Download Definite Integration: Properties, Theorems, and JEE Problems and more Study notes Mathematics in PDF only on Docsity!
D E F I N I T E I N T E G R AT I O N
1. INTRODUCTION
Let f(x) be a continuous function defined on a closed interval [a, b] and (^) ∫ f(x)dx^ =^ F(x)^ +c then
b b b a a a
f(x)dx = [F(x)] or f(x)dx = F(b) −F(a) ∫ ∫ is called the definite integral of f(x) within limits a and b. The interval
[a, b] is called the range of integration. Every definite integral has a unique solution.
Note:
b
a
f(x)dx = F(b) −F(a) ∫
also represents the net area of the curve f(x) with x-axis.
/ 2
0
sin x dx
π
∫
Sol:
/2 /2^ / 2
0 0 0
1 cos2x 1 sin2x 1 sin x dx dx x 0 2 2 2 2 2 4
π π π − π π = (^) = − = − =
∫ ∫
Illustration 1: If
1 2
0
(3x + 2x + k)dx =0, ∫
find the value of k. (JEE MAIN)
Sol: Here the answer of the definite integral
1 2
0
3x (^) + 2x +k dx ∫ (^)
is already given i.e. 0 hence by using simple integral
formulas we can solve it and by comparing it to 0, we will obtain the value of k.
Here, we have,
1 2
0
(3x + 2x + k)dx = 0 ∫
1 3 2
0
x x 3 2 kx 0 3 2
+^ +^ =
1 3 2
0
x (^) + x + kx = 0
(1 + 1 + k) – (0 + 0 + 0) = 0 ; 2 + k = 0 ⇒ k = –
Illustration 2: Evaluate:
4 2 3
0
(2sec x x 2)dx.
π
∫
(JEE MAIN)
Sol: As we know (^) { ( ) ( )} ( ) ( )
b b b
a a a
f x ± g x dx = f x dx ± g x dx ∫ ∫ ∫
. Hence by using this method we can solve the given
definite integral.
We have,
4 4 4 4 2 3 2 3
0 0 0 0
(2sec x x 2)dx 2 sec x dx x dx 2 dx
π π π π
∫ ∫ ∫ ∫
/ 4 4 / 4 (^) / 4
0 0 0
x 2 tanx 2[x] 4
π π ^ π (^) + (^) +
4 ( / 4) 2 tan tan0 0 2 0 4 4 4
π ^ π π − + (^) − (^) + −
23.2 | Definite Integration
4 4
5
(^) π π π π − + (^) − (^) + = + +
2. PROPERTIES OF DEFINITE INTEGRALS
Property 1
b b b
a a a
f(x)dx = f(t)dt = f(u)du ∫ ∫ ∫
Here x is a dummy variable; it can be replaced by any other variable t, u,……..
/2 /2 /
0 0 0
sin(x)dx sint dt sinudu
π π π
= = = ∫ ∫ ∫
This is similar to the summation property
10 10 10 2 2 2
T 1 T 1 U 1
r t u
= = =
∑ =^ ∑ =^ ∑ =……..
Property 2
b a
a b
f(x)dx = − f(x)dx ∫ ∫
i.e. the interchange of limits of a definite integral changes only its sign.
Property 3
b c b
a a c
f(x)dx = f(x)dx + f(x)dx ∫ ∫ ∫
(a < c < b)
Generally, this property is used when the integrand has two or more rules in the integration interval
b c c 1 2
a a c 1
f(x)dx = f(x)dx + f(x)dx + .......+ ∫ ∫ ∫
b
c n
f(x)dx ∫ where a < c 1 < c 2 < ……. c n < b.
Illustration 3: Evaluate:
4
1
f(x)dx, where ∫
2x 8, 1 x 2 f(x) 6x, 2 x 4
(JEE MAIN)
Sol: Here as we know,
b c b
a a c
f(x)dx = f(x)dx + f(x)dx ∫ ∫ ∫
where (a < c < b). Hence by using this property and solving by
using the integral formula we can solve it.
