Some important formulae

∫ 1dx = x + c ∫ cosec 2 x dx = − cot x + c

∫ sec x tan x dx = sec x + c

n xn+1
∫ x dx = n+1
+c ∫ cosec x cot x dx = − cosec x + c

(𝑎x+b)n+1 𝑓 ′ (𝑥)
∫(𝑎𝑥 + 𝑏) dx = n
+c ∫ 𝑑𝑥 = 𝑙𝑛|𝑓(𝑥)| + 𝑐
(n+1)a 𝑓(𝑥)

𝑓 ′ (𝑥)
mx
emx ∫ 𝑑𝑥 = 2√𝑓(𝑥) + 𝑐
∫e dx = +c √𝑓(𝑥)
m
[𝑓(𝑥)]𝑛+1
1 ∫ [𝑓(𝑥)]𝑛 . 𝑓 ′ (𝑥)𝑑𝑥 = +𝑐
𝑛+1
∫ dx = lnx + c
𝑥 𝑥√𝑎2 −𝑥2 𝑎2 𝑥
cos a x ∫ √𝑎2 − 𝑥 2 𝑑𝑥 = + 𝑠𝑖𝑛−1 (𝑎) + 𝑐
2 2
∫ sin a xdx = − +c
a 1
cos b x ∫ 1+𝑥 2 𝑑𝑥 = 𝑡𝑎𝑛−1 𝑥 + 𝑐
∫ cos b xdx = +c 𝑑𝑢
b ∫ 𝑢𝑣𝑑𝑥 = 𝑢 ∫ 𝑣𝑑𝑥 − ∫ {𝑑𝑥 ∫ 𝑣𝑑𝑥} 𝑑𝑥
∫ sec 2 x dx = tan x + c

Special formulae:
𝑓′ (𝑥)
∫ 𝑑𝑥 = 𝑙𝑛|𝑓(𝑥)| + 𝑐
𝑓(𝑥)

𝑓′ (𝑥)
∫ 𝑑𝑥 = 2√𝑓(𝑥) + 𝑐
√𝑓(𝑥)

[𝑓(𝑥)]𝑛+1
∫ [𝑓(𝑥)]𝑛 . 𝑓 ′ (𝑥)𝑑𝑥 = +𝑐
𝑛+1

(1) Evaluate the following indefinite integral

𝑥
(i) ∫ 𝑑𝑥
1−𝑥 2
𝑥
(ii) ∫ √1−𝑥2
𝑑𝑥
7
(iii) ∫𝑠𝑖𝑛 𝑥. 𝑐𝑜𝑠𝑥 𝑑𝑥
(iv) ∫cos 7 𝑥 . 𝑠𝑖𝑛𝑥 𝑑𝑥
(v) ∫(𝑐𝑜𝑡 7 𝑥 + 𝑐𝑜𝑡 5 𝑥 ) 𝑑𝑥
(vi) ∫(𝑡𝑎𝑛7 𝑥 + 𝑡𝑎𝑛5 𝑥 ) 𝑑𝑥
 Solution:
𝑥 1 −2𝑥 1
(i) ∫ 𝑑𝑥 = (− 2) ∫ 𝑑𝑥 = − 2 𝑙𝑛 |1 − 𝑥 2 | + 𝑐
1−𝑥 2 1−𝑥 2
𝑥 1 −2𝑥 1
(ii) ∫ √1−𝑥2
𝑑𝑥 = (− 2) ∫ √1−𝑥 2
𝑑𝑥 = (− 2) 2√1 − 𝑥 2 + 𝑐 = −√1 − 𝑥 2 + 𝑐
(𝑠𝑖𝑛 𝑥)7+1 1
(iii) ∫𝑠𝑖𝑛7 𝑥. cos 𝑑𝑥 = + 𝑐 = 8 𝑠𝑖𝑛8 𝑥 + 𝑐
7+1
1
(iv) ∫cos 𝑥 . 𝑠𝑖𝑛𝑥 𝑑𝑥=(−)∫cos7 𝑥 . (−𝑠𝑖𝑛𝑥)𝑑𝑥=− 8 cos 8 𝑥 + 𝑐
7

