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CHAPTER โ€“ 7
INTEGRALS
Basic Concepts and Formulae :
1.List of Some Standard Integrals :
(i) ๐‘ฅ ๐‘›
๐‘‘๐‘ฅ =
๐‘ฅ ๐‘›+1
๐‘›+1
+ ๐ถ (๐‘› โ‰  1) (ii)
๐‘‘๐‘ฅ
๐‘ฅ
= ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐ถ
(iii) ๐‘‘๐‘ฅ = ๐‘ฅ + ๐ถ (iv) ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ ๐‘–๐‘› ๐‘ฅ + ๐ถ
(v) ๐‘ ๐‘–๐‘›๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘๐‘œ๐‘  ๐‘ฅ + ๐ถ (vi) ๐‘ ๐‘’๐‘2
๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ก๐‘Ž๐‘› ๐‘ฅ + ๐ถ
(vii) ๐‘๐‘œ๐‘ ๐‘’๐‘2
๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘๐‘œ๐‘ก ๐‘ฅ + ๐ถ (viii) ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐ถ
(ix) ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐ถ (x) ๐‘’ ๐‘ฅ
๐‘‘๐‘ฅ = ๐‘’ ๐‘ฅ
+ ๐ถ
(xi) ๐‘Ž ๐‘ฅ
๐‘‘๐‘ฅ =
๐‘Ž ๐‘ฅ
๐‘™๐‘œ๐‘” ๐‘Ž
+ ๐ถ
(xii) (a)
1
1โˆ’๐‘ฅ2
๐‘‘๐‘ฅ = sinโˆ’1
๐‘ฅ + ๐ถ (b)
๐‘‘๐‘ฅ
๐‘Ž2โˆ’๐‘ฅ2
= sinโˆ’1 ๐‘ฅ
๐‘Ž
+ ๐ถ
(xiii) (a)
1
1+๐‘ฅ2 ๐‘‘๐‘ฅ = sinโˆ’1
๐‘ฅ + ๐ถ (b)
1
๐‘Ž2+๐‘ฅ2 dx =
1
๐‘Ž
tanโˆ’1 ๐‘ฅ
๐‘Ž
+ ๐ถ
(xiv)
1
๐‘ฅ ๐‘ฅ2โˆ’1
๐‘‘๐‘ฅ = secโˆ’1
๐‘ฅ + ๐ถ (xv) โˆ’
๐‘‘๐‘ฅ
๐‘ฅ ๐‘ฅ2โˆ’1
= cosecโˆ’1
๐‘ฅ + ๐ถ
(xvi) โˆ’
1
๐‘Ž2โˆ’๐‘ฅ2 ๐‘‘๐‘ฅ =
1
๐‘Ž
cotโˆ’1 ๐‘ฅ
๐‘Ž
+ ๐ถ (xvii)
1
๐‘ฅ ๐‘ฅ2โˆ’๐‘Ž2
๐‘‘๐‘ฅ =
1
๐‘Ž
secโˆ’1 ๐‘ฅ
๐‘Ž
+ ๐ถ
(xviii) โˆ’
1
๐‘ฅ ๐‘ฅ2โˆ’๐‘Ž2
๐‘‘๐‘ฅ =
1
๐‘Ž
cosecโˆ’1 ๐‘ฅ
๐‘Ž
+ ๐ถ
1. More Standard Results :
๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘™๐‘œ๐‘” ๐‘๐‘œ๐‘  ๐‘ฅ + ๐ถ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐ถ, provided x is not an odd multiple of
๐œ‹
2
.
๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ + ๐ถ.
๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘ก๐‘Ž๐‘› ๐‘ฅ + ๐ถ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘
๐œ‹
4
+
๐‘ฅ
2
+ ๐ถ .
๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ โˆ’ ๐‘๐‘œ๐‘ก ๐‘ฅ + ๐ถ = ๐‘™๐‘œ๐‘” ๐‘ก๐‘Ž๐‘›
๐‘ฅ
2
+ ๐ถ
2. Results of Some Special Integrals :
๐‘‘๐‘ฅ
๐‘Ž2 + ๐‘ฅ2
=
1
๐‘Ž
tanโˆ’1
๐‘ฅ
๐‘Ž
+ ๐ถ
๐‘‘๐‘ฅ
๐‘ฅ2 โˆ’ ๐‘Ž2
=
1
2๐‘Ž
log
๐‘ฅ โˆ’ ๐‘Ž
๐‘ฅ + ๐‘Ž
+ ๐ถ
๐‘‘๐‘ฅ
๐‘Ž2 โˆ’ ๐‘ฅ2
=
1
2๐‘Ž
log
๐‘Ž + ๐‘ฅ
๐‘Ž โˆ’ ๐‘ฅ
+ ๐ถ
1
๐‘Ž2+๐‘ฅ2
๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘”
๐‘ฅ+ ๐‘ฅ2+๐‘Ž2
๐‘Ž
+ ๐ถ ๐‘œ๐‘Ÿ ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 + ๐‘Ž2 +C
1
๐‘ฅ2โˆ’๐‘Ž2
๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘”
๐‘ฅ+ ๐‘ฅ2โˆ’๐‘Ž2
๐‘Ž
+ ๐ถ ๐‘œ๐‘Ÿ ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 โˆ’ ๐‘Ž2 +C
1
๐‘Ž2 + ๐‘ฅ2
๐‘‘๐‘ฅ = sinโˆ’1
๐‘ฅ
๐‘Ž
+ ๐ถ
๐‘Ž2 โˆ’ ๐‘ฅ2 ๐‘‘๐‘ฅ =
๐‘ฅ
2
๐‘Ž2 โˆ’ ๐‘ฅ2 +
๐‘Ž2
2
sinโˆ’1 ๐‘ฅ
๐‘Ž
+ ๐ถ
๐‘Ž2 + ๐‘ฅ2 ๐‘‘๐‘ฅ =
๐‘ฅ
2
๐‘ฅ2 + ๐‘Ž2 +
๐‘Ž2
2
๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 + ๐‘Ž2 +C
๐‘ฅ2 โˆ’ ๐‘Ž2 ๐‘‘๐‘ฅ =
๐‘ฅ
2
๐‘ฅ2 โˆ’ ๐‘Ž2 -
๐‘Ž2
2
๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 โˆ’ ๐‘Ž2 +C
3. Properties of Definite Integrals :
(i) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘ง ๐‘‘๐‘ง
๐‘
๐‘Ž
๐‘
๐‘Ž
(change of variable)
(ii) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’ ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ
๐‘Ž
๐‘
๐‘
๐‘Ž
(change of limits)
(iii) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ + ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ
๐‘
๐‘
๐‘
๐‘Ž
๐‘
๐‘Ž
where a < c < b
(iv) ๐‘Ž ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘Ž โˆ’ ๐‘ฅ ๐‘‘๐‘ฅ
๐‘Ž
0
๐‘Ž
0
(b) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘Ž + ๐‘ โˆ’ ๐‘ฅ ๐‘‘๐‘ฅ
๐‘
๐‘Ž
๐‘
๐‘Ž
(v) (a) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = 2 ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ,
๐‘Ž
0
2๐‘Ž
๐‘Ž
if f (2a โ€“ x) = f(x)
(b) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = 0
2๐‘Ž
๐‘Ž
, if f (2a โ€“ x) = - f(x)
1 Mark Questions
1. Integrate the following w.r.t.x.
(i)
1
๐‘ฅ3 ๐‘‘๐‘ฅ. (ii)
1
๐‘ฅ
dx. (iii) ๐‘ ๐‘–๐‘›2
๐‘ฅ ๐‘๐‘œ๐‘ ๐‘’๐‘2
๐‘ฅ ๐‘‘๐‘ฅ.
