DUIS

Differentiation under integral sign.

Weightage : 08 marks (optional)

𝜋 (1+𝑎 cos 𝑥)
Q. ST ∫0 log 𝑑𝑥 = 𝜋 sin−1 𝑎
cos 𝑥

𝜋 (1+cos 𝑥)
Hence evaluate ∫0 log 𝑑𝑥
cos 𝑥

Solution :-

𝜋 (1+𝑎 cos 𝑥)
I = ∫0 log 𝑑𝑥
cos 𝑥

Diff BS wrt a

𝑑𝐼 𝜋 1 1
= ∫0 cos 𝑥 𝑑𝑥
𝑑𝑎 cos 𝑥 1+𝑎 𝑐𝑜𝑠𝑥

𝑑𝐼 𝜋 𝑑𝑥
= ∫0
𝑑𝑎 1+𝑎 𝑐𝑜𝑠𝑥

𝑥
Subs t = tan 2

2𝑑𝑡 1−𝑡 2
𝑑𝑥 = 1+𝑡 2
cos 𝑥 = 1+𝑡 2

𝑑𝐼 ∞ 1 2𝑑𝑡
∴ = ∫0 1−𝑡2
𝑑𝑎 1+𝑎( ) 1+𝑡 2
1+𝑡2

𝑑𝐼 ∞ 2𝑑𝑡
𝑑𝑎
= ∫0 ( 1+𝑡 2 )+𝑎 (( 1−𝑡 2 )

𝑑𝐼 ∞ 2𝑑𝑡
𝑑𝑎
= ∫0 ( 1+𝑎 )+ (1−𝑎)𝑡 2

𝑑𝐼 1 ∞ 2𝑑𝑡
𝑑𝑎
= 1−𝑎
∫0 1+𝑎
( )+𝑡 2
1−𝑎

LMR NOTES BY RUCHI TAWANI-9870094748 Page 1

 2 1−𝑎 1−𝑎
= √ 𝑡𝑎𝑛−1 √ 𝑡 !∞
1−𝑎 1+𝑎 1+𝑎

2 𝜋
= 2
√1−𝑎 2

𝑑𝐼 𝜋
∴ = 2
𝑑𝑎 √1−𝑎

Int BS

I = 𝜋 𝑠𝑖𝑛−1 𝑎 + 𝑐

Subs a = 0

I (0) = 𝑠𝑖𝑛−1 (0) + 𝑐

∴ c = 0

∴ I = 𝜋 𝑠𝑖𝑛−1 𝑎

𝜋 ( 1 + 𝑎 𝑐𝑜𝑠𝑥 )
∴ ∫0 log 𝑑𝑥 = 𝜋 𝑠𝑖𝑛−1 𝑎
cos 𝑥

Subs a = 1

𝜋 ( 1 + 𝑎 𝑐𝑜𝑠𝑥 ) 𝜋2
∫0 log 𝑑𝑥 = 𝜋 𝑠𝑖𝑛−1 (1) = .
cos 𝑥 2

∞ 1−cos 𝑚𝑥 1
Q.2 PT ∫0 𝑒 −𝑥 𝑑𝑥 = log (𝑚2 + 1)
𝑥 2

∞ 1−cos 𝑥
Hence deduce that ∫0 𝑒 −𝑥 𝑑𝑥 = log √2
𝑥

Solution :-

∞ 1−cos 𝑚𝑥
𝐼 = ∫0 𝑒 −𝑥 𝑑𝑥
𝑥

diff BS wrt m

LMR NOTES BY RUCHI TAWANI-9870094748 Page 2

 𝑑𝐼 ∞ 𝑒 −𝑥
= ∫0 (𝑥 sin 𝑚𝑥)𝑑𝑥
𝑑𝑚 𝑥

𝑑𝐼 ∞
= ∫0 𝑒 −𝑥 sin 𝑚𝑥 𝑑𝑥
𝑑𝑚

𝑑𝐼 𝑒 −𝑥
= (− sin 𝑥 − 𝑚𝑐𝑜𝑠 𝑚𝑥) |∞
0
𝑑𝑚 𝑚2 + 1

𝑑𝐼 1 𝑚
= (+𝑚 ) =
𝑑𝑚 𝑚2 + 1 𝑚2 + 1

𝑏𝑢𝑡 𝐵𝑆

1
𝐼 = log (𝑚2 + 1) + 𝑐
2

Subs m = 0

𝐼 (0) = 𝐶

∴ 𝐶 = 0

∞ 1−cos 𝑚𝑥 𝑑𝑥 = 1
∴ ∫0 𝑒− 𝑥 log (𝑚2 + 1)
𝑥 2

𝑠𝑢𝑏𝑠 𝑚 = 1 𝑤𝑒 𝑔𝑒𝑡

∞ 1−cos 𝑥 1
∫0 ( ) 𝑒 −𝑥 𝑑𝑥 = log 2 = log √2
𝑥 2

∞ 2 +𝑎2 /𝑥 2 ) √𝜋
Q.3 ST ∫0 𝑒 −(𝑥 𝑑𝑥 = 2
𝑒 −2𝑎

∞ 2 √𝜋
𝑔𝑖𝑣𝑒𝑛 ∫0 𝑒 −𝑥 𝑑𝑥 = 2

∞ 2 +1 /𝑥 2 )
𝐻𝑒𝑛𝑐𝑒 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 ∫0 𝑒 −(𝑥 𝑑𝑥