(v) ∫(𝑐𝑜𝑡 7 𝑥 + 𝑐𝑜𝑡 5 𝑥 ) 𝑑𝑥

=∫𝑐𝑜𝑡 5 𝑥 (𝑐𝑜𝑡 2 𝑥 + 1) 𝑑𝑥
=∫𝑐𝑜𝑡 5 𝑥 𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑑𝑥
= (−)∫𝑐𝑜𝑡 5 𝑥 (−𝑐𝑜𝑠𝑒𝑐 2 𝑥) 𝑑𝑥
(𝑐𝑜𝑡𝑥)5+1
=− +𝑐
5+1
(𝑐𝑜𝑡𝑥)6
=− +𝑐
6

(vi) ∫( 𝑡𝑎𝑛 𝑥 + 𝑡𝑎𝑛5 𝑥)𝑑𝑥

7

=∫𝑡𝑎𝑛5 𝑥(𝑡𝑎𝑛2 𝑥 + 1)𝑑𝑥

=∫𝑡𝑎𝑛5 𝑥. 𝑠𝑒𝑐 2 𝑥𝑑𝑥

[𝑡𝑎𝑛 𝑥]5+1
= +𝑐
5+1

tan6 𝑥
= +𝑐
6

Method of Substitution:

Integration by substitution, also known as "u-substitution", is a technique that simplifies

integrals by making a substitution. The exact substitution depends on the integral's form, but
here are some steps you can follow:

1. Identify the integral in the form ∫ 𝑓(𝑔(𝑥))𝑔′(𝑥)𝑑𝑥

2. Substitute the independent variable as 𝑔(𝑥) = 𝑡
3. Differentiate the assumed function with respect to 𝑡
4. Substitute for the dependent variable, such as 𝑔′(𝑥)𝑑𝑥 = 𝑑𝑡
5. The resultant integral after substitution becomes ∫ 𝑓(𝑡)𝑑𝑡
6. Solve the integral using basic integration rules
7. Convert the result back to terms of 𝑥 by substituting 𝑡 with the original independent
variable

The substitution should be made for a function whose derivative also occurs in the
integrand. The ability to recognize an appropriate substitution comes from practicing many
different examples.

*** If integrand 𝑓(𝑥) = 𝑔(𝑥). 𝑔′ (𝑥) or [𝑔(𝑥)]𝑛 𝑔′ (𝑥) or 𝑒 𝑔(𝑥) . 𝑔′ (𝑥), 𝑇[𝑔(𝑥)]. 𝑔′ (𝑥) or
[𝑇( 𝑔(𝑥))]𝑛 . 𝑔′(𝑥) then substitute 𝑔(𝑥) = 𝑧 and 𝑔′(𝑥)𝑑𝑥 = 𝑑𝑧.

Here T means trigonometry function.

(2) Evaluate the following indefinite integral

𝑒 𝑥 (𝑥+1)
(i) ∫ 𝑐𝑜𝑠2 (𝑥 𝑒 𝑥 ) 𝑑𝑥
𝑒 𝑥 (𝑥+1)
(ii) ∫ 𝑑𝑥
𝑠𝑖𝑛2 (𝑥 𝑒 𝑥 )
9 10
(iii) ∫𝑥 𝑒 𝑑𝑥

Solution:
𝑒 𝑥 (𝑥+1) 𝑥
(i) ∫ 𝑑𝑥 Let, 𝑥. 𝑒 = 𝑧
𝑐𝑜𝑠2 (𝑥 𝑒 𝑥 )

𝑑𝑧
Then (𝑥. 𝑒 𝑥 + 𝑒 𝑥 . 1)𝑑𝑥 = 𝑑𝑧
=∫ 𝑐𝑜𝑠2 𝑧
⇒ 𝑒 𝑥 (𝑥 + 1)𝑑𝑥 = 𝑑𝑧
= ∫ 𝑠𝑒𝑐 2 𝑧 𝑑𝑧 𝑑 𝑑𝑣 𝑑𝑢
Formula: (𝑢𝑣) = 𝑢 +𝑣
𝑑𝑥 𝑑𝑥 𝑑𝑥
= 𝑡𝑎𝑛 𝑧 + 𝑐

= 𝑡𝑎𝑛 (𝑥 𝑒 𝑥 ) + 𝑐

𝑒 𝑥 (𝑥+1) 𝑥
(ii) ∫ 𝑑𝑥 Let, 𝑥. 𝑒 = 𝑧
𝑠𝑖𝑛2 (𝑥 𝑒 𝑥 )
𝑑𝑧
=∫ 𝑠𝑖𝑛2 𝑧
Then (𝑥. 𝑒 𝑥 + 𝑒 𝑥 . 1)𝑑𝑥 = 𝑑𝑧

= ∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑧 𝑑𝑧 ⇒ 𝑒 𝑥 (𝑥 + 1)𝑑𝑥 = 𝑑𝑧

= −𝑐𝑜𝑡 𝑧 + 𝑐

= −𝑐𝑜𝑡 (𝑥 𝑒 𝑥 ) + 𝑐

(iii) ∫ 𝑥9 𝑒 𝑥
10
𝑑𝑥 Let, 𝑥10 = 𝑧

= ∫ 𝑒𝑧
𝑑𝑧 ⇒ 10𝑥9 𝑑𝑥 = 𝑑𝑧
10
𝑑𝑧
1
= 10 ∫ 𝑒 𝑧 𝑑𝑧 ⇒ 𝑥9 𝑑𝑥 =
10
 1
= 10 . 𝑒 𝑧 + 𝑐
1 10
= 10 𝑒 𝑥 +𝑐

UV method:
𝑑𝑢
∫ 𝑢𝑣 𝑑𝑥 = 𝑢 ∫ 𝑣 𝑑𝑥 − ∫{𝑑𝑥 ∫ 𝑣 𝑑𝑥}𝑑𝑥

u L I A T E

𝑙𝑜𝑔𝑥,𝑙𝑛𝑥

𝑠𝑖𝑛−1 𝑥, 𝑐𝑜𝑠 −1 𝑥

𝑥 , 𝑥2, 𝑥3

𝐶𝑜𝑠𝑥, 𝑠𝑖𝑛𝑥

𝑒 𝑥 , 𝑎 𝑥 , 2𝑥

(3) Evaluate the following indefinite integral

(i) ∫ 𝑥𝑒 𝑥 𝑑𝑥
(ii) ∫ 𝑥𝑙𝑛𝑥 𝑑𝑥
(iii) ∫ 𝑙𝑛𝑥 𝑑𝑥

Solution:
(i) ∫ 𝑥𝑒 𝑥 𝑑𝑥
𝑑
= 𝑥 ∫ 𝑒𝑥 𝑑𝑥 − ∫ {𝑑𝑥 (𝑥) ∫ 𝑒𝑥 𝑑𝑥} 𝑑𝑥
= 𝑥𝑒𝑥 − ∫ 1. 𝑒𝑥 𝑑𝑥
= 𝑥 𝑒𝑥 − 𝑒𝑥 + 𝑐
(ii) ∫ 𝑥𝑙𝑛𝑥 𝑑𝑥
𝑑
= 𝑙𝑛𝑥 ∫ 𝑥𝑑𝑥 − ∫ { (𝑙𝑛𝑥) ∫ 𝑥𝑑𝑥} 𝑑𝑥
𝑑𝑥
𝑥2 1 𝑥2
= 𝑙𝑛𝑥 −∫ . 𝑑𝑥
2 𝑥 2
 𝑥2 1
= 𝑙𝑛𝑥 − ∫ 𝑥𝑑𝑥
2 2
𝑥 2 𝑥2
= 2
𝑙𝑛𝑥 − 4
+𝑐
(iii) ∫ 𝑙𝑛𝑥 𝑑𝑥
𝑑
= 𝑙𝑛𝑥 ∫ 1𝑑𝑥 − ∫ { (𝑙𝑛𝑥) ∫ 1𝑑𝑥} 𝑑𝑥
𝑑𝑥
1
= 𝑥𝑙𝑛𝑥 − ∫ . 𝑥𝑑𝑥
𝑥
= 𝑥𝑙𝑛𝑥 − ∫ 1𝑑𝑥
= 𝑥𝑙𝑛𝑥 − 𝑥 + 𝑐

The Fundamental Theorem of Calculus: If 𝑓is continuous on [𝑎, 𝑏] and 𝐹 is any anti-
𝑏
derivative of 𝑓on [𝑎, 𝑏], then ∫𝑎 𝑓(𝑥)𝑑𝑥 = [𝐹(𝑥)]𝑏𝑎 = 𝐹(𝑏) − 𝐹(𝑎)