(iv) ๐‘ก๐‘Ž๐‘›2
๐‘ฅ ๐‘‘๐‘ฅ (v) ๐‘๐‘œ๐‘ก2
๐‘ฅ ๐‘‘๐‘ฅ (vi)
๐‘‘๐‘ฅ
๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘๐‘œ๐‘ 2 ๐‘ฅ
.
(vii)
2 ๐‘๐‘œ๐‘  ๐‘ฅ
3๐‘ ๐‘–๐‘›2 ๐‘ฅ
๐‘‘๐‘ฅ. (viii) 2 ๐‘ก๐‘Ž๐‘› ๐‘ฅ โˆ’ 3 ๐‘๐‘œ๐‘ก ๐‘ฅ 2
๐‘‘๐‘ฅ.
(ix) ๐‘ฅ3
๐‘ ๐‘–๐‘› ๐‘ฅ4
๐‘‘๐‘ฅ (x)
๐‘ ๐‘’๐‘ 2 ๐‘™๐‘œ๐‘” ๐‘ฅ
๐‘ฅ
dx (xi)
๐‘ฅ
๐‘’ ๐‘ฅ2 dx
(xii)
1+๐‘™๐‘œ๐‘”๐‘ฅ 2
๐‘ฅ
dx (xiii)
๐‘ฅ2
1+๐‘ฅ3 dx (xiv)
๐‘ฅ+๐‘๐‘œ๐‘  6๐‘ฅ
3๐‘ฅ2+ ๐‘ ๐‘–๐‘› 6๐‘ฅ
dx
(xv) 2๐‘ฅ + 4 ๐‘ฅ2 + 4๐‘ฅ + 3 dx (xvi)
1+๐‘ ๐‘–๐‘›2๐‘ฅ
๐‘ฅ+๐‘ ๐‘–๐‘›2 ๐‘ฅ
dx
(xvii)
1+๐‘ก๐‘Ž๐‘›๐‘ฅ
๐‘ฅ+๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘๐‘ฅ
dx (xviii)
๐‘‘๐‘ฅ
๐‘ฅ+ ๐‘ฅ
(xix)
๐‘‘๐‘ฅ
4๐‘ฅ2โˆ’9
(xx)
๐‘ฅ3
1+๐‘ฅ8 dx (xxi)
๐‘ฅ2+4๐‘ฅ
๐‘ฅ3+6๐‘ฅ2+5
dx (xxii)
๐‘ ๐‘’๐‘ 2 ๐‘ฅ
3+๐‘ก๐‘Ž๐‘› ๐‘ฅ
dx
2. Evaluate the following integrals :
(๐‘–) ๐‘ ๐‘–๐‘›7
๐‘ฅ
๐œ‹
2
โˆ’
๐œ‹
2
dx (ii)
๐‘ ๐‘–๐‘› ๐‘ฅ
๐‘ฅ
dx (iii) ๐‘ ๐‘’๐‘2
(7 โˆ’ 4๐‘ฅ) ๐‘‘๐‘ฅ
(iv)
๐‘ฅ ๐‘’โˆ’1+๐‘’ ๐‘ฅโˆ’1
๐‘ฅ ๐‘’+๐‘’ ๐‘ฅ dx (v) ๐‘™๐‘œ๐‘”
3+5๐‘๐‘œ๐‘  ๐‘ฅ
3+5 ๐‘ ๐‘–๐‘› ๐‘ฅ
๐œ‹
2
0
dx (vi) ๐‘๐‘œ๐‘ 5
๐‘ฅ ๐‘‘๐‘ฅ
๐œ‹
0
(vii)
๐‘‘๐‘ฅ
1+๐‘ฅ2
1
0
(viii)
2๐‘ฅ
5๐‘ฅ2+1
1
0
dx (ix)
๐‘ฅ
๐‘ฅ
1
โˆ’2
dx
(x) ๐‘’3 ๐‘™๐‘œ๐‘” ๐‘ฅ
๐‘ฅ41
0
dx (xi) ๐‘ฅ
1.5
0
dx (xii) ๐‘ฅ ๐‘ฅ
2
0
dx
(xiii)
1
1+๐‘’ ๐‘ฅ dx (xiv) ๐‘ ๐‘–๐‘›2
๐‘ฅ
๐œ‹
2
0
dx (xv)
๐‘ฅ
๐‘ฅ2+1
4
2
dx
(xvi) ๐‘ฅ(1 โˆ’ ๐‘ฅ)2
๐‘‘๐‘ฅ
1
0
(xvii)
1โˆ’๐‘ก๐‘Ž๐‘› ๐‘ฅ
1+๐‘ก๐‘Ž๐‘› ๐‘ฅ
dx (xviii)
๐‘‘๐‘ฅ
๐‘ฅ2+1
โˆž
0
dx
(xix) ๐‘ ๐‘’๐‘4
๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ (xx) ๐‘ฅ
2
โˆ’1
dx (xxi)
๐‘’ ๐‘ฅ
4โˆ’๐‘’2๐‘ฅ
dx
(xxii) ๐‘™๐‘œ๐‘” ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ
๐œ‹
2
0
(xxiii) ๐‘ ๐‘–๐‘›75
๐‘ฅ + ๐‘ฅ125๐œ‹
โˆ’๐œ‹
dx (xxiv) ๐‘ ๐‘–๐‘› 2๐‘ฅ ๐‘‘๐‘ฅ
๐œ‹
2
0
(xv)
๐‘๐‘œ๐‘  ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ
dx (xvi) ๐‘Ž ๐‘ฅ
๐‘’ ๐‘ฅ
dx (xvii)
1โˆ’๐‘๐‘œ๐‘ก ๐‘ฅ
๐‘ฅ+ ๐‘™๐‘œ๐‘” ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ
dx
(xviii)
๐‘‘๐‘ฅ
9โˆ’๐‘ฅ2
3
0
(xix)
