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Indefinite Integration, Part-i
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Exercise-A
Integrate the following:
𝐀
2
2
𝐀
3
Solution: Let 𝐀 =
𝐀
2
2
𝐀
3
𝐀
4
+2𝐀
2
𝐀
3
𝐀
4
𝐀
3
2 𝐀
2
𝐀
3
1
𝐀
3
2
𝐀
1
𝐀
3
1
𝐀
1
𝐀
3
𦐀𝐀
𝐀
− 3
𝐀
2
2
𝐀
−3+
−3+
𝐀
2
2
𝐀
− 2
− 2
𝐀
2
2
1
2 𝐀
2
𦠀
4 𝐀
+𦠀
6 𝐀
𦠀
𝐀
+𦠀
−𝐀
Solution: Let
𦠀
4 𝐀
+𦠀
6 𝐀
𦠀
𝐀
+𦠀
−𝐀
𦠀
4 𝐀
1+𦠀
2 𝐀
𦠀
𝐀
1
𦠀
𝐀
𦠀
4 𝐀
1+𦠀
2 𝐀
𦠀
𝐀
𦠀
𝐀
𦠀
𝐀
𦠀
4 𝐀
1+𦠀
2 𝐀
𦠀
2 𝐀
𦠀
𝐀
𦠀
4 𝐀
1+𦠀
2 𝐀
1+𦠀
2 𝐀
𦠀
𝐀
4 𝐀
2 𝐀
𦠀
𝐀
1+𦠀
2 𝐀
Using identity
2
2
2
−𝐀
𝐀
8
1+𝐀
− 4
1 −𝐀
2
𝐀
Solution:
Let 𝐀 =
8
1+𝐀
− 4
1 −𝐀
2
𝐀
𝐀
−𝐀
2
𝐀
𝐀
2
𝐀
−𝐀
2
𝐀
3 𝐀
2
𝐀
− 2 𝐀
2
𝐀
3 𝐀−𝐀
− 2 𝐀−𝐀
2 𝐀
− 3 𝐀
2 𝐀
− 3 𝐀
2 𝐀
− 3 𝐀
Now,
1
2 log 2
𦐀𝐀
𝐀
1
− 3 log 2
𦐀𧀀
𧀀
Similarly,
Taking, 𧀀 = 2
− 3 𝐀
log 𧀀 = log 2
− 3 𝐀
log 𧀀 =− 3 𝐀 log 2
And
1
𧀀
𦐀𧀀
𦐀𝐀
=− 3 log 2
1
− 3 log 2
𦐀𧀀
𧀀
Taking, 2
2 𝐀
log 2
2 𝐀
= log 𝐀
2 𝐀 log 2 = log 𝐀
differentiating w.r.t 𝐀
2 log 2
𦐀𝐀
𦐀𝐀
𦐀
𦐀𝐀
log 𝐀
2 log 2 =
1
𝐀
𦐀𝐀
𦐀𝐀
1
2 log 2
𦐀𝐀
𝐀
8
2 log 2
𦐀𝐀
𝐀
4
3 log 2
𦐀𧀀
𧀀
8
2 log 2
4
3 log 2
8
2 log 2
4
3 log 2
2 𝐀
log 2
− 3 𝐀
3 log 2
2
2
2 𝐀
log 2
2
2
− 3 𝐀
3 log 2
2
2 𝐀+
log 2
2
2 − 3 𝐀
3 log 2
1
log 2
2 𝐀+
2
2 − 2 𝐀
3
sin 2 𝐀 cos 4 𝐀 𦐀𝐀
Solution:
Let 𝐀 = sin 2 𝐀 cos 4 𝐀 𦐀𝐀
We know that,
sin 𝐀 + 𝐀 = sin 𝐀 cos 𝐀 + cos 𝐀 sin 𝐀 And sin 𝐀 − 𝐀 = sin 𝐀 cos 𝐀 − cos 𝐀 sin 𝐀
sin 4 𝐀 + 2𝐀 = sin 4 𝐀 cos 2 𝐀 + cos 4 𝐀 sin 2 𝐀 ---------- (i)
And sin 4 𝐀 − 2 𝐀 = sin 4 𝐀 cos 2 𝐀 − cos 4 𝐀 sin 2 𝐀 ---------- (ii)
Subtracting (ii) from (i)