We have,
4
1
I = f(x)dx ∫
2 4 2 4
1 2 1 2
= f(x)dx + f(x)dx = (2x + 8)dx + 6x dx ∫ ∫ ∫ ∫
2 4 2 2 2 2
1 2
(^) x + 8x (^) + 3x (^) = (2) (^) + 8(2) − (1) −8(1)
2 2 3(4) (^) −3(2)
Illustration 4: Evaluate :
2
0
| 1 −x |dx ∫
(JEE MAIN)
23.4 | Definite Integration
( )
(^1 1 2 1 1 1 1 )
0 0 2 0 0
1 1 1 1 1 1
0 0 0
1 1 x x x (1 x) cot (1 x x )dx tan tan tan dx 1 x x^1 x 1^ x^1 x (1^ x)
tan x dx tan (1 x)dx 2 tan x .dx
− − − −
− − −
−^ + ^ −^ − −^ −
∫ ∫ ∫ ∫
∫ ∫ ∫
1 a^ b 1 1 tan tan a tan b 1 ab
− − −
=^ +
Illustration 6: Find the value of
1
0
log 1 dx x
∫
(JEE MAIN)
Sol: Here (^) ( ) ( )
1 x log log 1 x log x x
=^ −^ −
and
a a
0 0
f(x)dx = f(a −x)dx ∫ ∫ by using these two formulae we can solve it.
1
0
1 x log dx x
∫
1 1
0 0
log (1 − x)dx − log (x)dx ∫ ∫
= (^) ( )
1 1
0 0
log 1 − 1 − x dx − logx dx ∫ (^) ∫
1 1
0 0
= logx dx − logx dx ∫ ∫
1 1
0 0
log (x)dx − log (x)dx ∫ ∫
Illustration 7: Evaluate:
/
0
asinx bcos x dx sinx cos x
π
∫
(JEE MAIN)
Sol: As
a a
0 0
f(x)dx = f(a −x)dx ∫ ∫ therefore we can write
/
0
asinx bcos x dx sinx cos x
π
∫ in the form of
/
0
asin( / 2 x) bcos( / 2 x) dx sin( / 2 x) cos( / 2 x)
π π − + π −
π − + π −
∫
and then adding these two equations we can solve the given problem.
/
0
asinx bcos x I dx sinx cos x
π
=
∫
… (i)
/2 /
0 0
asin( / 2 x) bcos( / 2 x) acos x bsinx I dx dx sin( / 2 x) cos( / 2 x) sinx cos x
π π π − + π − + = = π − + π − +
∫ ∫
… (ii)
Adding (i) and (ii),
∴ 2I =
/2 /
0 0
(a b)(sinx cos x) dx (a b)dx (a b) / 2 sinx cos x
π π
= + = + π
∫ ∫ ⇒ I = (a + b) π/
Illustration 8: Show that
/2 (^2)
0
sin x 1 dx log( 2 1) sinx cos x 2
π
= +
∫
(JEE ADVANCED)
Sol: This problem is similar to the problem above.
Let
2 /
0
sin x I dx sinx cos x
π
∫
… (i)
By property 4, we have
( )
( ) ( )
(^2 ) /2 /
0 0
sin ( / 2) x (^) cos x I dx dx sin ( / 2) x cos ( / 2) x sinx cos x
π π^ − π = = π − + π − +
∫ ∫ … (ii)
Adding (i) and (ii), we get
Mathematics | 23.
2I =
2 2 /
0
sin x cos x dx sinx cos x
π (^) +
∫
/
0
1 dx I 2 sinx cos x
π
∫
/
0
dx
2 2 (1 / 2)sinx (1 / 2)cos x
π
∫
( )
/2 /
0 0
dx dx
2 2 cos(^ / 4)sinx^ sin(^ / 4)cos x^2 2 sin x^ (^ / 4)
π π
π + π (^) + π
∫ ∫
/ /
0 0
1 1 x cosec x dx logtan
2 2 4 2 2 2 8
π π (^) π (^) π
- = (^) + ^
∫
1 1 tan(3 / 8) 1 cot( / 8) log tan log tan log log
2 2 4 8 8 2 2 tan(^ / 8)^2 2 tan(^ / 8)
(^) π π (^) π (^) π (^) π +^ −^ =^ = (^) π^ π
logcot log( 2 1)
2 2 8 2
π = +
Illustration 9: Evaluate :
3 / 4
/ 4
tanx dx
1 tanx
π
−π
∫
(JEE ADVANCED)
Sol: By putting
sinx tanx cos x
= and using the property
b b
a a
f(x)dx = f(a + b −x)dx ∫ ∫
, we can solve the given problem.