๐‘’ ๐‘ฅ
1+๐‘’2๐‘ฅ
1
0
dx
4 Marks Questions.
3. Evaluate the following:
(i)
๐‘ฅ2+1
๐‘ฅ+1 2 dx (ii)
๐‘‘๐‘ฅ
1+๐‘ก๐‘Ž๐‘› ๐‘ฅ
(iii)
๐‘‘๐‘ฅ
1+๐‘๐‘œ๐‘ก ๐‘ฅ
(iv) ๐‘ ๐‘–๐‘›4
๐‘ฅ ๐‘‘๐‘ฅ (v) ๐‘๐‘œ๐‘ 4
๐‘ฅ ๐‘‘๐‘ฅ (vi)
๐‘๐‘œ๐‘  2๐‘ฅ
๐‘๐‘œ๐‘ ๐‘ฅ +๐‘ ๐‘–๐‘› ๐‘ฅ 2 dx
(vii)
๐‘๐‘œ๐‘ 5 ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ
dx (viii)
๐‘ก๐‘Ž๐‘› ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ
dx (ix)
๐‘‘๐‘ฅ
๐‘ ๐‘–๐‘›3 ๐‘ฅ ๐‘ ๐‘–๐‘› (๐‘ฅ+๐›ผ)
dx
(x)
๐‘ ๐‘–๐‘› ๐‘ฅ
๐‘ ๐‘–๐‘› (๐‘ฅ+๐‘Ž)
dx (xi)
๐‘๐‘œ๐‘  2๐‘ฅโˆ’๐‘๐‘œ๐‘  ๐›ผ
๐‘๐‘œ๐‘  ๐‘ฅโˆ’๐‘๐‘œ๐‘  ๐›ผ
dx (xii)
๐‘‘๐‘ฅ
๐‘๐‘œ๐‘  ๐‘ฅโˆ’๐‘Ž ๐‘๐‘œ๐‘  (๐‘ฅโˆ’๐‘)
(xiii) ๐‘๐‘œ๐‘  2๐‘ฅ ๐‘๐‘œ๐‘  4๐‘ฅ ๐‘๐‘œ๐‘  6๐‘ฅ ๐‘‘๐‘ฅ (xiv)
๐‘ ๐‘–๐‘› 2๐‘ฅ
๐‘Ž2 ๐‘ ๐‘–๐‘›2 ๐‘ฅ+๐‘2 ๐‘๐‘œ๐‘ 2 ๐‘ฅ
dx
(xv)
๐‘ฅ+2
2๐‘ฅ2+6๐‘ฅ+5
dx (xvi)
๐‘‘๐‘ฅ
7โˆ’6๐‘ฅโˆ’๐‘ฅ2
(xvii)
5๐‘ฅ+3
๐‘ฅ2+4๐‘ฅ+10)
dx
(xviii)
๐‘ ๐‘–๐‘›( ๐‘ฅโˆ’ ๐›ผ
๐‘ ๐‘–๐‘› (๐‘ฅ+ ๐›ผ)
dx (xix)
๐‘ฅ
๐‘ฅ4โˆ’๐‘ฅ2+1
dx (xx)
2๐‘ฅ
1โˆ’๐‘ฅ2โˆ’๐‘ฅ4
dx
(xxi)
๐‘’ ๐‘ฅ
5โˆ’4๐‘’ ๐‘ฅ โˆ’๐‘’2๐‘ฅ
dx (xxii)
๐‘๐‘œ๐‘ ๐‘ฅ
๐‘ ๐‘–๐‘›2 ๐‘ฅโˆ’2 ๐‘ ๐‘–๐‘› ๐‘ฅโˆ’3
dx (xxiii)
๐‘ฅ
1โˆ’๐‘ฅ2+๐‘ฅ4
dx
(xxiv)
2๐‘ฅโˆ’1
๐‘ฅโˆ’1 ๐‘ฅ+2 ๐‘ฅโˆ’3)
dx (xxv)
3๐‘ฅโˆ’2
๐‘ฅ+1 2(๐‘ฅ+3)
dx (xxvi)
๐‘ ๐‘–๐‘› ๐‘ฅ
1โˆ’๐‘๐‘œ๐‘  ๐‘ฅ 2โˆ’๐‘๐‘œ๐‘  ๐‘ฅ
dx
(xxvii)
๐‘ฅ
๐‘ฅ2+1 (๐‘ฅ+1)
dx (xxviii)
๐‘‘๐‘ฅ
๐‘ฅ ๐‘ฅ5+1
dx (xxix) ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ
(xxx)
2๐‘ฅ
๐‘ฅ2+1 (๐‘ฅ2+3)
dx (xxxi)
๐‘ฅ2
1+๐‘ฅ3 (2+๐‘ฅ3)
dx (xxxii)
๐‘‘๐‘ฅ
๐‘ฅ[6 ๐‘™๐‘œ๐‘” ๐‘ฅ 2+7 ๐‘™๐‘œ๐‘” ๐‘ฅ+2
dx
4. Integrate the following:
(i)
๐‘ฅ2+1
๐‘ฅ4+1
dx (ii)
๐‘‘๐‘ฅ
1+๐‘ฅ+๐‘ฅ2+๐‘ฅ3 dx (iii) ๐‘ฅ ๐‘™๐‘œ๐‘” 1 + ๐‘ฅ ๐‘‘๐‘ฅ
(iv) ๐‘ฅ tanโˆ’1
๐‘‘๐‘ฅ(v) sinโˆ’1
๐‘ฅ 2
dx (vi) ๐‘ ๐‘’๐‘3
๐‘ฅ ๐‘‘๐‘ฅ
(vii) ๐‘’ ๐‘ฅ
๐‘๐‘œ๐‘  ๐‘ฅ ๐‘‘๐‘ฅ (viii)
sin โˆ’1 ๐‘ฅ
๐‘ฅ2 dx (ix) ๐‘’ ๐‘ฅ
[tanโˆ’1
๐‘ฅ +
1
1+๐‘ฅ2]dx
(x)
๐‘ฅ
๐‘ฅ+1 2 ๐‘’ ๐‘ฅ
๐‘‘๐‘ฅ (xi)
1
๐‘™๐‘œ๐‘” ๐‘ฅ
โˆ’
1
๐‘™๐‘œ๐‘” ๐‘ฅ 2 dx (xii) ๐‘’ ๐‘ฅ 1
๐‘ฅ
โˆ’
1
๐‘ฅ2 dx
(xiii)
๐‘ฅ2+1
๐‘ฅ+1 2 ๐‘’ ๐‘ฅ
dx (xiv) ๐‘™๐‘œ๐‘” ๐‘™๐‘œ๐‘” ๐‘ฅ +
1
๐‘™๐‘œ๐‘” ๐‘ฅ
dx (xv) ๐‘’ ๐‘ฅ 1
๐‘ฅ2 โˆ’
1
๐‘ฅ3 dx
(xvi) ๐‘’ ๐‘ฅ 1+๐‘ ๐‘–๐‘›๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ
๐‘๐‘œ๐‘ 2 ๐‘ฅ
dx (xvii)
1โˆ’ ๐‘ฅ
1+ ๐‘ฅ
dx (xviii)
2+๐‘ ๐‘–๐‘› 2๐‘ฅ
1+๐‘๐‘œ๐‘  2๐‘ฅ
๐‘’ ๐‘ฅ
dx
(xix)
๐‘ ๐‘–๐‘›8 ๐‘ฅโˆ’๐‘๐‘œ๐‘ 8 ๐‘ฅ
1โˆ’2 ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘๐‘œ๐‘ 2 ๐‘ฅ
dx (xx)
sin โˆ’1 ๐‘ฅโˆ’cos โˆ’1 ๐‘ฅ
sin โˆ’1 ๐‘ฅ+cos โˆ’1 ๐‘ฅ
dx (xxi)
๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘ ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ
๐‘‘๐‘ฅ
(xxii)
๐‘‘๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ (5โˆ’4 ๐‘๐‘œ๐‘  ๐‘ฅ)
(xxiii)
๐‘ฅ
๐‘ฅ3โˆ’1
๐‘‘๐‘ฅ (xxiv) ๐‘ฅ sinโˆ’1
๐‘ฅ ๐‘‘๐‘ฅ
(xxv) ๐‘ฅ2
tanโˆ’1
๐‘ฅ ๐‘‘๐‘ฅ (xxvi)
1โˆ’๐‘ฅ
1+๐‘ฅ
dx (xxvii)
1โˆ’๐‘ฅ2
๐‘ฅ(1โˆ’2๐‘ฅ)
dx
4. Evaluate the following integrals :
(i)
๐‘ ๐‘–๐‘› ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ
๐œ‹
2
