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= 2 log 2 x
= 2 log 2 ( 2 1) = 2 log 2 1
YOU TRY IT Evaluate
x 2 sin x dx
We conclude this section by doing another definite integral, but we use a slightly different approach from that in Example 73
CALCULUS DeMYSTiFieD
EXAMPLE
Calculate the integral sin x cos x dx
SOLUTION
We use integration by parts, but we apply the technique to the corresponding indefinite integral We let u = sin x and dv = cos x dx Then u( x) = sin x v( x) = sin x So sin x cos x dx = u dv v du sin x cos x dx du = u ( x) dx = cos x dx dv = v ( x) dx = cos x dx
= uv
= ( sin x) ( sin x)
At first blush, it appears that we have accomplished nothing The new integral is just the same as the old integral But, in fact, we can move the new integral (on the right) to the left-hand side to obtain 2 sin x cos x dx = sin2 x
Throwing in the usual constant of integration, we obtain sin x cos x dx = 1 sin2 x + C 2
Now we complete our work by evaluating the definite integral:
sin x cos xdx =
1 sin2 x 2
1 1 [sin2 2 sin2 ( /2) ] = 2 2
7 M E T H O D S O F I N T E G R A T I O N
We see that there are two ways to treat a definite integral using integration by parts One is to carry the limits of integration along with the parts calculation The other is to do the parts calculation first (with an indefinite integral) and then plug in the limits of integration at the end Either method will lead to the same solution
YOU TRY IT Calculate the integral
e x cos 2x dx
72 Partial Fractions 721 Introductory Remarks
The method of partial fractions is used to integrate rational functions, or quotients of polynomials We shall treat here some of the basic aspects of the technique The first fundamental observation is that there are some elementary rational functions whose integrals we already know
I Integrals of Reciprocals of Linear Functions
An integral 1 dx ax + b with a = 0 is always a logarithmic function In fact we can calculate 1 1 dx = ax + b a 1 1 dx = log |x + b/a| x + b/a a
II Integrals of Reciprocals of Quadratic Expressions
An integral 1 dx , c + ax 2
CALCULUS DeMYSTiFieD
when a and c are positive, is an inverse trigonometric function In fact we can use what we learned in Section 663 to write 1 1 dx = 2 c c + ax = 1 c 1 dx 1 + (a/c)x 2 1 dx 1 + ( a/c x)2 1 dx 1 + ( a/c x)2
a 1 = ac c
1 = arctan( a/c x) + C ac
III More Integrals of Reciprocals of Quadratic Expressions
An integral 1 dx + bx + c
ax 2
with a > 0, and discriminant b2 4ac negative, will also be an inverse trigonometric function To see this, we notice that we can write ax 2 + bx + c = a x 2 + b x+ a +c b2 4a
= a x2 +
b2 b x+ 2 a 4a b 2a
+ c b2 4a
=a x+
+ c
Since b2 4ac < 0, the final expression in parentheses is positive For simplicity, let = b/2a and let = c b2 /(4a) Then our integral is 1 dx + a (x + )2
7 M E T H O D S O F I N T E G R A T I O N
Of course we can handle this using II above We find that 1 dx = ax 2 + bx + c 1 dx + a (x + )2
1 a = arctan (x + ) + C a
IV Even More on Integrals of Reciprocals of Quadratic Expressions
Evaluation of the integral 1 dx + bx + c
ax 2
when the discriminant b2 4ac is 0 will be a consequence of the work we do below with partial fractions We will say no more about it here
722 Products of Linear Factors
We illustrate the technique of partial fractions by way of examples
EXAMPLE
Here we treat the case of distinct linear factors Let us calculate 1 dx 3x + 2
SOLUTION
We notice that the integrand factors as 1 1 = ( x 1) ( x 2) 3x + 2 ( )
[Notice that the quadratic polynomial in the denominator will factor precisely when the discriminant is 0, which is case IV from Subsection 721] Our goal is to write the fraction on the right-hand side of ( ) as a sum of simpler fractions With this thought in mind, we write A B 1 = + , ( x 1) ( x 2) x 1 x 2
CALCULUS DeMYSTiFieD
where A and B are constants to be determined Let us put together the two fractions on the right by placing them over the common denominator ( x 1) ( x 2) Thus A B A( x 2) + B( x 1) 1 = + = ( x 1) ( x 2) x 1 x 2 ( x 1) ( x 2) The only way that the fraction on the far left can equal the fraction on the far right is if their numerators are equal This observation leads to the equation 1 = A( x 2) + B( x 1) or 0 = ( A + B) x + ( 2A B 1) Now this equation is to be identically true in x; in other words, it must hold for every value of x So the coefficients must be 0 At long last, then, we have a system of two equations in two unknowns: A+B =0
2A B 1 = 0 Of course this system is easily solved and the solutions found to be A = 1, B = 1 We conclude that 1 1 1 = + ( x 1) ( x 2) x 1 x 2 What we have learned, then, is that 1 dx = x 2 3x + 2 1 dx + x 1 1 dx x 2
Each of the individual integrals on the right may be evaluated using the information in I of Subsection 721 As a result, 1 dx = log |x 1| + log |x 2| + C 3x + 2
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