Example :
25 x dx
∫ ( x - 2 ) (4 x2 + 9 ) Use partial fractions to calculate the integral
Solution :
25 x
A
+
=
x-2
( x - 2 ) (4 x2 + 9 )
Bx+C
4 x2 + 9
(x-2)
3 pt
Multiply both sides with ( x - 2 ) and put x = 2; yield A = 2
4 pt
Evaluate the integrand at x = 0 ; 0 = -A / 2 + C / 9 , yields C = 9
4 pt
25 x
A
+
=
2
x-2
( x - 2 ) (4 x + 9 )
Bx+C
4 x2 + 9
(x)
Multiply both sides with ( x ) and take limit as x goes to ∞ ;
that yields 0 = A + B / 4 ; which gives B = -8
Assoc. Prof. Dr. Ali Çınar
MEF University, Faculty of Engineering, Department of Mechanical Engineering
4 pt
∫
25 x dx
=
2
( x - 2 ) (4 x + 9 )
2
+
x-2
∫
-8 x + 9
4x2+9
dx
= 2 ln | x -2 | - ln ( 4 x 2 + 9 ) + ( 3 / 2 ) tan -1 ( 2 x / 3 ) + C
3 pt
2 pt
3 pt
2 pt
= ln [ ( x - 2 ) 2 / ( 4 x 2 + 9 ) + ( 3 / 2 ) tan -1 ( 2 x / 3 ) + C
Assoc. Prof. Dr. Ali Çınar
MEF University, Faculty of Engineering, Department of Mechanical Engineering
Use u = x 2 substitution and integration by parts to calculate the integral
1/√2
∫
2 x sin
-1
3 pt
Let u = x 2 ; du = 2 x dx ;
x=0,u=0,x=1/√2,u=1/2
2
( x ) dx
0
3 pt
1/2
=
∫
sin -1 u du
= u sin -1 u -
0
3 pt
∫
u du
√1-u2
3 pt
3 pt
1/2
= u sin -1 u + √ 1 - u 2
3 pt
0
3 pt
3 pt
1 pt
= ( 1 / 2 ) ( π / 6 ) + √ 3 / 2 - 1 = π / 12 + √ 3 / 2 - 1
Alternative Solution : sin -1 ( x 2 ) = u ; x 2 = sin u ;
x = 0 , u = 0 , x = 1 / √ 2 , u = π / 6 ; 2 x dx = cos u du
1/√2
∫
2 x sin -1 ( x 2 ) dx =
0
1/√2
∫
π/6
∫
u cos u du = u sin u - ∫ sin u du = u sin u + cos u
0
2 x sin -1 ( x 2 ) dx = π / 12 + √ 3 / 2 - 1
0
Assoc. Prof. Dr. Ali Çınar
MEF University, Faculty of Engineering, Department of Mechanical Engineering
π/6
0
1/√2
Alternative Solution :
∫2x
1/√2
sin -1 ( x 2 ) dx =
0
Recall :
d
sin -1 u =
du
x
2
sin -1
2
(x )-
∫
0
x2
sin -1 ( x 2 ) =
0
1
√1-u2
1/√2
∫ 2 x dx
2 x dx
√1-x4
du
dx
1√2
= x 2 sin -1 ( x 2 ) + √ 1 - x 4
0
since , sin -1 ( 1 / 2 ) = π / 6
1/√2
∫
2 x sin -1 ( x 2 ) dx = π / 12 + √ 3 / 2 - 1
0
Assoc. Prof. Dr. Ali Çınar
MEF University, Faculty of Engineering, Department of Mechanical Engineering
2a ) Find the derivative of the function y = 2 x sin -1 ( 2 x ) + √ 1 - 4 x 2
Solution : y´= 2 sin
-1
(2x)+
4 pt
2x(2)
√1-4x2
+
-8x
2√1-4x2
4 pt
4 pt
y´= 2 sin -1 ( 2 x )
3 pt
2b ) Fin the derivative of the function y = cosh -1 ( sec x ) = cosh - 1 u
Solution : let u = sec x , du / dx = tan x sec x
dy/ dx = [ 1 / √ ( u 2 - 1 ) ] du / dx ;
2 pt
u 2 -1 = sec 2 x - 1 = tan 2 x
4pt
dy/ dx = ( tan x sec x ) / tan x = sec x
2 pt
4 pt
3 pt
Assoc. Prof. Dr. Ali Çınar
MEF University, Faculty of Engineering, Department of Mechanical Engineering
2
∫
2c . Use u = sec -1 x substitution to evaluate the integral
2/ √ 3
cos ( sec -1 x ) dx
x√ x2-1)
u = sec -1 x ; x = sec u = 1 / cos u
cos u = 1 / x ; x = 2 / √ 3 , u = π / 6 ; x = 2 , u = π / 3
du = dx / [ x √ ( x 2 -1 )
2
∫
2/ √ 3
cos ( sec
-1
x ) dx
x√ x2-1)
4 pt
3 pt
π/3
=
∫ cos u du = sin u
π/6
3 pt
4 pt
π/3
π/6
4 pt
= ( √ 3 -1 ) / 2
2 pt
Assoc. Prof. Dr. Ali Çınar
MEF University, Faculty of Engineering, Department of Mechanical Engineering