0
๐‘‘๐‘ฅ (ii)
๐‘ ๐‘–๐‘› ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ+ ๐‘๐‘œ๐‘  ๐‘ฅ
๐œ‹
2
0
๐‘‘๐‘ฅ (iii)
๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ
1+ ๐‘๐‘œ๐‘ 2 ๐‘ฅ
๐œ‹
2
0
๐‘‘๐‘ฅ
(iv)
๐‘‘๐‘ฅ
1+ ๐‘ก๐‘Ž๐‘› ๐‘ฅ
๐œ‹
3
๐œ‹
6
(v) ๐‘ฅ + 2
5
โˆ’5
๐‘‘๐‘ฅ (vi) ๐‘™๐‘œ๐‘” 1 + ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ
๐œ‹
4
0
(vii) (2 ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ โˆ’ ๐‘™๐‘œ๐‘” ๐‘ ๐‘› 2๐‘ฅ)๐‘‘๐‘ฅ
๐œ‹
2
0
(viii) ๐‘ ๐‘–๐‘›2
๐‘ฅ ๐‘‘๐‘ฅ
๐œ‹
4
โˆ’
๐œ‹
4
(ix)
๐‘‘ ๐‘ฅ
1+ ๐‘ก๐‘Ž๐‘› 3 ๐‘ฅ
๐œ‹
2
0
(x)
๐‘ฅ
1+๐‘ ๐‘–๐‘› ๐‘ฅ
๐œ‹
0
๐‘‘๐‘ฅ (xi)
๐‘ ๐‘–๐‘› ๐‘ฅโˆ’๐‘๐‘œ๐‘  ๐‘ฅ
1+๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ
๐œ‹
2
0
๐‘‘๐‘ฅ
(xii)
๐‘ฅ
๐‘ฅ+ ๐‘Žโˆ’๐‘ฅ
๐‘Ž
0
dx (xiii) ๐‘ฅ โˆ’ 1
4
0
๐‘‘๐‘ฅ (xiv)
๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ
๐‘ ๐‘–๐‘› 2 ๐‘ฅ
๐œ‹
3
๐œ‹
6
dx
(xv)
๐‘ฅ
1+๐‘๐‘œ๐‘ 2 ๐‘ฅ
๐œ‹
0
๐‘‘๐‘ฅ (xvi)
๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ
๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ
๐œ‹
0
๐‘‘๐‘ฅ (xvii) ๐‘ ๐‘–๐‘› ๐‘ฅ
๐œ‹
4
โˆ’
๐œ‹
4
๐‘‘๐‘ฅ
(xviii)
๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ
๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘ก๐‘Ž๐‘› ๐‘ฅ
๐œ‹
0
๐‘‘๐‘ฅ (xix) ๐‘ฅ ๐‘๐‘œ๐‘  ๐œ‹ ๐‘ฅ
1
โˆ’1
๐‘‘๐‘ฅ (xx) ๐‘’ ๐‘ฅ ๐‘ ๐‘–๐‘› 4๐‘ฅโˆ’4
1โˆ’๐‘๐‘œ๐‘  4๐‘ฅ
๐‘‘๐‘ฅ
6 Marks Questions :
5. Using properties of definite, evaluate.
(i)
๐‘ฅ ๐‘‘๐‘ฅ
4โˆ’๐‘๐‘œ๐‘ 2 ๐‘ฅ
๐œ‹
0
(ii)
๐‘’ ๐‘๐‘œ๐‘  ๐‘ฅ
๐‘’ ๐‘๐‘œ๐‘  ๐‘ฅ +๐‘’โˆ’๐‘๐‘œ๐‘  ๐‘ฅ
๐œ‹
0
dx (iii)
๐‘ฅ ๐‘‘๐‘ฅ
๐‘Ž2 ๐‘๐‘œ๐‘ 2 ๐‘ฅ+ ๐‘2 ๐‘ ๐‘–๐‘›2 ๐‘ฅ
๐œ‹
0
(iv)
๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ
9+16 ๐‘ ๐‘–๐‘› 2๐‘ฅ
๐œ‹
4
0
๐‘‘๐‘ฅ (v) ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘‘๐‘ฅ
๐œ‹
2
0
(vi) sinโˆ’1 2๐‘ฅ
1+๐‘ฅ2
1
0
dx
(vii)
๐‘‘ ๐‘ฅ
3+2 ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ
๐œ‹
2
0
(viii) Show that ๐‘ก๐‘Ž๐‘› ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘‘๐‘ฅ = 2 ๐œ‹
๐œ‹
2
0
(ix) sinโˆ’1 ๐‘ฅ
๐‘Ž+๐‘ฅ
๐‘Ž
0
(x)
๐‘ ๐‘–๐‘› 2 ๐‘ฅ
๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ
๐œ‹
2
0
dx
(xi) ๐‘ฅ + ๐‘ฅ + 2 + ๐‘ฅ + 5
0
โˆ’5
dx (xii)
๐‘™๐‘œ๐‘” ๐‘ฅ
1โˆ’๐‘ฅ
1
0
dx (xiii) ๐‘ก๐‘Ž๐‘› ๐‘ฅ
๐œ‹
4
0
๐‘‘๐‘ฅ
6. Evaluate as the limit of sums :
(i) 2๐‘ฅ2
โˆ’ 5 ๐‘‘๐‘ฅ
3
0
(ii) ๐‘ฅ2
+ 5๐‘ฅ ๐‘‘๐‘ฅ
3
1
(iii) ๐‘ฅ2
+ ๐‘ฅ + 1 ๐‘‘๐‘ฅ
2
0
(v) 3๐‘ฅ2
+ 2๐‘ฅ ๐‘‘๐‘ฅ
3
1
(vi)
๐‘ฅ2
๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ 2 dx
Class 12 chapter 7

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ๅ“ช้‡Œๅฏไปฅ่ดญไนฐๆ—ฅๆœฌ็ญ‘ๆณขๅญฆ้™ขๅคงๅญฆๅญฆไฝ่ฎฐ/ๅšไธชๅ‡็š„ๆ–‡ๅ‡ญๅฏ่ฎค่ฏๅ—/ไปฟๅˆถๆ—ฅๆœฌๅคงๅญฆๆฏ•ไธš่ฏ/ๆ„ๅคงๅˆฉ่ฏญCELI่ฏไนฆๅฎšๅˆถ
ย 
1.๐ŸŽ‰โ€œๅ…ฅไพตๅคงๅญฆๅ…ฅๅญฆ่€ƒ่ฏ•ไธญๅฟƒไฟฎๆ”นๆˆ็ปฉโ€ๆฅ่ขญ๏ผALEVELๆ›ฟ่€ƒๅคงๆญ็ง˜๏ผŒ่ฝปๆพๆžๅฎš่€ƒ่ฏ•ๆˆ็ปฉ๏ผ ๐Ÿ’ฅไฝ ่ฟ˜ๅœจไธบๆ— ๆณ•่ฟ›ๅ…ฅๅคงๅญฆๆ‹›็”Ÿ็ณป็ปŸ่€Œ็ƒฆๆผๅ—๏ผŸๆƒณ็Ÿฅ้“ๅฆ‚ไฝ•้€š่ฟ‡ๆŠ€ๆœฏๆ‰‹ๆฎตๆ›ดๆ”น...
