Home > calculus, maths > Intergration using a Reduction Formula – with example

## 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

0pi/2sin4x 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

∫sinnx dx

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

∫sinx sinn-1x dx

Then we differentiate sinn-1x to get (n-1)sinn-2xcosx
and intergrate sinx to get -cosx

Using integration by parts we now get

∫sinnx dx = -sinn-1x cosx + ∫cosx (n-1)sinn-2x cosx dx

Which gives

∫sinnx dx = -sinn-1x cosx + (n-1)∫cos2x sinn-2x dx

and substituting the identity sin2 = 1 – cos2

gives

∫sinnx dx = -sinn-1x cosx + (n-1)∫(1-sin2x )sinn-2x dx

which simplifies to

∫sinnx dx = -sinn-1x cosx + (n-1)∫sinn-2x dx – (n-1)∫sinnx 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)∫sinnx dx = -sinn-1x cosx + (n-1)∫sinn-2x dx

and finally get our reduction formula

∫sinnx dx = 1/n [-sinn-1x cosx + (n-1)∫sinn-2x dx]

This is useful as it tells us ane expression for the integral of sinnx in terms of sinn-2x. This formula can then be applied to sinn-2 untill we have reduced the poser to one which we can solve the integral more easily such as sin0x or sin1x

In our original example we had
0pi/2sin4x dx

using our reduction formula we can work backwards up from sin0 to fin the integral of sin8

so

0pi/2sin0dx = 0pi/21dx = π/2 – 0 = π/2

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

0pi/2sin2dx =1/2 [ -sinxcosx + π/2] = π/4

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

0pi/2sin4dx= 1/2[-sin3x cosx + 3π/4] = 3π/8

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

Categories: calculus, maths
1. October 9, 2009 at 5:42 am

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”

• October 9, 2009 at 1:21 pm

For the first one i would try writing
$\int \! (ln(x))^n \cdot ln(x) \, dx$ and then using integration by parts to solve it.

Id then try using integration by parts again on the second one – differentiating $x^n$ and integrating $e^x$.

Hope that helps,
david

1. April 4, 2018 at 7:04 am