Let
3 / 4
/ 4
tanx I dx
1 tanx
π
−π
∫
3 / 4
/ 4
sinx I dx
cos x sinx
π
−π
∫
… (i)
On applying
b b
a a
f(x)dx = f(a + b −x)dx ∫ ∫
we get
( )
( ) ( )
3 / 4
/ 4
sin (3 / 4) ( / 4) x
I dx
cos (3 / 4) ( / 4) x sin (3 / 4) ( / 4) x
π
−π
π − π −
=
π − π − + π − π −
∫
( )
( ) ( )
3 / 4
/ 4
sin ( / 2) x dx
cos ( / 2) x sin ( / 2) x
π
−π
π −
π − + π −
∫
3 / 4
/ 4
cos x dx
sinx cos x
π
−π
∫ … (ii)
Adding (i) and (ii), we get
3 / 4 3 / 4
/ 4 / 4
sinx cos x 2I dx dx
sinx cos x sinx cos x
π π
−π −π
∫ ∫
3 / 4
/ 4
sinx cos x dx
sinx cos x
π
−π
∫
3 / 4 (^3) / 4
/ 4 / 4
dx [x] 4 4 4 4
π (^) π
−π −π
(^) π (^) π (^) π π = = (^) − (^) − (^) = + ^ ^ ^
∫
= π ⇒ I 2
π
Illustration 10: The value of
/
0
4 3sinx log dx is 4 3cos x
π (^) +
+
∫ (JEE ADVANCED)
Sol: Similar to the problems above, we can write
/
0
4 3sinx log dx 4 3cos x
π (^) + +
∫
as
( )
( )
/
0
4 3sin ( / 2) x log dx 4 3cos ( / 2) x
π ^ +^ π^ − (^) + π −
∫
and then by adding these two equations we can solve the given problem.
Mathematics | 23.
Property 6
a a
0 a
2 f(x) dx if f( x) f(x) (even function) f(x)dx
0 if f( x) f(x) (odd function)
−
−^ = −
∫ ∫
Note: This property is to be used if the integrand is either an even or odd function of x
Illustration 13:
/ 2
/
cos x dx
π
∫−π
is equal to (JEE MAIN)
Sol: As
/ 2
/
cos x dx
π
∫−π
/ 2
0
2 cos x dx
π
∫
, therefore using property 7 we can solve it.
Here I = 2
/ 2
0
cos x dx { f( x) f(x)}
π
− = ∫
/2^ /
0 0
sin2x (1 cos2x)dx x 2 2
π π π
∫
Illustration 14:
3 2 1
1 2
x sin(1 x ) dx
1 x
−
∫
is equal to (JEE ADVANCED)
Sol: Here by using the property
a a
0 a
2 f(x) dx if f( x) f(x) (even function) f(x)dx
0 if f( x) f(x) (odd function)
−
−^ = −
∫ ∫
Here f(x) =
3 2
2
x sin(1 x )
1 x
& f(–x) = –
3 2
2
x sin(1 x )
1 x
f(x) = – f(x)
∴ I = 0
Property 7:
a
2a
0 0
2 f(x)dx, if f(2a x) f(x) f(x)dx
0, if f(2a x) f(x)
−^ = −
∫ ∫
Note: The above property is used to halve the limits
Illustration 15: Evaluate :
2
0
sin d a bcos
π (^) θ θ − θ
∫
(JEE MAIN)
Sol: Let
a
2a
0 0
2 f(x)dx, if f(2a x) f(x) f(x)dx
0, if f(2a x) f(x)
−^ = −
∫ ∫
. Hence by using this property we can solve the given problem.
