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Indefinite Integral Part-II
{Method of Substitution}
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Indefinite Integral Part-II
{Method of Substitution}
Change of Variable – The most important method of solving integration problem is the introduction of a new
variable i.e, the method of substitution or transformation.
The Substitution formula: – We suppose that a new variable 𝐀 is introduced into a function 𝐀(𝐀) by means of the
equation 𝐀 = 𝐀(𝐀) by the formula
𝐀𝐀
𝐀𝐀
. 𝐀𝐀 , where 𝐀 is any function of 𝐀.
Proof: Let 𝐀 = 𝐀 𝐀𝐀
𝐀𝐀
𝐀𝐀
But
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
Exercise−A
- Evaluate : 𝐀 𝐀
2
Solution; Let 𝐀 = 𝐀 𝐀
2
Taking 𝐀 = 𝐀
2
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
2
𝐀𝐀
𝐀𝐀
𝐀𝐀
2
𝐀𝐀
𝐀 1
𝐀𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
2
Now
2
𝐀𝐀
2
𝐀𝐀
𝐀𝐀
1
1+𝐀
2
𝐀𝐀
1+𝐀
2
Now
𝐀𝐀
1+𝐀
2
tan
− 1
𝐀+
𝐀𝐀
𝐀
𝐀𝐀
𝐀
1
2
−
1
2
𝐀
−
1
2
−
1
2
𝐀
1
2
1
2
1
2
𝐀 = 2 tan
− 1
u = tan
− 1
- Integrate:
𝐀
𝐀
1+𝐀 𝐀𝐀
tan 𝐀𝐀
𝐀
Solution: Let 𝐀 =
𝐀
𝐀
1+𝐀 𝐀𝐀
tan 𝐀𝐀
𝐀
Taking 𝐀 = 𝐀𝐀
𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
𝐀𝐀
𝐀 𝐀
𝐀
𝐀𝐀
𝐀
𝐀
𝐀
𝐀
𝐀
𝐀
Now
𝐀
𝐀
1+𝐀 𝐀𝐀
tan 𝐀𝐀
𝐀
𝐀𝐀
tan 𝐀
𝐀 = cot 𝐀 𝐀𝐀
𝐀 = log sin 𝐀 + 𝐀
𝐀 = log sin 𝐀𝐀
𝐀
𝐀
- Evaluate: sec
2
𝐀 tan 𝐀 𝐀𝐀
Solution: Let 𝐀 = sec
2
𝐀 tan 𝐀 𝐀𝐀
Taking
𝐀 = tan 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
tan 𝐀 = sec
2
𝐀𝐀 = sec
2
Now
𝐀 = sec
2
𝐀 tan 𝐀 𝐀𝐀
𝐀
1+
1+
𝐀
2
2
1
2
2
1
2
tan 𝐀
2
1
2
tan
2
- Integrate:
tan sin
− 1
𝐀
1 −𝐀
2
Solution: Let 𝐀 =
tan sin
− 1
𝐀
1 −𝐀
2
Taking 𝐀 = sin
− 1
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
sin
− 1
𝐀𝐀
𝐀𝐀
1
1 −𝐀
2
1
1 −𝐀
2
1
32
2
7
32
2
127
32
7
32
𝐀
2+
2+
127
32
7
32
𝐀
3