(4) Evaluate the following integral

𝜋
(i) ∫04 (𝑡𝑎𝑛7 𝑥 + 𝑡𝑎𝑛5 𝑥) 𝑑𝑥
1
(ii) ∫0 𝑥 3 √1 + 3𝑥 4 𝑑𝑥
𝜋
(iii) ∫02 𝑐𝑜𝑠 7 𝑥 𝑠𝑖𝑛 𝑥 𝑑𝑥
𝜋
(iv) ∫0 𝑠𝑖𝑛7 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥
2

1 x
(v) ∫0 𝑑𝑥
4−x2
1 x
(vi) ∫0 √1−𝑥 2 𝑑𝑥
1
(vii) ∫0 𝑥𝑒 𝑥 𝑑𝑥

Solution:
𝜋
(i) ∫04 (𝑡𝑎𝑛7 𝑥 + 𝑡𝑎𝑛5 𝑥) 𝑑𝑥
𝜋
= ∫04 (𝑡𝑎𝑛5 𝑥(𝑡𝑎𝑛2 𝑥 + 1) 𝑑𝑥
𝜋
= ∫04 𝑡𝑎𝑛5 𝑥. 𝑠𝑒𝑐 2 𝑥 𝑑𝑥
𝜋
[𝑡𝑎𝑛 𝑥]5+1 4
=[ ]
5+1 0

𝜋
1
= 6 [𝑡𝑎𝑛6 𝑥]04

1 𝜋
= [𝑡𝑎𝑛6 ( ) − 𝑡𝑎𝑛6 (0)]
6 4
 1
= 6 [1 − 0]

1
=6

1
(ii) ∫0 𝑥 3 √1 + 3𝑥 4 𝑑𝑥

= ∫1 √𝑧
4 𝑑𝑧 Let, 1 + 3𝑥 4 = 𝑧
12
⇒ (0 + 3.4𝑥3 )𝑑𝑥 = 𝑑𝑧
14 1
= ∫1
𝑧 2 𝑑𝑧
12
𝑑𝑧
4 ⇒ 𝑥3 𝑑𝑥 =
1
+1 12
1 𝑧2
= 12 [ 1 ]
+1 Limit:
2 1
4 x 0 1
1 𝑧 3/2
= 12 [ 3 ]
2 1 z 1 4
1 2
= 12 × 3 [43/2 − 13/2 ]
1
= [8 − 1]
18

7
= 18

𝜋
(iii) ∫02 𝑐𝑜𝑠 7 𝑥 𝑠𝑖𝑛 𝑥 𝑑𝑥
𝜋
= (−) ∫02 𝑐𝑜𝑠 7 𝑥 . (−𝑠𝑖𝑛𝑥)𝑑𝑥
𝜋
(𝑐𝑜𝑠 𝑥)7+1 2
=− [ ]
7+1 0
𝜋
1 8 2
= − [8 𝑐𝑜𝑠 𝑥]
0
1 𝜋
= − 8 {cos8 2 − cos8 0}
1
=8

𝜋
(iv) ∫0 𝑠𝑖𝑛7 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥
2

𝜋
(𝑠𝑖𝑛 𝑥)7+1 2
=[ ]
7+1 0
 𝜋
1 8 2
= [8 𝑠𝑖𝑛 𝑥]
0
1 𝜋
= {sin8 2 − sin8 0}
8
1
=8
1 x
(v) ∫0 𝑑𝑥
4−x2
1 1 −2x
= (− 2) ∫0 𝑑𝑥
4−x2
1
=− 2 [𝑙𝑛 |4 − 𝑥 2 |]10
1
= − {ln(4 − 12 ) − ln(4 − 0)}
2
1
= − 2 {𝑙𝑛3 − 𝑙𝑛4}
1 3
= − 2 ln(4)

1 x
(vi) ∫0 √1−𝑥 2
𝑑𝑥
1 1 −2x
= (− 2) ∫0 𝑑𝑥
√1−x2
1 1
=− 2 [2√1 − 𝑥 2 ]0
= −{√1 − 1 − √1 − 0}
=0
1
(vii) ∫0 𝑥𝑒 𝑥 𝑑𝑥
Here ∫ 𝑥𝑒 𝑥 𝑑𝑥
𝑑
= 𝑥 ∫ 𝑒𝑥 𝑑𝑥 − ∫ { (𝑥) ∫ 𝑒𝑥 𝑑𝑥} 𝑑𝑥
𝑑𝑥
= 𝑥𝑒𝑥 − ∫ 1. 𝑒𝑥 𝑑𝑥
= 𝑥 𝑒𝑥 − 𝑒𝑥 + 𝑐
1
∴ ∫0 𝑥𝑒𝑥 𝑑𝑥 = [𝑥 𝑒𝑥 − 𝑒𝑥 ]10 = (1. 𝑒 1 − 𝑒 1 ) − (0 − 𝑒 0 ) = 1