Let
2
0
sin I d a bcos
π (^) θ = θ − θ
∫ → Let
sin f( ) a bcos
θ θ = − θ
sin2(2 ) sin f(2 ) f( ) a bcos(2 ) a bcos
π − θ − θ π − θ = = = − θ − π − θ − θ
By property 7, we have
2
0
sin d 0 a bcos
π (^) θ θ = − θ
∫
23.8 | Definite Integration
Illustration 16: Evaluate
2 4 6
0
x sin x cos x dx
π
∫
(JEE ADVANCED)
Sol: Similar to the problem above.
(^2 4 6 24 )
0 0
I x sin x cos x dx (2 x) sin x cos x dx
π π = = π − ∫ ∫
2 4 6
0
2I 2 sin x cos x dx
π = π ∫
4 6
0
I 2 sin x cos x dx
π = π ∫
/ 4 6
0
I 4 sin x cos x dx
π = π ∫
/ 4 6
0
I 4 cos x sin x
π = π ∫
/ 4
0
I (sin2x) dx 16
π^ π
∫ ⇒ 2x = t ⇒
dt dx 2
/ 4 4
0 0
I sin t dt sin t dt 16 8
π π^ π π = = ∫ ∫
⇒ (^) ( )
2 /2 (^4 )
0
I sin t sin t dt.. 8 2 8 2 8 128
π π π π π = + = =
∫
Property 8: If f(x) = f(x + a) (i.e. f(x) is a function with period a), then
na a
0 0
f(x)dx =n f(x)dx ∫ ∫
Illustration 17: Evaluate:
(^4 )
0
sin x dx
π
∫
(JEE MAIN)
Sol: Here sin 8 (π – x) = sin 8 x, therefore by using this property, we can solve the given problem.
8
0
I 4 sin x dx
π
∫
/ 8
0
8 sin x dx 8. 8.6.4.2 2 32
π π π = = = ∫
Illustration 18: Evaluate:
2 5
0
cos x dx
π
∫
(JEE ADVANCED)
Sol: Let
2 5
0
I cos x dx
π
∫
Let f(x) = cos
5 x
f(2π – x) = cos 5 (2π – x) = cos 5 x = f(x)
Then
(^2 5 )
0 0
cos x x dx 2 cos x dx
π π
∫ ∫
Now, f(π – x) = cos
5 (π – x) = (– cos x)
3 = – cos
5 x
= – f(x) ;
5
0
cos x dx 0
π
∫
Hence
2 5
0
cos x dx 0
π
∫
Property 9
a nT T
a 0
f(x)dx n f(x)dx
= ∫ ∫ (if f(x + T) = f(x), and nÎN i.e. f(x) is a function with period T)
b nT T b
a mT 0 a
f(x)dx (n m) f(x)dx f(x)dx
∫ ∫ ∫
m,n ∈ I
Illustration 19:
200
0
I 1 cos x dx
π
= + ∫
(JEE MAIN)
23.10 | Definite Integration
Change of variables: If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously
differentiable on the interval [t 1 , t 2 ] and a = φ(t 1 ), b = φ(t 2 ), then
b t 2
a t 1
f(x)dx = f( φ( ))φ ( )t ' t dt ∫ ∫
Nitish Jhawar (JEE 2009 AIR 7)
3. SOME SPECIAL INTEGRALS
3.1 Walli’s Formula
/2 / n n
0 0
(n 1) (n 3).... sin x dx cos x dx n(n 2)....
π π − − = = −
∫ ∫ (if n is odd positive integer)
(n 1) (n 3).....
n(n 2).....2 2
− − π − (^)
(if n is even positive integer)
Illustration 20: Evaluate
/ 7
0
cos x dx
π
∫
(JEE MAIN)
Sol: By using Walli’s formula we can solve the given problem.
