]> Methods

Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 1: Indefinite Integrals; page 2

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One very general method for converting an integral to a familiar form is called integration by substitution, which actually uses a change of the variables. It is obtained from the relation ( from the previous page namely   df= df dx dx  :

F( x ) dx= F[ x( u ) ] dx du du (

where the expression on the left is a function of  x  and the one on the right is a function of  u , which can be converted after integration to a function of  x . Very often when a method different from this one is used for integration, an additional substitution is needed, which shows its generality.

A simple example of substitution was given in exercise 4 from the previous page:

u( x )=bxwhere   b   is   a   constant f( bx )dx= f( u ) du b = 1 b f( u )du }    

Another example is:

( a+bx ) p dxwhere   a,b,p    are   constants    

In this case after doing a binomial expansion under the integral, one obtains terms that can be integrated one by one by the rules that we already know. This is not very convenient, in particular when the power  p  is not a natural number, and the binomial expansion is a series. There is a simpler way by substituting   u=a+bx  :

( a+bx ) p dx = 1 b u p du = =C+{ u ( p+1 ) b( p+1 ) = ( a+bx ) ( p+1 ) b( p+1 ) ( p1 ) ln| a+bx | b ( p=1 ) } (

Here are two more useful examples of linear transformations:

a 2 ± x 2 = a 2 ( 1± u 2 ) with{ x=au dx=adu } (

which transforms for instance the following integral to a familiar form:

dx a 2 ± x 2 = 1 a du 1± u 2    

and the second is:

x 2 +px+q= u 2 +( q p 2 4 ) with{ x=u p 2 dx=du } (

which can be furthermore transformed by use of (

The following example shows two ways of substitution. The first uses   u( x )=sinx  :

sinxcosxdx = sinx d( sinx )= = udu = u 2 2 +C= sin 2 x 2 +C } (

An alternative way of substitution is   u( x )=2x  :

sinxcosxdx = 1 2 sin( 2x )dx = 1 4 sinudu= = cosu 4 +C= cos2x 4 +C= cos 2 x sin 2 x 4 +C= = 12 sin 2 x 4 +C= sin 2 x 2 1 4 +C } (

Are the two solutions ( and ( equivalent? Yes they are, since the additional term of 1 4 is irrelevant due to the arbitrary additive constant. On many occasions an integral can be solved by more than one (even two) ways, like this example, but of course the results should be equivalent.

The integral of tanx can be obtained by substitution, as can be seen in the following example:

tanxdx = sinx cosx dx= d( cosx ) cosx =ln| cosx |+C (

Here is the list of the integrals of few elementary functions that can be obtained similarly to (

{ tanxdx =ln| cosx |+C cotxdx =ln| sinx |+C tanhxdx =ln( coshx )+C cothxdx =ln| sinhx |+C } (

Tips. Before trying to do an integral, bring it to the simplest form possible! If its form is not familiar you can try substitution. Use your experience from differentials and your ingenuity to find a suitable substitution!

By Parts

The integration by parts results from the rule of the differential of a product as presented in ( namely   d( uv )=udv+vdu  . By reshuffling the terms and integrating, one obtains:

udv =uv vdu (

The following example with   m1   uses  d(lnx)  in order to obtain a familiar integral:

x m lnxdx= lnxd( x m+1 m+1 ) = = x m+1 lnx m+1 x m+1 m+1 dx x = = x m+1 lnx m+1 x m+1 ( m+1 ) 2 +C= = x m+1 ( m+1 ) 2 [ ( m+1 )lnx1 ]+C } (

The case of  m=0  is included in ( and was obtained by:

lnxdx =xlnx x dx x =x( lnx1 )+C (

where  u = lnx  and  v=x .

The following example appears similar to ( but the use of the parts is different:

x m expxdx= x m d( expx )= = x m expxm x m1 expxdx } (

It is obvious that this procedure lowers the power of  x  by one unit each time. In order to obtain a finite number of steps one has to restrict  m  to natural numbers.