sin 4 𝐀 + 2𝐀 = sin 4 𝐀 cos 2 𝐀 + cos 4 𝐀 sin 2 𝐀 ---------- (i)
sin 4 𝐀 − 2 𝐀 = sin 4 𝐀 cos 2 𝐀 − cos 4 𝐀 sin 2 𝐀 -------- (ii)
sin 4 𝐀 + 2𝐀 − sin 4 𝐀 − 2 𝐀 = 2 cos 4 𝐀 sin 2 𝐀
Or sin 4 𝐀 + 2𝐀 − sin 4 𝐀 − 2 𝐀 = 2 sin 2 𝐀 cos 4 𝐀
sin 4 𝐀+2𝐀 −sin 4 𝐀− 2 𝐀
2
= sin 2 𝐀
cos 4 𝐀
Now,
𝐀 = sin 2 𝐀 cos 4 𝐀 𦐀𝐀
sin 4 𝐀+2𝐀 −sin 4 𝐀− 2 𝐀
2
1+tan
2
𝐀
1+cot
2
𝐀
Solution:
Let 𝐀 =
1+tan
2
𝐀
1+cot
2
𝐀
1+tan
2
𝐀
1+
1
tan
2
𝐀
1+tan
2
𝐀
tan
2
𝐀 +
tan
2
𝐀
1+tan
2
𝐀
1+tan
2
𝐀
tan
2
𝐀
𝐀 = 1 + tan
2
tan
2
𝐀
1+tan
2
𝐀
𝐀 = tan
2
𝐀 = 1 − sec
2
𝐀 = 1 𦐀𝐀 − sec
2
𝐀 = 𝐀 − tan 𝐀 + 𝐀
1
1+cos 𝐀
Solution:
Let 𝐀 =
1
1+cos 𝐀
1
1+cos 𝐀
1 −cos 𝐀
1 −cos 𝐀
1 −cos 𝐀
1 −cos
2
𝐀
1 −cos 𝐀
sin
2
𝐀
1
sin
2
𝐀
cos 𝐀
sin
2
𝐀
𝐀 = cosec
2
cos 𝐀
sin 𝐀 sin 𝐀
𝐀 = cosec
2
cos 𝐀
sin 𝐀
1
sin 𝐀
cot 𝐀 =
tan 𝐀
𝐀 = cosec
2
𝐀 − cot 𝐀 cosec 𝐀 𦐀𝐀
𝐀 = cosec
2
𝐀 𦐀𝐀 − cot 𝐀 cosec 𝐀 𦐀𝐀
𝐀 =− cot 𝐀 − ( − cosec 𝐀 ) + 𝐀
𝐀 =− cot 𝐀 + cosec 𝐀 + 𝐀
𝐀 = cosec 𝐀 – cot 𝐀 + 𝐀
- 1 + sin 2 𝐀 𦐀𝐀
Solution:
Let 𝐀 = 1 + sin 2 𝐀 𦐀𝐀
𝐀 = 1 + 2 sin 𝐀 cos 𝐀 𦐀𝐀
𝐀 = sin
2
𝐀 + cos
2
𝐀 + 2 sin 𝐀 cos 𝐀
𝐀 = sin 𝐀 + cos 𝐀
2
𝐀 = ( sin 𝐀 + cos 𝐀 ) 𦐀𝐀
𝐀 = sin 𝐀 𦐀𝐀 + cos 𝐀 𦐀𝐀
𝐀 =− sin 𝐀 + cos 𝐀 + 𝐀
- 1 + cos 𝐀𦐀𝐀
Solution: Let 𝐀 = 1 +
cos 𝐀 𦐀𝐀
𝐀 = 2 cos
2
𝐀
2
𝐀 = 2 cos
𝐀
2
sin
𝐀
2
1
2
𝐀 = 2 2 sin
𝐀
2
- sin
2
𝐀 cos 4 𝐀 𦐀𝐀
sin 𝐀 + 𝐀 = sin 𝐀 cos 𝐀 + cos 𝐀 sin 𝐀
sin 2 𝐀 = 2 sin 𝐀 cos 𝐀
And
sin
2
𝐀 + cos
2
cos
= cos
cos
− sin
sin
cos 𝐀 = cos
2
𝐀
2
− sin
2
𝐀
2
cos 𝐀 = cos
2
𝐀
2
− 1 − cos
2
𝐀
2
cos 𝐀 = cos
2
𝐀
2
− 1 + cos
2
𝐀
2
cos 𝐀 = 2 cos
2
𝐀
2
1 + cos 𝐀 = 2 cos
2
𝐀
2
𝐀
𝐀
+𝐀
2 𝐀
+𝐀
3 𝐀
𝐀
