## Intergration using a Reduction Formula – with example

Reduction formulae are used to help intergrate complex expressions. They are useful because they can make a link between a complex expression and slightly simpler one. Then applying the same formula to the simpler one we can make it into an even simpler one untill we find an expression that we are able to intergrate easily.

For example consider

_{0}∫^{pi/2}sin^{4}x dx

Without a reduction formula this would be hard to intergrate however we can use a reduction formula to try and reduce the number 8 to something we can intergrate.

In order to do this we should consider the general integral

∫sin^{n}x dx

WE can use integration by parts to reduce this if we right it as

∫sinx sin^{n-1}x dx

Then we differentiate sin^{n-1}x to get (n-1)sin^{n-2}xcosx

and intergrate sinx to get -cosx

Using integration by parts we now get

∫sin^{n}x dx = -sin^{n-1}x cosx + ∫cosx (n-1)sin^{n-2}x cosx dx

Which gives

∫sin^{n}x dx = -sin^{n-1}x cosx + (n-1)∫cos^{2}x sin^{n-2}x dx

and substituting the identity sin^{2} = 1 – cos^{2}

gives

∫sin^{n}x dx = -sin^{n-1}x cosx + (n-1)∫(1-sin^{2}x )sin^{n-2}x dx

which simplifies to

∫sin^{n}x dx = -sin^{n-1}x cosx + (n-1)∫sin^{n-2}x dx – (n-1)∫sin^{n}x dx

But the last integral on the right is equal to the one on the left so these can be grouped together to give

(1+n-1)∫sin^{n}x dx = -sin^{n-1}x cosx + (n-1)∫sin^{n-2}x dx

and finally get our reduction formula

∫sin^{n}x dx = 1/n [-sin^{n-1}x cosx + (n-1)∫sin^{n-2}x dx]

This is useful as it tells us ane expression for the integral of sin^{n}x in terms of sin^{n-2}x. This formula can then be applied to sin^{n-2} untill we have reduced the poser to one which we can solve the integral more easily such as sin^{0}x or sin^{1}x

In our original example we had

_{0}∫^{pi/2}sin^{4}x dx

using our reduction formula we can work backwards up from sin^{0} to fin the integral of sin^{8}

so

_{0}∫^{pi/2}sin^{0}dx = _{0}∫^{pi/2}1dx = π/2 – 0 = π/2

Then we find it for sin<sup>2</sup> x using our reduction formula

_{0}∫^{pi/2}sin^{2}dx =1/2 [ -sinxcosx + π/2] = π/4

and then finally for sin<sup>4</sup> we get

_{0}∫^{pi/2}sin^{4}dx= 1/2[-sin^{3}x cosx + 3π/4] = 3π/8

Which is our final answer

If you have any questions or comments please leave them using the form below

Hey can u help me with these questions?

“Use integration by parts to prove the reduction formula:

∫(lnx)^n dx= x(lnx)^n – n∫(lnx)^n-1 dx

Then use the above to evaluate the integral:

∫ (lnx)^3 dx

Prove:

“∫ x^n e^x dx= x^n e^x – n∫x^n-1 e^x dx

Then use the above to evaluate:

∫x^3 e^x dx”

For the first one i would try writing

and then using integration by parts to solve it.

Id then try using integration by parts again on the second one – differentiating and integrating .

Hope that helps,

david