There is a similarity in the next example:

x m sinxdx= x m d( cosx )= = x m cosx+m x m1 cosxdx= = x m cosx+m x m1 d( sinx )= = x m cosx+m x m1 sinxm( m1 ) x m2 sinxdx } (

Here again the power  m  should be restricted to natural numbers.

Another example with trigonometric functions is:

cos 2 xdx= cosxd( sinx )= =cosxsinx+ sin 2 xdx= =cosxsinx+ dx cos 2 xdx } (

which looks as a boomerang since   cos 2 xdx   appears on both sides of the equation. A careful look shows that this is actually the solution and after moving it from the right to the left one obtains:

cos 2 xdx = 1 2 ( cosxsinx+x )+C (

It was stressed in Chapter 1,Derivatives that the derivative of a periodic function is also periodic. Could one extend this also to integrals? No! ( shows that the indefinite integral of a periodic function does not have to be periodic. This is illustrated at Fig. ∫cos2xdx.

In the case of expressions containing  (1−x2)  it could be useful to apply the substitution  x=sinθ  as shown in the following example:

x=sinθdx=cosθdθ 1 x 2 dx= cos 2 θ dθ= = 1 2 ( sinθcosθ+θ )+C= = 1 2 ( x 1 x 2 +asinx )+C } (

where ( obtained previously with integration by parts, was used.

Tips. For a given integral the number of options for doing integration by parts is limited, therefore trial and error can be applied. On the other hand acquired experience is always helpful.

Parametric functions

In order to express an integral of a parametric function one has to substitute the independent variable by the parameter :

Fory=y( t )andx=x( t ) y( x )dx= y( t ) dx dt dt } (

The cycloid ( will be used as an example:

{ x=r( φsinφ ) y=r( 1cosφ ) }    

According to (

ydx= r( 1cosφ ) d dφ [ r( φsinφ ) ]dφ= = r 2 ( 1cosφ ) 2 dφ= r 2 ( 12cosφ+ cos 2 φ )dφ= = r 2 [ φ2sinφ+ 1 2 ( cosφsinφ+φ ) ]+C= = r 2 2 ( 3φ4sinφ+cosφsinφ )+C } (

where we used the elementary integrals and (

As already stated, it is a very good habit to check an integral by differentiating it. The expression obtained at ( is a function of φ , namely  I( φ )  . Here we have to take its derivative with respect to  x  in order to obtain  y . According to the known rules for differentiation of a parametric function:

d dx I( φ )= d dφ I( φ ) d dφ x( φ ) =y( φ )    

where y should correspond to that of the cycloid above. Try it!


Exercise 1. Calculate the following indefinite integrals! (hint: substitution)

  1. lnx x dx
  3. x 2 dx x 3 1
  5. cos 2 x sin 5 xdx

Exercise 2. Calculate the following indefinite integrals! (hint: substitution)

  1. dx x 2 2x+5
  3. ( x+1 )dx 1 x 2
  5. ( x+1 )dx 2+2x3 x 2
Remark.The last integral of this group contains elements of the previous two. It is not advisable to solve it before the others.

Exercise 3. Calculate the following indefinite integrals! (hint: by parts)

  1. atanx dx
  3. asinhx dx
  5. xatanx dx

Exercise 4. For any natural number n show that

  1. x 2n1 exp( x 2 2 ) dx could be solved as a finite sum of additive terms expressed by elementary functions!
  2. x 2n exp( x 2 2 ) dx could be solved as a finite sum of additive terms expressed by elementary functions except one term that includes exp( x 2 2 ) dx that cannot be expressed in terms of elementary functions.

Exercise 5. Calculate the indefinite integral I(x)= ydx for the elliptic curve

y=b 1 x 2 a 2

  1. directly from this formula.
  2. by use of the parametric expression of the ellipse: x=acosθ y=bsinθ }
  3. Calculate dI( x ) dx in both cases and compare them!

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Next topic: page 3 Special Methods

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