Integration by Substitution

Current UK exam textbooks pass over proofs and mathematical discussions in a hurry to show the ‘how to’ of exam questions.

Integration by substitution is a little more than just backwards chain-rule and deserves a fuller treatment.

Try this,

Let,

y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x

then,

\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}(x)

Suppose that there exists a function g, of another variable u, such that x=\textnormal{g}(u) and let, \textnormal{f}(x)=\textnormal{f}(\textnormal{g}(u))=\textnormal{F}(u). So that,

\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{F}(u)

Now, by the chain rule,

\dfrac{\textnormal{d}y}{\textnormal{d}u}=\dfrac{\textnormal{d}y}{\textnormal{d}x}\times \dfrac{\textnormal{d}x}{\textnormal{d}u}=\textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u}

Hence,

y=\displaystyle \int \dfrac{\textnormal{d}y}{\textnormal{d}u} \ \textnormal{d}u=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u

i.e.

y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u

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