Solution :-

LMR NOTES BY RUCHI TAWANI-9870094748 Page 3

 𝑎2
∞ −(𝑥 2 + 2 )
𝐼 = ∫0 𝑒 𝑥 𝑑𝑥

𝑑𝑖𝑓𝑓 𝐵𝑆 𝑤𝑟𝑡 𝑎

𝑎2
𝑑𝐼 ∞ −(𝑥 2 + 2 ) 2𝑎
= ∫0 𝑒 𝑥 (− 𝑥 2 ) 𝑑𝑥
𝑑𝑎

𝑎 −𝑎
𝑠𝑢𝑏𝑠 = 𝑦 𝑑𝑥 = 𝑑𝑦
𝑥 𝑥2

𝑎2
𝑑𝐼 0 −(𝑦 2 + 2 )
∴ = ∫∞ 𝑒 𝑦 2𝑑𝑦
𝑑𝑎

𝑎2
𝑑𝐼 ∞ −(𝑦 2 + 2 )
= −2 ∫0 𝑒 𝑦 𝑑𝑦
𝑑𝑎

𝑑𝐼
= −2I
𝑑𝑎

𝑑𝐼
= −2 𝑑𝑎
𝐼

𝐼𝑛𝑡 𝐵𝑆

𝑑𝐼
∫ = ∫ −2 𝑑𝑎
𝐼

log 𝐼 = −2𝑎 + log 𝑐

𝐼
log (𝑐) = −2𝑎

𝐼
= 𝑒 −2𝑎
𝑐

∴ 𝐼 = 𝑐𝑒 −2𝑎

𝑠𝑢𝑏𝑠 𝑎 = 0

∴ 𝐼 (0) = 𝑐

LMR NOTES BY RUCHI TAWANI-9870094748 Page 4

 ∞ 2 𝜋
∴ 𝐼 (0) = ∫0 𝑒 −𝑥 𝑑𝑥 = √ 2

𝜋
∴ 𝐼 = √2 𝑒 −2𝑎

Subs a = 1

∞ 2 +1 /𝑥 2 ) √𝜋 √𝜋
∴ 𝐼 = ∫0 𝑒 −(𝑥 𝑑𝑥 = 𝑒 −2 =
2 2𝑒 2

𝜋
Q. Evaluate ∫02 log (𝑎2 𝑐𝑜𝑠 2 𝜃 + 𝑏 2 𝑠𝑖𝑛2 𝜃) 𝑑𝜃

𝜋
𝐼 = ∫02 log (𝑎2 𝑐𝑜𝑠 2 𝜃 + 𝑏 2 𝑠𝑖𝑛2 𝜃) 𝑑𝜃

𝑑𝑖𝑓𝑓 𝐵𝑆 𝑤𝑟𝑡 𝑎

𝜋
𝑑𝐼 1 [2𝑎 𝑐𝑜𝑠2 𝜃] 𝑑𝜃
= ∫0 2
𝑑𝑎 𝑎2 𝑐𝑜𝑠2 𝜃 +𝑏2 𝑠𝑖𝑛2 𝜃

𝜋
𝑑𝐼 2𝑎 𝑑𝜃
= ∫02 [𝐷𝑖𝑣 𝑁 & 𝐷 𝑏𝑦 𝑐𝑜𝑠 2 𝜃]
𝑑𝑎 𝑎2 +𝑏2 𝑡𝑎𝑛2 𝜃

𝑆𝑢𝑏𝑠 𝑡 = 𝑡𝑎𝑛 𝜃

𝑑𝐼 ∞ 2𝑎 𝑑𝑡
= ∫0
𝑑𝑎 (𝑎2 +𝑏 2 𝑡 2 ) 1+𝑡 2

𝑑𝑡
𝑑𝐼 2𝑎 ∞ 1 𝑏2
= ∫0 (1+𝑡 2 − 𝑎2 + 𝑏2 𝑡 2 )
𝑑𝑎 𝑎2 − 𝑏2

( perform PF )

𝑑𝐼 2𝑎 𝑏 𝑏
= [𝑡𝑎𝑛−1 𝑡 − 𝑡𝑎𝑛−1 𝑡 𝑎]
𝑑𝑎 𝑎2 − 𝑏2 𝑎

𝑑𝐼 2𝑎 𝜋 𝑏 𝜋 𝜋
𝑑𝑎
= 𝑎2 − 𝑏2
(2−𝑎 2
) = 𝑎+𝑏

LMR NOTES BY RUCHI TAWANI-9870094748 Page 5

 𝑑𝐼 𝜋
=
𝑑𝑎 𝑎+𝑏

𝑖𝑛𝑡 𝐵𝑆 𝑤𝑟𝑡 𝑎

𝐼 = 𝜋 log(𝑎 + 𝑏) + 𝑐

Subs a = b

𝜋
𝜋
𝐼 (𝑏) = ∫02 log 𝑏 2 𝑑𝜃 = log 𝑏 2 ( 2 − 0)

= 𝜋 log 𝑏 𝐼𝐼

𝐼 (𝑏) = 𝜋 log 2 𝑏 + 𝑐

∴ 𝜋 log 𝑏 = 𝜋 log 2 𝑏 + 𝑐

C = 𝜋 log(1⁄2)

1
∴ 𝐼 = 𝜋 log(𝑎 + 𝑏) + 𝜋 log (2)

𝑎+𝑏
𝐼 = 𝜋 log ( )
2

LMR NOTES BY RUCHI TAWANI-9870094748 Page 6