# Methods of Integration

Page Contents

## Integration by Parts

From the properties of derivatives we have:

fract{d ( ~f ~g ),d~x} _ = _ ~f _ fract{d ~g,d~x} _ + _ ~g fract{d ~f,d~x}

rearranging

_ ~f _ fract{d ~g,d~x} _ = _ fract{d ( ~f ~g ),d~x} - ~g fract{d ~f,d~x}

integrating

 int{ _ ~f,,,}fract{d ~g,d~x} d~x _ = _ [ ~f ~g ] _ - _ int{,,,}~g fract{d ~f,d~x} d~x

#{Example}

Integrate _ ~x ~e ^~x.

_

int{,,,} ~x ~e ^~x d~x _ = _ int{,,,}~x fract{d ~e ^~x,d~x} d~x _ = _ [ ~x ~e ^~x ] _ - _ int{,,,} ~e ^~x d~x _ = _ [ ( ~x - 1 ) ~e ^~x ]

## Change of Variable

We want to integrate a function ~f of ~x, i.e. find

int{ _ ~f,~a,~b,d~x} _ = _ script{sqrb{ _ ~F ( ~x ) _ },,,~b,~a}

Now suppose _ ~u = ~u ( ~x ) _ is such that ~u ( ) is a monotone function. Then we can find a function ~H such that _ ~F ( ~x ) _ = _ ~H ( ~u ) , _ simply by defining _ ~H ( ~u ) = ~F ( ~u^{-1} ( ~x ) ) . _ So we have

script{sqrb{ _ ~F ( ~x ) _ },,,~b,~a} _ = _ script{sqrb{ _ ~H ( ~u ( ~x ) ) _ },,,~u ( ~b ),~u ( ~a )} _ = _ int{,~u(~a),~u(~b),} fract{d~H,d~u} d~u

From the properties of derivatives ("function of a function") we have:

fract{d~F,d~x} _ = _ fract{d ~H,d~u} fract{d ~u,d~x}

So

fract{d ~H,d~u} _ = _ fract{d~F,d~x} ./ fract{d ~u,d~x} _ = fract{ _ ~f _ , d~u/d~x}

i.e.

 int{ _ ~f ( ~x ),~a,~b,d~x} _ = _ int{,~u(~a),~u(~b),} fract{ _ ~f _ , d~u/d~x} d~u

This is sometimes written in the following form as a #{mnemonic}:

 int{ _ ~f ( ~x ),~a,~b,d~x} _ = _ int{,~u(~a),~u(~b),} _ ~f _ fract{d~x,d~u} d~u

#{Example}

Calculate the integral: _ _ _ I _ = _ int{,0,~a,} fract{cos ( &sqrt.\$~x ),&sqrt.\$~x} d~x

Consider _ ~u _ = _ &sqrt.\$~x . _ This is monotone in the range _ 0 < ~x < &pi. ./ 2 , _ and _ d~u ./ d~x _ = _ 1 ./ 2 &sqrt.\$~x , so

I _ = _ int{,0,&sqrt.\$~a,} fract{cos ( &sqrt.\$~x ),&sqrt.\$~x} 2 &sqrt.\$~x _ d~u _ = _ int{,0,&sqrt.\$~a,} 2 cos ( ~u ) d~u _ = _ script{sqrb{2 sin ( ~u )},,,&sqrt.\$~a,0}_ = _ 2 sin ( &sqrt.\$~a )

Of course we could have integrated directly by noticing that _ d ( 2 sin( &sqrt.\$~x ) ) ./ d~x _ = _ cos( &sqrt.\$~x ) ./ &sqrt.\$~x , _ but the method was useful in coming to this conclusion, and can be used in more complicated cases.