1.๐ŸŽ‰โ€œๅ…ฅไพตๅคงๅญฆๅ…ฅๅญฆ่€ƒ่ฏ•ไธญๅฟƒไฟฎๆ”นๆˆ็ปฉโ€ๆฅ่ขญ๏ผALEVELๆ›ฟ่€ƒๅคงๆญ็ง˜๏ผŒ่ฝปๆพๆžๅฎš่€ƒ่ฏ•ๆˆ็ปฉ๏ผ ๐Ÿ’ฅไฝ ่ฟ˜ๅœจไธบๆ— ๆณ•่ฟ›ๅ…ฅๅคงๅญฆๆ‹›็”Ÿ็ณป็ปŸ่€Œ็ƒฆๆผๅ—๏ผŸๆƒณ็Ÿฅ้“ๅฆ‚ไฝ•้€š่ฟ‡ๆŠ€ๆœฏๆ‰‹ๆฎตๆ›ดๆ”น...1.๐ŸŽ‰โ€œๅ…ฅไพตๅคงๅญฆๅ…ฅๅญฆ่€ƒ่ฏ•ไธญๅฟƒไฟฎๆ”นๆˆ็ปฉโ€ๆฅ่ขญ๏ผALEVELๆ›ฟ่€ƒๅคงๆญ็ง˜๏ผŒ่ฝปๆพๆžๅฎš่€ƒ่ฏ•ๆˆ็ปฉ๏ผ ๐Ÿ’ฅไฝ ่ฟ˜ๅœจไธบๆ— ๆณ•่ฟ›ๅ…ฅๅคงๅญฆๆ‹›็”Ÿ็ณป็ปŸ่€Œ็ƒฆๆผๅ—๏ผŸๆƒณ็Ÿฅ้“ๅฆ‚ไฝ•้€š่ฟ‡ๆŠ€ๆœฏๆ‰‹ๆฎตๆ›ดๆ”น...
1.๐ŸŽ‰โ€œๅ…ฅไพตๅคงๅญฆๅ…ฅๅญฆ่€ƒ่ฏ•ไธญๅฟƒไฟฎๆ”นๆˆ็ปฉโ€ๆฅ่ขญ๏ผALEVELๆ›ฟ่€ƒๅคงๆญ็ง˜๏ผŒ่ฝปๆพๆžๅฎš่€ƒ่ฏ•ๆˆ็ปฉ๏ผ ๐Ÿ’ฅไฝ ่ฟ˜ๅœจไธบๆ— ๆณ•่ฟ›ๅ…ฅๅคงๅญฆๆ‹›็”Ÿ็ณป็ปŸ่€Œ็ƒฆๆผๅ—๏ผŸๆƒณ็Ÿฅ้“ๅฆ‚ไฝ•้€š่ฟ‡ๆŠ€ๆœฏๆ‰‹ๆฎตๆ›ดๆ”น...
ย 

Class 12 chapter 7

  • 1. CHAPTER โ€“ 7 INTEGRALS Basic Concepts and Formulae : 1.List of Some Standard Integrals : (i) ๐‘ฅ ๐‘› ๐‘‘๐‘ฅ = ๐‘ฅ ๐‘›+1 ๐‘›+1 + ๐ถ (๐‘› โ‰  1) (ii) ๐‘‘๐‘ฅ ๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐ถ (iii) ๐‘‘๐‘ฅ = ๐‘ฅ + ๐ถ (iv) ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ ๐‘–๐‘› ๐‘ฅ + ๐ถ (v) ๐‘ ๐‘–๐‘›๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘๐‘œ๐‘  ๐‘ฅ + ๐ถ (vi) ๐‘ ๐‘’๐‘2 ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ก๐‘Ž๐‘› ๐‘ฅ + ๐ถ (vii) ๐‘๐‘œ๐‘ ๐‘’๐‘2 ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘๐‘œ๐‘ก ๐‘ฅ + ๐ถ (viii) ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐ถ (ix) ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐ถ (x) ๐‘’ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘’ ๐‘ฅ + ๐ถ (xi) ๐‘Ž ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘Ž ๐‘ฅ ๐‘™๐‘œ๐‘” ๐‘Ž + ๐ถ (xii) (a) 1 1โˆ’๐‘ฅ2 ๐‘‘๐‘ฅ = sinโˆ’1 ๐‘ฅ + ๐ถ (b) ๐‘‘๐‘ฅ ๐‘Ž2โˆ’๐‘ฅ2 = sinโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ (xiii) (a) 1 1+๐‘ฅ2 ๐‘‘๐‘ฅ = sinโˆ’1 ๐‘ฅ + ๐ถ (b) 1 ๐‘Ž2+๐‘ฅ2 dx = 1 ๐‘Ž tanโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ (xiv) 1 ๐‘ฅ ๐‘ฅ2โˆ’1 ๐‘‘๐‘ฅ = secโˆ’1 ๐‘ฅ + ๐ถ (xv) โˆ’ ๐‘‘๐‘ฅ ๐‘ฅ ๐‘ฅ2โˆ’1 = cosecโˆ’1 ๐‘ฅ + ๐ถ (xvi) โˆ’ 1 ๐‘Ž2โˆ’๐‘ฅ2 ๐‘‘๐‘ฅ = 1 ๐‘Ž cotโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ (xvii) 1 ๐‘ฅ ๐‘ฅ2โˆ’๐‘Ž2 ๐‘‘๐‘ฅ = 1 ๐‘Ž secโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ (xviii) โˆ’ 1 ๐‘ฅ ๐‘ฅ2โˆ’๐‘Ž2 ๐‘‘๐‘ฅ = 1 ๐‘Ž cosecโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ 1. More Standard Results : ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’๐‘™๐‘œ๐‘” ๐‘๐‘œ๐‘  ๐‘ฅ + ๐ถ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐ถ, provided x is not an odd multiple of ๐œ‹ 2 . ๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ + ๐ถ. ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘ ๐‘ฅ + ๐‘ก๐‘Ž๐‘› ๐‘ฅ + ๐ถ = ๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘ ๐œ‹ 4 + ๐‘ฅ 2 + ๐ถ . ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ โˆ’ ๐‘๐‘œ๐‘ก ๐‘ฅ + ๐ถ = ๐‘™๐‘œ๐‘” ๐‘ก๐‘Ž๐‘› ๐‘ฅ 2 + ๐ถ 2. Results of Some Special Integrals : ๐‘‘๐‘ฅ ๐‘Ž2 + ๐‘ฅ2 = 1 ๐‘Ž tanโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ ๐‘‘๐‘ฅ ๐‘ฅ2 โˆ’ ๐‘Ž2 = 1 2๐‘Ž log ๐‘ฅ โˆ’ ๐‘Ž ๐‘ฅ + ๐‘Ž + ๐ถ ๐‘‘๐‘ฅ ๐‘Ž2 โˆ’ ๐‘ฅ2 = 1 2๐‘Ž log ๐‘Ž + ๐‘ฅ ๐‘Ž โˆ’ ๐‘ฅ + ๐ถ 1 ๐‘Ž2+๐‘ฅ2 ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ฅ+ ๐‘ฅ2+๐‘Ž2 ๐‘Ž + ๐ถ ๐‘œ๐‘Ÿ ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 + ๐‘Ž2 +C 1 ๐‘ฅ2โˆ’๐‘Ž2 ๐‘‘๐‘ฅ = ๐‘™๐‘œ๐‘” ๐‘ฅ+ ๐‘ฅ2โˆ’๐‘Ž2 ๐‘Ž + ๐ถ ๐‘œ๐‘Ÿ ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 โˆ’ ๐‘Ž2 +C 1 ๐‘Ž2 + ๐‘ฅ2 ๐‘‘๐‘ฅ = sinโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ
  • 2. ๐‘Ž2 โˆ’ ๐‘ฅ2 ๐‘‘๐‘ฅ = ๐‘ฅ 2 ๐‘Ž2 โˆ’ ๐‘ฅ2 + ๐‘Ž2 2 sinโˆ’1 ๐‘ฅ ๐‘Ž + ๐ถ ๐‘Ž2 + ๐‘ฅ2 ๐‘‘๐‘ฅ = ๐‘ฅ 2 ๐‘ฅ2 + ๐‘Ž2 + ๐‘Ž2 2 ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 + ๐‘Ž2 +C ๐‘ฅ2 โˆ’ ๐‘Ž2 ๐‘‘๐‘ฅ = ๐‘ฅ 2 ๐‘ฅ2 โˆ’ ๐‘Ž2 - ๐‘Ž2 2 ๐‘™๐‘œ๐‘” ๐‘ฅ + ๐‘ฅ2 โˆ’ ๐‘Ž2 +C 3. Properties of Definite Integrals : (i) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘ง ๐‘‘๐‘ง ๐‘ ๐‘Ž ๐‘ ๐‘Ž (change of variable) (ii) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = โˆ’ ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ ๐‘Ž ๐‘ ๐‘ ๐‘Ž (change of limits) (iii) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ + ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ ๐‘ ๐‘ ๐‘ ๐‘Ž ๐‘ ๐‘Ž where a < c < b (iv) ๐‘Ž ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘Ž โˆ’ ๐‘ฅ ๐‘‘๐‘ฅ ๐‘Ž 0 ๐‘Ž 0 (b) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = ๐‘“ ๐‘Ž + ๐‘ โˆ’ ๐‘ฅ ๐‘‘๐‘ฅ ๐‘ ๐‘Ž ๐‘ ๐‘Ž (v) (a) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = 2 ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ, ๐‘Ž 0 2๐‘Ž ๐‘Ž if f (2a โ€“ x) = f(x) (b) ๐‘“ ๐‘ฅ ๐‘‘๐‘ฅ = 0 2๐‘Ž ๐‘Ž , if f (2a โ€“ x) = - f(x) 1 Mark Questions 1. Integrate the following w.r.t.x. (i) 1 ๐‘ฅ3 ๐‘‘๐‘ฅ. (ii) 1 ๐‘ฅ dx. (iii) ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘๐‘œ๐‘ ๐‘’๐‘2 ๐‘ฅ ๐‘‘๐‘ฅ. (iv) ๐‘ก๐‘Ž๐‘›2 ๐‘ฅ ๐‘‘๐‘ฅ (v) ๐‘๐‘œ๐‘ก2 ๐‘ฅ ๐‘‘๐‘ฅ (vi) ๐‘‘๐‘ฅ ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘๐‘œ๐‘ 2 ๐‘ฅ . (vii) 2 ๐‘๐‘œ๐‘  ๐‘ฅ 3๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘‘๐‘ฅ. (viii) 2 ๐‘ก๐‘Ž๐‘› ๐‘ฅ โˆ’ 3 ๐‘๐‘œ๐‘ก ๐‘ฅ 2 ๐‘‘๐‘ฅ. (ix) ๐‘ฅ3 ๐‘ ๐‘–๐‘› ๐‘ฅ4 ๐‘‘๐‘ฅ (x) ๐‘ ๐‘’๐‘ 2 ๐‘™๐‘œ๐‘” ๐‘ฅ ๐‘ฅ dx (xi) ๐‘ฅ ๐‘’ ๐‘ฅ2 dx (xii) 1+๐‘™๐‘œ๐‘”๐‘ฅ 2 ๐‘ฅ dx (xiii) ๐‘ฅ2 1+๐‘ฅ3 dx (xiv) ๐‘ฅ+๐‘๐‘œ๐‘  6๐‘ฅ 3๐‘ฅ2+ ๐‘ ๐‘–๐‘› 6๐‘ฅ dx (xv) 2๐‘ฅ + 4 ๐‘ฅ2 + 4๐‘ฅ + 3 dx (xvi) 1+๐‘ ๐‘–๐‘›2๐‘ฅ ๐‘ฅ+๐‘ ๐‘–๐‘›2 ๐‘ฅ dx (xvii) 1+๐‘ก๐‘Ž๐‘›๐‘ฅ ๐‘ฅ+๐‘™๐‘œ๐‘” ๐‘ ๐‘’๐‘๐‘ฅ dx (xviii) ๐‘‘๐‘ฅ ๐‘ฅ+ ๐‘ฅ (xix) ๐‘‘๐‘ฅ 4๐‘ฅ2โˆ’9 (xx) ๐‘ฅ3 1+๐‘ฅ8 dx (xxi) ๐‘ฅ2+4๐‘ฅ ๐‘ฅ3+6๐‘ฅ2+5 dx (xxii) ๐‘ ๐‘’๐‘ 2 ๐‘ฅ 3+๐‘ก๐‘Ž๐‘› ๐‘ฅ dx 2. Evaluate the following integrals : (๐‘–) ๐‘ ๐‘–๐‘›7 ๐‘ฅ ๐œ‹ 2 โˆ’ ๐œ‹ 2 dx (ii) ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘ฅ dx (iii) ๐‘ ๐‘’๐‘2 (7 โˆ’ 4๐‘ฅ) ๐‘‘๐‘ฅ