I
3.2 Gamma Function
( ) ( )
( )
/ m n
0
(m 1) / 2 (n 1) / 2 sin x cos x dx 2 (m n 2) / 2
π Γ + Γ + = Γ + +
∫
where Γ(n) is called the gamma function
OR
/ m n
0
((m 1) (m 3).....(2 or 1)) (n 1) ((n 3)...(2 or 1)) sin cos x dx (m n) (m n 2).......(2 or 1)
π − − − − =
∫
(if m and n both are not simultaneously even positive integers)
((m 1) (m 3)....(1))((n 1)(n 3)....(1))
(m n)(m n 2)....(2) 2
− − − − (^) π
(if m and n are both even positive integers)
Illustration 21: Evaluate I =
/ 4 5
0
sin x cos x dx.
π
∫
(JEE MAIN)
Sol: Using the gamma function formula i.e.
MASTERJEE CONCEPTS
Mathematics | 23.
( ) ( )
( )
/ m n
0
(m 1) / 2 (n 1) / 2 sin x cos x dx 2 (m n 2) / 2
π Γ + Γ + = Γ + +
∫
We can solve it.
( ) ( )
( )
( )
( )
( )( )
( )
r (4 1) / 2 (5 1) / 2 5 / 2 (3) (3 / 2).(1 / 2) 2.1 (^8) I 2 (4 5 2) / 2 2 11 / 2 2 (9 / 2).(7 / 2).(5 / 2).(3 / 2).(1 / 2)^315
4. NEWTON LEIBNITZ FORMULA
In calculus, Leibnitz’s rule for differentiation under the integral sign named after Gottfried Leibnitz tells us that if
we have an integral
y 1
y 0
f(x, y)dy ∫ then for x in (x 0 , x 1 ) the derivative of this integral is thus expressible as
y y 1 1
y y x 0 0
d f(x, y)dy f (x, y)dy dx
∫ ∫
provided that f and its partial derivative f x
are both continuous over a region in the form [x 0
. x 1
] × [y 0
, y 1
].
5. SUMMATION OF SERIES BY INTEGRATION (LIMIT AS A SUM)
To find the sum of an infinite series with the help of definite integration, the following formula is used
n 1^1
n r (^0 )
r 1 lim f f(x)dx n n
−
→ ∞
∑ (^) ∫
The following method is used to solve the questions on summation of series.
(i) After writing (r – 1)th or rth term of the series, express it in the form
1 r f. n n
Therefore the given series will take the form as
n 1
n r 0
1 r lim f n n
−
→ ∞
∑
(ii) Now write ∫
in place of n
lim →∞
∑ and x in place of
r
n
and dx in place of n. We get summation in the form of
integral
1
0
f(x)dx ∫
Also we can write (^) ( )
b
n a
b a f(x)dx lim [f(a) f(a h) .... f(a n 1 h)] → ∞ n
∫
b a h re h n
w e
Illustration 22: Evaluate n
lim ..... → ∞ n 1 n 2 2n
+^ +^ +
+^ +
(JEE MAIN)
Sol: By using the summation of series by integration formula i.e
n 1^1
n r (^0 )
r 1 lim f f(x)dx n n
−
→ ∞
∑ (^) ∫ we can solve it.
Limit =
n 1 1 0 n n r (^1 )
lim lim. dx [log(1 x)] log → ∞ (^) n r → ∞ 1 (r / n) n 1 x =
∑ ∑ (^) ∫
Illustration 23:
100 100 100 100
n^101
1 2 3 ........n lim
n → ∞
(JEE MAIN)
Sol: By observing the given problem, we can say that it’s a sum of an infinite series so by using the summation of
series by integration formula we can solve it.
Mathematics | 23.