3
127
32
7
96
3
127
32
7
96
2
127
32
7
96
1
2
127
32
1
2
2
1
2
7
96
3
2
127
32
1
2
- Evaluate:
sin 2 𝐀
𝐀 sin
2
𝐀+𝐀 cos
2
𝐀
Solution: Let 𝐀 =
sin 2 𝐀
𝐀 sin
2
𝐀+𝐀 cos
2
𝐀
Take, 𝐀 = 𝐀 sin
2
𝐀 + 𝐀 cos
2
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
𝐀 sin
2
𝐀 + 𝐀 cos
2
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
sin
2
𝐀
𝐀𝐀
cos
2
𝐀𝐀
𝐀𝐀
= 𝐀 2 sin 𝐀 cos 𝐀 + 𝐀 2 cos 𝐀( − sin 𝐀)
𝐀𝐀
𝐀𝐀
= 2𝐀 sin 𝐀 cos 𝐀 − 2 𝐀 cos 𝐀 sin 𝐀
𝐀𝐀
𝐀𝐀
= 𝐀 − 𝐀 2 sin 𝐀 cos 𝐀
𝐀𝐀
𝐀𝐀
= 𝐀 − 𝐀 sin 2 𝐀
𝐀𝐀
𝐀−𝐀
sin 2 𝐀 𝐀𝐀
Now,
sin 2 𝐀
𝐀 sin
2
𝐀+𝐀 cos
2
𝐀
sin 2 𝐀 = 2 sin 𝐀 cos 𝐀
𝐀𝐀
𝐀−𝐀
𝐀
1
𝐀−𝐀
𝐀𝐀
𝐀
1
𝐀−𝐀
log 𝐀 + 𝐀
1
𝐀−𝐀
log 𝐀 sin
2
𝐀 + 𝐀 cos
2
𝐀 + 𝐀 Since, 𝐀 = 𝐀 sin
2
𝐀 + 𝐀 cos
2
- Integrate:
cos 𝐀 𝐀𝐀
𝐀+𝐀 sin 𝐀
2
Solution: Let 𝐀 =
cos 𝐀 𝐀𝐀
𝐀+𝐀 sin 𝐀
2
Take 𝐀 = 𝐀 + 𝐀 sin 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
sin 𝐀
𝐀𝐀
𝐀𝐀
= 𝐀 cos 𝐀
𝐀𝐀
𝐀
= cos 𝐀 𝐀𝐀
Now,
cos 𝐀 𝐀𝐀
𝐀+𝐀 sin 𝐀
2
𝐀𝐀
𝐀
𝐀
2
1
𝐀
𝐀𝐀
𝐀
2
1
𝐀
− 2
1
𝐀
𝐀
−2+
−2+
1
𝐀
𝐀
− 1
− 1
1
𝐀
1
𝐀
1
𝐀
1
𝐀+𝐀 sin 𝐀
- Integrate:
𝐀𝐀
sin 𝐀+cos 𝐀
Solution: Let 𝐀 =
𝐀𝐀
sin 𝐀+cos 𝐀
𝐀
4
. 𝐀
3
1 −𝐀
4
3
𝐀
4
−
𝐀𝐀
4
.
𝐀
3
1
4
(1−𝐀)
𝐀
3
1
4
1
𝐀
3
1
4
𝐀
𝐀
3
1
4
− 3
1
4
1
𝐀
2
1
4
− 3
1
4
− 2
1
4
𝐀
−3+
−3+
1
4
𝐀
−2+
−2+
1
4
𝐀
− 2
− 2
1
4
𝐀
− 1
− 1
1
8
1
𝐀
2
1
4
1
𝐀
1
8
1
1 −𝐀
4
2
1
4
1
1 −𝐀
4
4
- Integrate:
𝐀
2
sec
2
𝐀+tan
2
𝐀
1+𝐀
2
Solution: Let 𝐀 =
𝐀
2
sec
2
𝐀+tan
2
𝐀
1+𝐀
2
𝐀
2
1+tan
𝐀
𝐀
2
𝐀
1+𝐀
2
𝐀
2
2
tan
2
𝐀 +tan
2
𝐀
1+𝐀
2
𝐀
2
+tan
2
𝐀 1+𝐀
2
1+𝐀
2
𝐀
2
1+ 𝐀
2
tan
2
𝐀 1+𝐀
2
1+ 𝐀
2
1+𝐀
2
− 1
1+ 𝐀
2
𝐀𝐀 + tan
2
1+𝐀
2
1+ 𝐀
2
1
1+ 𝐀
2
𝐀𝐀+ sec
2
𝐀𝐀
1+𝐀
2
+ sec
2
𝐀 = 𝐀 − tan
− 1
𝐀 + tan 𝐀 − 𝐀 + 𝐀
𝐀 = tan 𝐀 − tan
− 1
4
4
sec
2
𝐀 = 1 + tan
2
tan
2
𝐀 = sec
2
- Integrate:
𝐀𝐀
𝐀 sin 𝐀+𝐀 cos 𝐀
Solution: Let 𝐀 =
𝐀𝐀
𝐀 sin 𝐀+𝐀 cos 𝐀
Take, 𝐀 = 𝐀 cos 𝐀 and 𝐀 = 𝐀 sin 𝐀