  • 3. (iv) ๐‘ฅ ๐‘’โˆ’1+๐‘’ ๐‘ฅโˆ’1 ๐‘ฅ ๐‘’+๐‘’ ๐‘ฅ dx (v) ๐‘™๐‘œ๐‘” 3+5๐‘๐‘œ๐‘  ๐‘ฅ 3+5 ๐‘ ๐‘–๐‘› ๐‘ฅ ๐œ‹ 2 0 dx (vi) ๐‘๐‘œ๐‘ 5 ๐‘ฅ ๐‘‘๐‘ฅ ๐œ‹ 0 (vii) ๐‘‘๐‘ฅ 1+๐‘ฅ2 1 0 (viii) 2๐‘ฅ 5๐‘ฅ2+1 1 0 dx (ix) ๐‘ฅ ๐‘ฅ 1 โˆ’2 dx (x) ๐‘’3 ๐‘™๐‘œ๐‘” ๐‘ฅ ๐‘ฅ41 0 dx (xi) ๐‘ฅ 1.5 0 dx (xii) ๐‘ฅ ๐‘ฅ 2 0 dx (xiii) 1 1+๐‘’ ๐‘ฅ dx (xiv) ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐œ‹ 2 0 dx (xv) ๐‘ฅ ๐‘ฅ2+1 4 2 dx (xvi) ๐‘ฅ(1 โˆ’ ๐‘ฅ)2 ๐‘‘๐‘ฅ 1 0 (xvii) 1โˆ’๐‘ก๐‘Ž๐‘› ๐‘ฅ 1+๐‘ก๐‘Ž๐‘› ๐‘ฅ dx (xviii) ๐‘‘๐‘ฅ ๐‘ฅ2+1 โˆž 0 dx (xix) ๐‘ ๐‘’๐‘4 ๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ (xx) ๐‘ฅ 2 โˆ’1 dx (xxi) ๐‘’ ๐‘ฅ 4โˆ’๐‘’2๐‘ฅ dx (xxii) ๐‘™๐‘œ๐‘” ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ ๐œ‹ 2 0 (xxiii) ๐‘ ๐‘–๐‘›75 ๐‘ฅ + ๐‘ฅ125๐œ‹ โˆ’๐œ‹ dx (xxiv) ๐‘ ๐‘–๐‘› 2๐‘ฅ ๐‘‘๐‘ฅ ๐œ‹ 2 0 (xv) ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ dx (xvi) ๐‘Ž ๐‘ฅ ๐‘’ ๐‘ฅ dx (xvii) 1โˆ’๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘ฅ+ ๐‘™๐‘œ๐‘” ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ dx (xviii) ๐‘‘๐‘ฅ 9โˆ’๐‘ฅ2 3 0 (xix) ๐‘’ ๐‘ฅ 1+๐‘’2๐‘ฅ 1 0 dx 4 Marks Questions. 3. Evaluate the following: (i) ๐‘ฅ2+1 ๐‘ฅ+1 2 dx (ii) ๐‘‘๐‘ฅ 1+๐‘ก๐‘Ž๐‘› ๐‘ฅ (iii) ๐‘‘๐‘ฅ 1+๐‘๐‘œ๐‘ก ๐‘ฅ (iv) ๐‘ ๐‘–๐‘›4 ๐‘ฅ ๐‘‘๐‘ฅ (v) ๐‘๐‘œ๐‘ 4 ๐‘ฅ ๐‘‘๐‘ฅ (vi) ๐‘๐‘œ๐‘  2๐‘ฅ ๐‘๐‘œ๐‘ ๐‘ฅ +๐‘ ๐‘–๐‘› ๐‘ฅ 2 dx (vii) ๐‘๐‘œ๐‘ 5 ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ dx (viii) ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ dx (ix) ๐‘‘๐‘ฅ ๐‘ ๐‘–๐‘›3 ๐‘ฅ ๐‘ ๐‘–๐‘› (๐‘ฅ+๐›ผ) dx (x) ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘ ๐‘–๐‘› (๐‘ฅ+๐‘Ž) dx (xi) ๐‘๐‘œ๐‘  2๐‘ฅโˆ’๐‘๐‘œ๐‘  ๐›ผ ๐‘๐‘œ๐‘  ๐‘ฅโˆ’๐‘๐‘œ๐‘  ๐›ผ dx (xii) ๐‘‘๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅโˆ’๐‘Ž ๐‘๐‘œ๐‘  (๐‘ฅโˆ’๐‘) (xiii) ๐‘๐‘œ๐‘  2๐‘ฅ ๐‘๐‘œ๐‘  4๐‘ฅ ๐‘๐‘œ๐‘  6๐‘ฅ ๐‘‘๐‘ฅ (xiv) ๐‘ ๐‘–๐‘› 2๐‘ฅ ๐‘Ž2 ๐‘ ๐‘–๐‘›2 ๐‘ฅ+๐‘2 ๐‘๐‘œ๐‘ 2 ๐‘ฅ dx (xv) ๐‘ฅ+2 2๐‘ฅ2+6๐‘ฅ+5 dx (xvi) ๐‘‘๐‘ฅ 7โˆ’6๐‘ฅโˆ’๐‘ฅ2 (xvii) 5๐‘ฅ+3 ๐‘ฅ2+4๐‘ฅ+10) dx
  • 4. (xviii) ๐‘ ๐‘–๐‘›( ๐‘ฅโˆ’ ๐›ผ ๐‘ ๐‘–๐‘› (๐‘ฅ+ ๐›ผ) dx (xix) ๐‘ฅ ๐‘ฅ4โˆ’๐‘ฅ2+1 dx (xx) 2๐‘ฅ 1โˆ’๐‘ฅ2โˆ’๐‘ฅ4 dx (xxi) ๐‘’ ๐‘ฅ 5โˆ’4๐‘’ ๐‘ฅ โˆ’๐‘’2๐‘ฅ dx (xxii) ๐‘๐‘œ๐‘ ๐‘ฅ ๐‘ ๐‘–๐‘›2 ๐‘ฅโˆ’2 ๐‘ ๐‘–๐‘› ๐‘ฅโˆ’3 dx (xxiii) ๐‘ฅ 1โˆ’๐‘ฅ2+๐‘ฅ4 dx (xxiv) 2๐‘ฅโˆ’1 ๐‘ฅโˆ’1 ๐‘ฅ+2 ๐‘ฅโˆ’3) dx (xxv) 3๐‘ฅโˆ’2 ๐‘ฅ+1 2(๐‘ฅ+3) dx (xxvi) ๐‘ ๐‘–๐‘› ๐‘ฅ 1โˆ’๐‘๐‘œ๐‘  ๐‘ฅ 2โˆ’๐‘๐‘œ๐‘  ๐‘ฅ dx (xxvii) ๐‘ฅ ๐‘ฅ2+1 (๐‘ฅ+1) dx (xxviii) ๐‘‘๐‘ฅ ๐‘ฅ ๐‘ฅ5+1 dx (xxix) ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ (xxx) 2๐‘ฅ ๐‘ฅ2+1 (๐‘ฅ2+3) dx (xxxi) ๐‘ฅ2 1+๐‘ฅ3 (2+๐‘ฅ3) dx (xxxii) ๐‘‘๐‘ฅ ๐‘ฅ[6 ๐‘™๐‘œ๐‘” ๐‘ฅ 2+7 ๐‘™๐‘œ๐‘” ๐‘ฅ+2 dx 4. Integrate the following: (i) ๐‘ฅ2+1 ๐‘ฅ4+1 dx (ii) ๐‘‘๐‘ฅ 1+๐‘ฅ+๐‘ฅ2+๐‘ฅ3 dx (iii) ๐‘ฅ ๐‘™๐‘œ๐‘” 1 + ๐‘ฅ ๐‘‘๐‘ฅ (iv) ๐‘ฅ tanโˆ’1 ๐‘‘๐‘ฅ(v) sinโˆ’1 ๐‘ฅ 2 dx (vi) ๐‘ ๐‘’๐‘3 ๐‘ฅ ๐‘‘๐‘ฅ (vii) ๐‘’ ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘‘๐‘ฅ (viii) sin โˆ’1 ๐‘ฅ ๐‘ฅ2 dx (ix) ๐‘’ ๐‘ฅ [tanโˆ’1 ๐‘ฅ + 1 1+๐‘ฅ2]dx (x) ๐‘ฅ ๐‘ฅ+1 2 ๐‘’ ๐‘ฅ ๐‘‘๐‘ฅ (xi) 1 ๐‘™๐‘œ๐‘” ๐‘ฅ โˆ’ 1 ๐‘™๐‘œ๐‘” ๐‘ฅ 2 dx (xii) ๐‘’ ๐‘ฅ 1 ๐‘ฅ โˆ’ 1 ๐‘ฅ2 dx (xiii) ๐‘ฅ2+1 ๐‘ฅ+1 2 ๐‘’ ๐‘ฅ dx (xiv) ๐‘™๐‘œ๐‘” ๐‘™๐‘œ๐‘” ๐‘ฅ + 1 ๐‘™๐‘œ๐‘” ๐‘ฅ dx (xv) ๐‘’ ๐‘ฅ 1 ๐‘ฅ2 โˆ’ 1 ๐‘ฅ3 dx (xvi) ๐‘’ ๐‘ฅ 1+๐‘ ๐‘–๐‘›๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘๐‘œ๐‘ 2 ๐‘ฅ dx (xvii) 1โˆ’ ๐‘ฅ 1+ ๐‘ฅ dx (xviii) 2+๐‘ ๐‘–๐‘› 2๐‘ฅ 1+๐‘๐‘œ๐‘  2๐‘ฅ ๐‘’ ๐‘ฅ dx (xix) ๐‘ ๐‘–๐‘›8 ๐‘ฅโˆ’๐‘๐‘œ๐‘ 8 ๐‘ฅ 1โˆ’2 ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘๐‘œ๐‘ 2 ๐‘ฅ dx (xx) sin โˆ’1 ๐‘ฅโˆ’cos โˆ’1 ๐‘ฅ sin โˆ’1 ๐‘ฅ+cos โˆ’1 ๐‘ฅ dx (xxi) ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘ ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘‘๐‘ฅ (xxii) ๐‘‘๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ (5โˆ’4 ๐‘๐‘œ๐‘  ๐‘ฅ) (xxiii) ๐‘ฅ ๐‘ฅ3โˆ’1 ๐‘‘๐‘ฅ (xxiv) ๐‘ฅ sinโˆ’1 ๐‘ฅ ๐‘‘๐‘ฅ (xxv) ๐‘ฅ2 tanโˆ’1 ๐‘ฅ ๐‘‘๐‘ฅ (xxvi) 1โˆ’๐‘ฅ 1+๐‘ฅ dx (xxvii) 1โˆ’๐‘ฅ2 ๐‘ฅ(1โˆ’2๐‘ฅ) dx 4. Evaluate the following integrals : (i) ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ ๐œ‹ 2 0 ๐‘‘๐‘ฅ (ii) ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ+ ๐‘๐‘œ๐‘  ๐‘ฅ ๐œ‹ 2 0 ๐‘‘๐‘ฅ (iii) ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ 1+ ๐‘๐‘œ๐‘ 2 ๐‘ฅ ๐œ‹ 2 0 ๐‘‘๐‘ฅ
  • 5. (iv) ๐‘‘๐‘ฅ 1+ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐œ‹ 3 ๐œ‹ 6 (v) ๐‘ฅ + 2 5 โˆ’5 ๐‘‘๐‘ฅ (vi) ๐‘™๐‘œ๐‘” 1 + ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘‘๐‘ฅ ๐œ‹ 4 0 (vii) (2 ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ โˆ’ ๐‘™๐‘œ๐‘” ๐‘ ๐‘› 2๐‘ฅ)๐‘‘๐‘ฅ ๐œ‹ 2 0 (viii) ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐‘‘๐‘ฅ ๐œ‹ 4 โˆ’ ๐œ‹ 4 (ix) ๐‘‘ ๐‘ฅ 1+ ๐‘ก๐‘Ž๐‘› 3 ๐‘ฅ ๐œ‹ 2 0 (x) ๐‘ฅ 1+๐‘ ๐‘–๐‘› ๐‘ฅ ๐œ‹ 0 ๐‘‘๐‘ฅ (xi) ๐‘ ๐‘–๐‘› ๐‘ฅโˆ’๐‘๐‘œ๐‘  ๐‘ฅ 1+๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘๐‘œ๐‘  ๐‘ฅ ๐œ‹ 2 0 ๐‘‘๐‘ฅ (xii) ๐‘ฅ ๐‘ฅ+ ๐‘Žโˆ’๐‘ฅ ๐‘Ž 0 dx (xiii) ๐‘ฅ โˆ’ 1 4 0 ๐‘‘๐‘ฅ (xiv) ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ ๐‘ ๐‘–๐‘› 2 ๐‘ฅ ๐œ‹ 3 ๐œ‹ 6 dx (xv) ๐‘ฅ 1+๐‘๐‘œ๐‘ 2 ๐‘ฅ ๐œ‹ 0 ๐‘‘๐‘ฅ (xvi) ๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘ ๐‘’๐‘ ๐‘ฅ ๐‘๐‘œ๐‘ ๐‘’๐‘ ๐‘ฅ ๐œ‹ 0 ๐‘‘๐‘ฅ (xvii) ๐‘ ๐‘–๐‘› ๐‘ฅ ๐œ‹ 4 โˆ’ ๐œ‹ 4 ๐‘‘๐‘ฅ (xviii) ๐‘ฅ ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐‘ ๐‘’๐‘ ๐‘ฅ+๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐œ‹ 0 ๐‘‘๐‘ฅ (xix) ๐‘ฅ ๐‘๐‘œ๐‘  ๐œ‹ ๐‘ฅ 1 โˆ’1 ๐‘‘๐‘ฅ (xx) ๐‘’ ๐‘ฅ ๐‘ ๐‘–๐‘› 4๐‘ฅโˆ’4 1โˆ’๐‘๐‘œ๐‘  4๐‘ฅ ๐‘‘๐‘ฅ 6 Marks Questions : 5. Using properties of definite, evaluate. (i) ๐‘ฅ ๐‘‘๐‘ฅ 4โˆ’๐‘๐‘œ๐‘ 2 ๐‘ฅ ๐œ‹ 0 (ii) ๐‘’ ๐‘๐‘œ๐‘  ๐‘ฅ ๐‘’ ๐‘๐‘œ๐‘  ๐‘ฅ +๐‘’โˆ’๐‘๐‘œ๐‘  ๐‘ฅ ๐œ‹ 0 dx (iii) ๐‘ฅ ๐‘‘๐‘ฅ ๐‘Ž2 ๐‘๐‘œ๐‘ 2 ๐‘ฅ+ ๐‘2 ๐‘ ๐‘–๐‘›2 ๐‘ฅ ๐œ‹ 0 (iv) ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ 9+16 ๐‘ ๐‘–๐‘› 2๐‘ฅ ๐œ‹ 4 0 ๐‘‘๐‘ฅ (v) ๐‘™๐‘œ๐‘” ๐‘ ๐‘–๐‘› ๐‘ฅ ๐‘‘๐‘ฅ ๐œ‹ 2 0 (vi) sinโˆ’1 2๐‘ฅ 1+๐‘ฅ2 1 0 dx (vii) ๐‘‘ ๐‘ฅ 3+2 ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ ๐œ‹ 2 0 (viii) Show that ๐‘ก๐‘Ž๐‘› ๐‘ฅ + ๐‘๐‘œ๐‘ก ๐‘ฅ ๐‘‘๐‘ฅ = 2 ๐œ‹ ๐œ‹ 2 0 (ix) sinโˆ’1 ๐‘ฅ ๐‘Ž+๐‘ฅ ๐‘Ž 0 (x) ๐‘ ๐‘–๐‘› 2 ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ ๐œ‹ 2 0 dx (xi) ๐‘ฅ + ๐‘ฅ + 2 + ๐‘ฅ + 5 0 โˆ’5 dx (xii) ๐‘™๐‘œ๐‘” ๐‘ฅ 1โˆ’๐‘ฅ 1 0 dx (xiii) ๐‘ก๐‘Ž๐‘› ๐‘ฅ ๐œ‹ 4 0 ๐‘‘๐‘ฅ 6. Evaluate as the limit of sums : (i) 2๐‘ฅ2 โˆ’ 5 ๐‘‘๐‘ฅ 3 0 (ii) ๐‘ฅ2 + 5๐‘ฅ ๐‘‘๐‘ฅ 3 1 (iii) ๐‘ฅ2 + ๐‘ฅ + 1 ๐‘‘๐‘ฅ 2 0 (v) 3๐‘ฅ2 + 2๐‘ฅ ๐‘‘๐‘ฅ 3 1 (vi) ๐‘ฅ2 ๐‘ฅ ๐‘ ๐‘–๐‘› ๐‘ฅ+๐‘๐‘œ๐‘  ๐‘ฅ 2 dx