𝛼 𝛼
Q. If ∫1 {3 + 𝑥𝑙𝑛(𝑥 2 + 5)}𝑑𝑥 + ∫1 {2 − 𝑥𝑙𝑛(𝑥 2 + 5)}𝑑𝑥 = 30 then find the value of 𝛼.

Solution: Given that

𝛼 𝛼
∫1 {3 + 𝑥𝑙𝑛(𝑥 2 + 5)}𝑑𝑥 + ∫1 {2 − 𝑥𝑙𝑛(𝑥 2 + 5)}𝑑𝑥 = 30
𝛼
⇒ ∫1 {3 + 𝑥𝑙𝑛(𝑥2 + 5) + 2 − 𝑥𝑙𝑛(𝑥2 + 5)}𝑑𝑥 = 30

𝛼
⇒ ∫1 5𝑑𝑥 = 30

⇒ 5[𝑥]𝛼1 = 30

⇒ 𝛼−1=6
⇒ 𝛼=7

Properties of definite integral:

𝑏 𝑏
(i) ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑧)𝑑𝑧
𝑏 𝑐 𝑏
(ii) ∫𝑎 𝑓(𝑥)𝑑𝑥 = ∫𝑎 𝑓(𝑥)𝑑𝑥 + ∫𝑐 𝑓(𝑥)𝑑𝑥 𝑤ℎ𝑒𝑛 𝑎 < 𝑒 < 𝑏.
𝑎 𝑎
(iii) ∫0 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥)𝑑𝑥 .
𝑛𝑎 𝑎
(iv) ∫0 𝑓(𝑥)𝑑𝑥 = 𝑛 ∫0 𝑓(𝑥)𝑑𝑥 , 𝑤ℎ𝑒𝑛 𝑓(𝑎 + 𝑥) = 𝑓(𝑥).
𝑎
𝑎 2 ∫0 𝑓(𝑥)𝑑𝑥 𝑖𝑓 𝑓(𝑥)𝑖𝑠 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
(v) ∫−𝑎 𝑓(𝑥)𝑑𝑥 ={
0 𝑖𝑓 𝑓(𝑥)𝑖𝑠 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝜋
(𝑡𝑎𝑛𝑥) 2024
Q. Determine the value of ∫02 (𝑡𝑎𝑛𝑥)2024 +(𝑐𝑜𝑡𝑥)2024 𝑑𝑥

Solution:
𝜋
(𝑡𝑎𝑛𝑥)2024
Let, 𝐼 = ∫0 2
(𝑐𝑜𝑡𝑥)2024 +(𝑡𝑎𝑛𝑥)2024
𝑑𝑥 … … . (𝑖)

𝜋
𝜋 {tan( −𝑥)} 2024
2
= ∫0 2
𝜋 𝜋 dx
{cot( 2 −𝑥)} 2024 +{tan( 2 −𝑥)} 2024

𝜋
(cot 𝑥)2024
= ∫0 2 𝑑𝑥 … … … (𝑖𝑖)
(𝑡𝑎𝑛𝑥)2024+(𝑐𝑜𝑡𝑥)2024

Now (𝑖) + (𝑖𝑖) we get

𝜋 𝜋
(𝑡𝑎𝑛𝑥)2024
2 2 (cot 𝑥)2024
𝐼+𝐼 =∫ 2024 + (𝑡𝑎𝑛𝑥)2024
𝑑𝑥 + ∫ 2024 + (𝑐𝑜𝑡𝑥)2024
𝑑𝑥
0 (cot 𝑥) 0 (𝑡𝑎𝑛𝑥)
𝜋
(𝑡𝑎𝑛𝑥)2024 (cot 𝑥)2024
⇒ 2𝐼 = ∫02 [(cot 2024 + ]𝑑𝑥
𝑥) +(𝑡𝑎𝑛𝑥)2024 (𝑡𝑎𝑛𝑥)2024 +(𝑐𝑜𝑡𝑥)2024