2 2
1
(x +x)dx ∫
n
lim [f(1) f(1 h) .... f(1 (n 1)h)] → ∞n
2 2 2
n
lim [(1 1) {(1 h) (1 h)} .... {(1 (n 1)h) (1 (n 1)h)} → ∞n
2 2 2 2 2
n
lim [1 .n h(1 2 ... (n 1)) 1. n 2h(1 2 ... (n 1)) h (1 2 ...(n 1) )] → ∞ (^) n
Here h =
n
n^2
1 1 (n^ 1)(n) 2 n(n 1) 1 (n 1)n(2n 1) lim n n. → ∞n n 2 n (^2) n 6
+^ +^ +^ +
( ) ( ) ( ) ( )
n
1 (1 / n) (1) 2 1 (1 / n) 1 (1 / n) (1) 2 (1 / n) lim 1 1 → ∞ 2 2 6
+ + + + =^
6. INTEGRAL WITH INFINITE LIMITS
If a function f(x) is continuous for a ≤ x < ∞, then by definition,
b
a (^) b a
f(x)dx lim f(x)dx
∞
→ ∞
∫ ∫
… (i)
If there exists a finite limit on the right-hand side of (i), then the improper integral is said to be convergent;
otherwise it is divergent.
Geometrically, the improper integral (i) for f(x) > 0, is the area of the figure bounded by the graph of the function
y = f(x), the straight line x = a, and the x-axis. Similarly, we can define
b b
a^ a
f(x)dx lim f(x)dx −∞ (^) →− ∞
∫ ∫
and
a
a
f(x)dx f(x)dx f(x)dx
∞ ∞
−∞ −∞
∫ ∫ ∫
7. IMPORTANT RESULTS
If f(x) ≥ 0 and a < b, then
b /
a 0
f(x)dx 0, e.g. sinx dx 1
π
≥ = ∫ ∫
If f(x) ≥ 0 and a < b, then
a 0
b /
f(x)dx 0, e.g. cos x dx 1
π
∫ ∫
If f(x) ≤ 0 and a < b, then
a 0
b /
f(x)dx 0, e.g. sinx dx 1
π
∫ ∫
x 1 2 3 x
0 0 1 2 [x]
[x]dx = (0)dx + (1)dx + 2dx + ... + [x]dx, ∫ ∫ ∫ ∫ ∫
where [ ] denotes the greatest integer of x.
/2 /
0 0
log(sinx)dx log(cos x)dx log 2
π π π = = − ∫ ∫
/2 /
0 0
log(tanx)dx (cot x)dx 0
π π
= = ∫ ∫
2a a a a a
0 0 0 0 0
f(x)dx = f(x)dx + f(2a − x)dx = f(x)dx + f(a +x)dx ∫ ∫ ∫ ∫ ∫
23.14 | Definite Integration
b 1
a 0
[x]dx = (b −a) x dx, where [ ] denotes thefractional part of x. ∫ ∫
e.g.,
5 1
0 0
[x]dx 5 x dx 2
∫ ∫
Integral of an inverse function is given by
f(b) b 1
f(a) a
f (y)dy bf(b) af(a) f(x)dx
− = − − ∫ ∫
Derivation of the given formula is given in the solved examples
8. GEOMETRICAL APPLICATION
The area of the figure bounded by the graphs of two continuous functions y = f 1 (x) and
y = f (x) 2
y = f (x) 1
Figure 23.
y = f 2 (x), f 1 (x) ≤ f 2 (x), and two straight lines x= a and x = b is determined by the formula
b
a^2
S = (f (x) −f (x))dx ∫
. It is sometimes convenient to use formulae analogous to x.with
respect to y, i.e., regarding x as a function of y. In particular, the area bounded by the curve
x =f(y), the y-axis and the two abscissae y = c and y = d is given by
d
c
f ∫ (y)dy. The area of
the figure bounded by the graphs of two continuous functions x = f 1 (y) and f 2 (y) (with f 1 (y)
≤ f 2 (y)), and the two straight lines y = c, y = d is given by
d
c 2 1
(f (y) −f (y))dy ∫
From the view of geometry we get an important inequality as if m ≤ f(x) ≤ M for a ≤ x ≤ b, then m(b – a) ≤
b
a
f(x)dx ∫ ≤ M(b – a)
FORMULAE SHEET
Important results
- (^) { ( ) ( )}
b b b b
a a a a
f(x) ± g x ± h x dx = f(x)dx ± g(x)dx + h(x)dx ∫ ∫ ∫ ∫ 2.