2
2
cos
2
𝐀 and 𝐀
2
2
sin
2
Thus,
2
2
2
cos
2
2
sin
2
2
2
2
cos
2
𝐀 + sin
2
𝐀 b
2
2
2
2
2
Now,
𝐀𝐀
𝐀 sin 𝐀+𝐀 cos 𝐀
𝐀𝐀
𝐀 cos 𝐀 sin 𝐀+𝐀 sin 𝐀 cos 𝐀
1
𝐀
𝐀𝐀
sin 𝐀 cos 𝐀+ cos 𝐀 sin 𝐀
1
𝐀
𝐀𝐀
sin 𝐀+𝐀
1
𝐀
cosec 𝐀 + 𝐀 𝐀𝐀
1
𝐀
log tan
𝐀+𝐀
2
1
𝐀
2
+𝐀
2
log tan
𝐀
2
𝐀
2
1
𝐀
2
+𝐀
2
log tan
𝐀
2
tan
− 1
𝐀
𝐀
2
1
𝐀
2
+𝐀
2
log tan
𝐀
2
1
2
tan
− 1
𝐀
𝐀
- Evaluate: 1 −
1
𝐀
2
tan 𝐀 +
1
𝐀
Solution: let 𝐀 = 1 −
1
𝐀
2
tan 𝐀 +
1
𝐀
2
2
And we have
𝐀 = 𝐀 cos 𝐀 and 𝐀 = 𝐀 sin 𝐀
𝐀
𝐀
sin 𝐀
cos 𝐀
= tan 𝐀
θ = tan
− 1
𝐀
𝐀
𝐀 = tan
2
𝐀 1 + tan
2
𝐀 sec
2
Take
𝐀 = tan 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
tan 𝐀
𝐀𝐀
𝐀𝐀
= sec
2
𝐀𝐀 = sec
2
Now,
𝐀 = tan
2
𝐀 1 + tan
2
𝐀 sec
2
2
2
2
4
2
4
𝐀
2+
2+
𝐀
4+
4+
𝐀
3
3
𝐀
5
5
tan 𝐀
3
3
tan 𝐀
6
6
tan
3
𝐀
3
tan
5
𝐀
5
- Integrate:
sec
4
𝐀
tan 𝐀
Solution: Let 𝐀 =
sec
4
𝐀
tan 𝐀
sec
2
𝐀 sec
2
𝐀
tan 𝐀
1+tan
2
𝐀 sec
2
𝐀
tan 𝐀
Take, 𝐀 = tan 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
tan 𝐀
𝐀𝐀
𝐀𝐀
= sec
2
𝐀𝐀 = sec
2
1+𝐀
2
𝐀
1
𝐀
𝐀
2
𝐀
1
𝐀
1
2
𝐀
2
𝐀
1
2
−
1
2 𝐀𝐀 + 𝐀
2 −
1
2 𝐀𝐀
−
1
2 𝐀𝐀 + 𝐀
3
2
𝐀
−
1
2
−
1
2
𝐀
3
2
3
2
𝐀
1
2
1
2
𝐀
5
2
5
2
1
2
2
5
5
2
2
5
5
2
𝐀 = 2 tan 𝐀 +
2
5
tan 𝐀
5
2
- Integrate:
sec
2
𝐀
1+tan 𝐀
2
Solution: Let 𝐀 =
sec
2
𝐀
1+tan 𝐀
2
Take 𝐀 = 1 + tan 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
1 + tan 𝐀
𝐀𝐀
𝐀𝐀
= sec
2
𝐀𝐀 = sec
2
Now,
sec
2
𝐀
1+tan 𝐀
2
𝐀𝐀
𝐀
2
− 2
𝐀
−2+
−2+
𝐀
− 1
− 1
− 1
Take, 𝐀 = 2 + 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
1
2 𝐀
𝐀𝐀
𝐀
𝐀𝐀
𝐀 2+ 𝐀
3
2 𝐀𝐀
𝐀
3
− 3
𝐀
−3+
−3+
𝐀
− 2
− 2
− 2
1
𝐀
2
1
2+ 𝐀
2
- Integrate: tan 𝐀 tan 2 𝐀 tan 3 𝐀 𝐀𝐀
Solution: Let 𝐀 = tan 𝐀 tan 2 𝐀 tan 3 𝐀 𝐀𝐀
We know that, tan 3 𝐀 = tan (𝐀 + 2𝐀)
tan 3 𝐀 =
tan 𝐀+tan 2 𝐀
1 −tan 𝐀 tan 2 𝐀
tan 3 𝐀 1 − tan 𝐀 tan 2 𝐀 = tan 𝐀 + tan 2 𝐀