𝜋
(𝑡𝑎𝑛𝑥)2024 +(cot 𝑥)2024
= ∫0 [ 2 ]𝑑𝑥
(cot 𝑥)2024 +(𝑡𝑎𝑛𝑥)2024

𝜋
⇒ 2𝐼 = ∫0 1 𝑑𝑥 2

𝜋
2
⇒ 2𝐼 =[𝑥]0
𝜋
⇒ 2𝐼 =
2
 𝜋
⇒𝐼=
4
𝜋
(𝑡𝑎𝑛𝑥) 2024 𝜋
∴ ∫02 ( 𝑑𝑥 =
𝑡𝑎𝑛𝑥)2024 +(𝑐𝑜𝑡𝑥)2024 4

𝜋 𝑥 sin 𝑥
Q. Find the value of ∫0
1+𝑐𝑜𝑠2 𝑥
𝑑𝑥.

Solution:
𝜋 𝑥 sin 𝑥
Let, 𝐼 = ∫0 1+𝑐𝑜𝑠2 𝑥
𝑑𝑥 . . . . . . . (𝑖) let, cos 𝑥 = 𝑧

𝜋 (𝜋−𝑥) sin(𝜋−𝑥)
= ∫0 1+𝑐𝑜𝑠2 (𝜋−𝑥)
𝑑𝑥 ∴ sin 𝑥𝑑𝑥 = −𝑑𝑧

𝜋 (𝜋−𝑥) sin(𝑥)
= ∫0 𝑑𝑥 . . . . . . . . . . (𝑖𝑖) if 𝑥 = 0 then 𝑧 = 1 and if 𝑥 = 𝜋 then 𝑧 = −1
1+𝑐𝑜𝑠2 (𝑥)

𝜋 𝜋 sin(𝑥)
Now (𝑖) + (𝑖𝑖) ⟹ 2𝐼 = ∫0 𝑑𝑥
1+𝑐𝑜𝑠2 (𝑥)

−1 𝜋𝑑𝑧
= − ∫1 𝑧2 +1

1 𝑑𝑧
= 𝜋 ∫−1 𝑧2 +1

1
= 𝜋[tan−1 𝑧]
−1
= 𝜋[tan−1 ( 1) − tan−1(−1)]
𝜋 𝜋
= 𝜋( 4 + 4 )

𝜋2
⇒ 2𝐼 = 2

𝜋2
⇒𝐼= 2

𝜋 𝑥 sin 𝑥 𝜋2
∴ ∫0 𝑑𝑥 . =
1+𝑐𝑜𝑠2 𝑥 4

Exercise:
𝜋
(𝑠𝑖𝑛𝑥) 2024
(i) ∫0 2
(𝑐𝑜𝑠𝑥)2024+(𝑠𝑖𝑛𝑥)2024
𝑑𝑥
𝜋
(𝑡𝑎𝑛𝑥) 2024
(ii) ∫02 1+(𝑡𝑎𝑛𝑥)2024 𝑑𝑥
 Walle’s theorem: If n is a positive integer then
𝑛−1 𝑛−3 𝑛−5 5 3 𝜋
𝜋 𝜋 . .
𝑛 𝑛−2 𝑛−4
…………. . .
3 2 2
when n is even
𝑛 𝑛
I = ∫0 sin 𝑥 𝑑𝑥 = ∫0 cos 𝑥 𝑑𝑥
2 2 ={𝑛−1 𝑛−3 𝑛−5 6 4 2
. .
𝑛 𝑛−2 𝑛−4
………….7.5.3.1 when n is odd

Q. Evaluate by walle’s formula

𝜋
(i) ∫0 cos 7 𝑥 𝑑𝑥
2

𝜋
(ii) ∫02 cos10 𝑥 𝑑𝑥
𝜋
(iii) ∫0 sin8 𝑥 𝑑𝑥

Solution:
𝜋
6.4.2 16
(i) ∫02 cos 7 𝑥 𝑑𝑥 = 7.5.3.1 = 35
𝜋
9.7.5.3.1 𝜋 63𝜋
(ii) ∫02 cos10 𝑥 𝑑𝑥 = 10.8.6.4.2 . 2 = 512
𝜋
𝜋 7 5 3 𝜋 35𝜋
(iv) ∫0 sin8 𝑥 𝑑𝑥 = 2 ∫02 sin8 𝑥 𝑑𝑥 = 8 . 6 . 4 . 2 = 128