b a
a b
f(x)dx = − f(x)dx ∫ ∫
b c b
a a c
f(x)dx = f(x)dx + f(x)dx ∫ ∫ ∫ (a < c < b)^ 4.
a a
0 0
f(x)dx = f(a −x)dx ∫ ∫
a a
0 a
2 f(x) dx if f( x) f(x) (even function) f(x)dx
0 if ( x) f(x) (odd function)
−
∫ ∫
b b
a a
f(x)dx = f(a + b −x)dx ∫ ∫
a
2a
0 0
2 f(x)dx, if f(2a x) f(x) f(x)dx
0, if f(2a x) f(x)
−^ = −
∫ ∫
h(x)
g(x)
d f(t) dx
∫
dt = h’(x) f(h(x)) – g’(x) f(g(x))
a nT T
a 0
f(x)dx n f(x)dx
= ∫ ∫ (if f(x + T) = f(x), and n∈N i.e. f(x) is a
function with period T)
- If f(x) = f(x + a) then
na a
0 0
f(x)dx =n f(x)dx ∫ ∫
23.16 | Definite Integration
2 m ax
0 (m 1)/
m 1
x e dx
2a
∞ −
∫
2
0 x
x dx
e 1 6
∞ (^) π
−
∫
n 1
0 x^ n^ n^ n
x 1 1 1 dx (n) ..... e 1 1 2 3
− ∞ (^) = Γ (^) + + + − (^)
∫
2
0 x
x dx
e 1 12
∞ π
∫
n 1
0 x n n n
x 1 1 1 dx (n) .......
e 1 1 2 3
− ∞ (^) = Γ − + −
∫ 10.
ax bx 2 2
0 2 2
e e 1 b p dx ln x sec (px) (^2) a p
− − ∞ (^) − ^ + = (^) +
∫
ax bx
0
e e b a dx arctan arctan x csc(px) p p
− − ∞ (^) − = − ∫
ax 2
0 2
e (1 cos x) a dx arccot a ln(a 1)
x^2
− ∞ (^) − = − + ∫
Solved Examples
JEE Main/Boards
Example 1: Evaluate:
(i)
( )
a
2 2 0
dx
(a / 4) − x −(a / 2)
∫ (ii)
a
a
a x dx a x −
∫
Sol: (i) As we know
1
2 2
dx x sin a a x
−
−
∫
, therefore by
using this formula we can solve the given problem.
(ii) Put x = a cos θ : θ ∈ [0, p] and solve it using the
appropriate formula.
(i)
( )
a
2 0 2
dx
(a / 4) − x −(a / 2)
∫
a 1
0
x (a / 2) sin (a / 2)
a 1
0
2x a sin a
= [sin
- 1–sin - (–1)] = 2 sin - (1) = 2 × 2
π = π. (ii)
Then dx = –a sin θ dθ. Hence,
a
a
a x dx a x −
∫
0 1 cos ( asin )d 1 cos π
− θ − θ θ
∫
2
2 0
2sin ( / 2) a. 2sin cos d 2cos ( / 2)^2
π θ θ θ θ θ
∫
2
0 0
a 2sin d a (1 cos )d 2
π π θ θ = − θ θ ∫ ∫
0 a( sin ) a( ) a.
π θ − θ = π = π
Example 2: Evaluate
/
0
sinx dx sinx cos x
π
∫
Sol: Let
a a
0 0
f(x)dx = f(a −x)dx ∫ ∫
By using this we can write
/
0
sinx dx sinx cos x
π
∫
as
/
0
sin ( / 2) x dx sin ( / 2) x cos ( / 2) x
π (^) π −
(^) π − (^) + (^) π −
∫ and by adding
we can get the result.
I =
/
0
sin ( / 2) x dx sin ( / 2) x cos ( / 2) x
π π −
π − + π −
∫
/
0
cos x dx cos x sinx
π
=
∫
/2 /
0 0
sinx cos x 2I dx dx sinx cos x 2
π π
∫ ∫
∴ I^
π
