Reduction formulae are used to help integrate 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 until we find an expression that we are able to integrate easily.

For example consider

₀∫^pi/2sin⁴x dx

Without a reduction formula this would be hard to integrate however we can use a reduction formula to try and reduce the number 8 to something we can integrate.
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 sin^n-1x dx

Then we differentiate sin^n-1x to get (n-1)sin^n-2xcosx
and integrate sinx to get -cosx

Using integration by parts we now get

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

Which gives

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

and substituting the identity sin² = 1 – cos²

gives

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

which simplifies to

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

and finally get our reduction formula

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

This is useful as it tells us ane expression for the integral of sinⁿx in terms of sin^n-2x. 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 sin⁰x or sin¹x

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

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

so

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

Then we find it for sin² x using our reduction formula

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

and then finally for sin⁴ we get

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

Which is our final answer

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