hey,i got it on net

rakesh61

What is reductional formula in Integration

it will be better if any one can explain this concept with example rates assured

mukulss

hey,i got it on net

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

$\displaystyle \int_{0}^{1}$	= $\displaystyle \left[\vphantom{x^6e^x}\right.$ x⁶e^x $\displaystyle \left.\vphantom{x^6e^x}\right]_{0}^{1}$ - $\displaystyle \int_{0}^{1}$ 6x⁵e^x dx
	= e - 6 $\displaystyle \int_{0}^{1}$ x⁵e^x dx

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 = $\displaystyle \int_{0}^{1}$ xⁿe^x dx

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

I_n	= $\displaystyle \left[\vphantom{x^ne^x}\right.$ xⁿe^x $\displaystyle \left.\vphantom{x^ne^x}\right]_{0}^{1}$ - $\displaystyle \int$ nx^n-1e^x dx
	= e - n $\displaystyle \int_{0}^{n}$ x^n-1e^x dx
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).

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

Therefore

I_n	= - sin^n-1cos x - $\displaystyle \int$ (n - 1)sin^n-2x cos x(- cos x) dx
	= - sin^n-1cos x + (n - 1) $\displaystyle \int$ sin^n-2x cos²x dx
	= - sin^n-1cos x + (n - 1) $\displaystyle \int$ sin^n-2x(1 - sin²x) dx
	= - sin^n-1cos x + (n - 1) $\displaystyle \int$ $\displaystyle \left(\vphantom{\sin^{n-2}x-\sin^nx}\right.$ sin^n-2x - sinⁿx $\displaystyle \left.\vphantom{\sin^{n-2}x-\sin^nx}\right)$ dx
	= - sin^n-1cos x + (n - 1) $\displaystyle \left[\vphantom{\int\sin^{n-2}x dx-\int\sin^nx dx}\right.$ $\displaystyle \int$ sin^n-2x dx - $\displaystyle \int$ sinⁿx dx $\displaystyle \left.\vphantom{\int\sin^{n-2}x dx-\int\sin^nx dx}\right]$
	= - sin^n-1x cos x + (n - 1) $\displaystyle \left[\vphantom{I_{n-2}-I_n}\right.$ I_n-2 - I_n $\displaystyle \left.\vphantom{I_{n-2}-I_n}\right]$

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

nI_n	= $\displaystyle \left[\vphantom{-\sin^{n-1}x\cos x}\right.$ -sin^n-1x cos x $\displaystyle \left.\vphantom{-\sin^{n-1}x\cos x}\right]_{0}^{{\pi/2}}$ + (n - 1)I_n-2
	= (n - 1)I_n-2 provided n $\displaystyle \geq$ 2

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

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

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

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

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

and

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

rakesh61

thanks

Ask & Discuss Questions with Community & Experts

hey,i got it on net

Reduction Formulas