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There are five methods that can help you integrate a function. All these methods do is simplify the expression you're trying to integrate until you are left with something that you can recall what it integrates to.
1. Trigonometric Identities
This is where you have an expression made up of trigonometric functions (i.e. sin, cos, tan, sec, cosec, cot). You can use one or more of the trig identities, that you should be able to recall from memor,y to rewrite the expression into something that you can integrate with more ease.
Identities you'll need to memorise:
sin2(x)+cos2(x)=1 (here you can derive identies involving cot, sec and cosec by diving by sin or cos)
These are the simplest forms of trig identities that you'll need to know. Using them will make it a lot easier to integrate a tricky expression.
2. Partial Fractions
The method of partial fractions can be used when you have to integrate a fraction where the denominator is a product of two different functions.
i.e (a+b)/cd ,
you begin with supposing that
(a+b)/cd = e/c + f/d (then multiply by cd)
a + b = ed +fc (at this stage you should be able to solve for e and f)
It may then be easier to integrate this simpler expression of e/c + f/d.
This method is probably the most common and the best to try if you are still stuck after using trig identities and/or partial fractions.
This involves picking a part of the function say x2 and letting it equal an arbitary letter, for example u. You then differentiate this picked out part: u = x2 , du/dx =2x , thus dx= 1/2x du
You can then substitute this 1/2x du for dx in the original expression. This should hopefully simplify the expression and make it easy to integrate with respect to u.
Knowing what substitution to use in different questions comes from experience, so do a lot of practice and you'll get the hang of it.
4. By Parts
This method is derived from the product rule however it is easiest to just memorise it in this form:
∫ u(dv/dx) dx = uv − ∫ v(du/dx) dx .
where u and v are any functions.
This may be the trickiest method but there will be cases where it is the only one you can use.
This involves trying to 'guess' what functions could be differentiated to get your original expression.
With practice this becomes easier.
Integration can be really tricky but with a lot practice you can begin to spot the best ways to simplify expressions into formats that you know the integral of.see more