tan 3 𝐀 − tan 3 𝐀 tan 𝐀 tan 2 𝐀 = tan 𝐀 + tan 2 𝐀
tan 3 𝐀 − tan 𝐀 + tan 2 𝐀 = tan 3 𝐀 tan 𝐀 tan 2 𝐀
tan 𝐀 tan 2 𝐀 tan 3 𝐀 = tan 3 𝐀 − tan 𝐀 + tan 2 𝐀
Now,
𝐀 = tan 𝐀 tan 2 𝐀 tan 3 𝐀 𝐀𝐀
𝐀 = tan 3 𝐀 − tan 𝐀 + tan 2 𝐀 𝐀𝐀
𝐀 = tan 3 𝐀 𝐀𝐀− tan 𝐀𝐀𝐀 + tan 2 𝐀 𝐀𝐀
1
3
log sec 3 𝐀 − log sec 𝐀 +
1
2
log sec 2 𝐀 + 𝐀
- Integrate:
sin
5
𝐀
cos 𝐀
Solution: Let 𝐀 =
sin
5
𝐀
cos 𝐀
sin
4
𝐀
cos 𝐀
sin 𝐀 𝐀𝐀
sin
2
𝐀
2
cos 𝐀
sin 𝐀 𝐀𝐀
1 −cos
2
𝐀
2
cos 𝐀
sin 𝐀 𝐀𝐀
Take, 𝐀 = cos 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
cos 𝐀
𝐀𝐀
𝐀𝐀
=− sin 𝐀
−𝐀𝐀 = sin 𝐀 𝐀𝐀
Now,
1 −cos
2
𝐀
2
cos 𝐀
sin 𝐀 𝐀𝐀
1 −𝐀
2
2
𝐀
1 − 2 𝐀
2
+𝐀
4
𝐀
1
2
1
𝐀
1
2
𝐀
2
𝐀
1
𝐀
𝐀
4
𝐀
1
2
−
1
2 𝐀𝐀 + 2 𝐀
2 −
1
2 𝐀𝐀 − 𝐀
4 −
1
2 𝐀𝐀
−
1
2 𝐀𝐀 + 2 𝐀
3
2 𝐀𝐀 − 𝐀
7
2 𝐀𝐀
𝐀
−
1
2
−
1
2
𝐀
3
2
3
2
𝐀
7
2
7
2
𝐀
1
2
1
2
𝐀
5
2
5
2
𝐀
9
2
9
2
2
5
5
2
−
2
9
9
2
𝐀 = 2 cos 𝐀 +
4
5
cos 𝐀
5
2 −
2
9
cos 𝐀
9
2
𝐀 = cos 𝐀
Solution: Let
cos (𝐀𝐀+𝐀)
sin 𝐀𝐀 + 𝐀 𝐀𝐀
Take, 𝐀 = cos (𝐀𝐀 + 𝐀)
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
cos (𝐀𝐀 + 𝐀)
𝐀𝐀
𝐀𝐀
=− sin (𝐀𝐀 + 𝐀)
𝐀
𝐀𝐀
𝐀𝐀
𝐀𝐀
=− sin (𝐀𝐀 + 𝐀) 𝐀
𝐀𝐀
𝐀
= sin (𝐀𝐀 + 𝐀) 𝐀𝐀
Now,
cos (𝐀𝐀+𝐀)
sin 𝐀𝐀 + 𝐀 𝐀𝐀
𝐀
𝐀𝐀
𝐀
1
𝐀
𝐀
1
𝐀
𝐀
1
𝐀
cos (𝐀𝐀+𝐀)
- 𝐀 Since, 𝐀 = cos (𝐀𝐀 + 𝐀)
- Evaluate :
1
𝐀
2
1
𝐀
Solution: Let 𝐀 =
1
𝐀
2
1
𝐀
Take, 𝐀 = 2 −
1
𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
1
𝐀
𝐀𝐀
𝐀𝐀
1
𝐀
2
1
𝐀
2
Now,
1
𝐀
2
1
𝐀
1
2 𝐀𝐀
𝐀
1
2
1
2
𝐀
3
2
3
2
2
3
3
2
2
3
1
𝐀
3
2
1
𝐀
- Integrate :
𝐀𝐀
𝐀 𝐀
2
−𝐀
2
Solution: Let 𝐀 =
𝐀𝐀
𝐀 𝐀
2
−𝐀
2
Take, 𝐀 = 𝐀 sec 𝐀
𝐀𝐀
𝐀𝐀
𝐀
𝐀𝐀
sec 𝐀
𝐀𝐀
𝐀𝐀
= 𝐀 sec 𝐀 tan 𝐀
𝐀𝐀 = 𝐀 sec 𝐀 tan 𝐀 𝐀𝐀
Now,
𝐀𝐀
𝐀 𝐀
2
−𝐀
2
𝐀 sec 𝐀 tan 𝐀 𝐀𝐀
𝐀 sec 𝐀 𝐀
2
sec
2
𝐀−𝐀
2
tan 𝐀 𝐀𝐀
𝐀
2
sec
2
𝐀− 1
tan 𝐀 𝐀𝐀
𝐀
2
tan
2
𝐀
tan 𝐀 𝐀𝐀
𝐀 tan 𝐀
𝐀𝐀
𝐀
1
𝐀
1
𝐀
1
𝐀
sec
− 1
𝐀
𝐀
𝐀 = 𝐀 sec 𝐀
𝐀
𝐀
= sec 𝐀
𝐀 = sec
− 1
𝐀
𝐀
