\sin^{n-1}x\cos x}\right]_{0}^{{\pi/2}}$$ nI_n

$$\geq$$

Repeated application of this formula will bring you down to I₁ = $\int$sin x dx or I₀ = $\int$ dx, both of which you can do without difficulty.

For example, with the definite integral and n = 6, we have

I₆ = $${\frac{{5}}{{6}}}$$I₄ = $${\frac{{5}}{{6}}}$$$${\frac{{3}}{{4}}}$$I₂ = $${\frac{{5}}{{6}}}$$$${\frac{{3}}{{4}}}$$$${\frac{{1}}{{2}}}$$I₀

and since $$\int_{0}^{{\pi/2}}$$ dx = $${\frac{{\pi}}{{2}}}$$, this leads to

I₆ = $${\frac{{5}}{{6}}}$$$${\frac{{3}}{{4}}}$$$${\frac{{1}}{{2}}}$$I₀ = $${\frac{{5}}{{6}}}$$$${\frac{{3}}{{4}}}$$$${\frac{{1}}{{2}}}$$$${\frac{{\pi}}{{2}}}$$ = $${\frac{{5\pi}}{{32}}}$$

Reduction formulas for $\int$cosⁿx dx and $\int$xⁿe^ax dx are to be found in your handbook. Others that are sometimes useful are

$$\int$$sin^mx cosⁿx dx = $${\frac{{\sin^{m+1}x\cos^{n-1}x}}{{m+n}}}$$ + $${\frac{{n-1}}{{m+n}}}$$$$\int$$sin^mx cos^n-2x dx

and

$$\int$$sin^mx cosⁿx dx = - $${\frac{{\sin^{m-1}x\cos^{n+1}x}}{{m+n}}}$$ + $${\frac{{m-1}}{{m+n}}}$$$$\int$$sin^m-2x cosⁿx dx

Faced with an integral such as $\int$sin⁸x cos⁴x dx you use the first of these to bring down the powers of cos x and the second to bring down those of sin x.