4 𝐀
Solution:
Let
𝐀
𝐀
+𝐀
2 𝐀
+𝐀
3 𝐀
𝐀
4 𝐀
𝐀
𝐀
𝐀
4 𝐀
𝐀
2 𝐀
𝐀
4 𝐀
𝐀
3 𝐀
𝐀
4 𝐀
𝐀− 4 𝐀
2 𝐀− 4 𝐀
3 𝐀− 4 𝐀
− 3 𝐀
− 2 𝐀
−𝐀
− 3 𝐀
− 2 𝐀
−𝐀
1
− 3 log 𝐀
𦐀𝐀
𝐀
1
− 2 log 𝐀
𦐀𧀀
𧀀
1
−log 𝐀
𦐀𝐀
𝐀
1
3 log 𝐀
1
2 log 𝐀
1
log 𝐀
1
3 log 𝐀
− 3 𝐀
1
2 log 𝐀
− 2 𝐀
1
log 𝐀
−𝐀
1
log 𝐀
𝐀
− 3 𝐀
3
𝐀
− 2 𝐀
2
−𝐀
sin 𝐀 + cosec 𝐀
tan 𝐀
Solution:
Let 𝐀 =
sin 𝐀 + cosec 𝐀
tan 𝐀
Similarly,
− 2 𝐀
log 𧀀 = log 𝐀
− 2 𝐀
log 𧀀 =− 2 𝐀 log 𝐀
1
𧀀
𦐀𧀀
𦐀𝐀
=− 2 log 𝐀
1
− 2 log 𝐀
𦐀𝐀
𧀀
And,
−𝐀
log 𝐀 = log 𝐀
− 2 𝐀
log 𝐀 =− 𝐀 log 𝐀
1
𝐀
𦐀𝐀
𦐀𝐀
=− log 𝐀
1
−log 𝐀
𦐀𝐀
𝐀
Let 𝐀 = 𝐀
− 3 𝐀
log 𝐀 = log 𝐀
− 3 𝐀
log 𝐀 =− 3 𝐀 log 𝐀
1
𝐀
𦐀𝐀
𦐀𝐀
=− 3 log 𝐀
1
− 3 log 𝐀
𦐀𝐀
𝐀
sin 𝐀 +
tan 𝐀
cosec 𝐀
tan 𝐀
sin 𝐀 +
sin 𝐀
cos 𝐀
cosec 𝐀
sin 𝐀
cos 𝐀
𝐀 = sin 𝐀
cos 𝐀
sin 𝐀
+ cosec 𝐀
cos 𝐀
𝐀𨀀𝐀𝐀
𝐀 = cos 𝐀 + cosec 𝐀 cot 𝐀 𦐀𝐀
𝐀 = cos 𝐀 𦐀𝐀 + cosec 𝐀cot 𝐀 𦐀𝐀
𝐀 = sin 𝐀 − cosec 𝐀 + 𝐀
cos 𝐀−cos 2 𝐀
1 −cos 𝐀
Solution:
Let 𝐀 =
cos 𝐀−cos 2 𝐀
1 −cos 𝐀
cos 𝐀−cos 2 𝐀
1 −cos 𝐀
cos 𝐀− 2cos
2
𝐀− 1
1 −cos 𝐀
cos 𝐀−2cos
2
𝐀+
1 −cos 𝐀
cos 𝐀−cos
2
𝐀−cos
2
𝐀+
1 −cos 𝐀
cos 𝐀 1 −cos 𝐀 −cos
2
𝐀+
1 −cos 𝐀
cos 𝐀 1 −cos 𝐀 +1−cos
2
𝐀
1 −cos 𝐀
cos 𝐀 1 −cos 𝐀 + 1 −cos
2
𝐀
1 −cos 𝐀
cos 𝐀
1 −cos 𝐀 + 1 −cos 𝐀 1+cos 𝐀
1 −cos 𝐀
1 −cos 𝐀 cos 𝐀 +1+cos 𝐀
1 −cos 𝐀
𝐀 = cos 𝐀 + 1 + cos 𝐀 𦐀𝐀
𝐀 = 2cos 𝐀 + 1 𦐀𝐀
cos 2 𝐀 = cos 𝐀 cos 𝐀 − sin 𝐀 sin 𝐀
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = cos
2
𝐀 − 1 − cos
2
cos 2 𝐀 = cos
2
𝐀 − 1 + cos
2
cos 2 𝐀 = 2cos
2
cos
2
= (1 + cos 𝐀)(1 − cos 𝐀)
𝐀 = sin
𝐀
2
+cos
𝐀
2
𝐀 = sin
𝐀
2
𦐀𝐀 + cos
𝐀
2
−cos
𝐀
2
1
2
sin
𝐀
2
1
2
𝐀 =− 2 cos
𝐀
2
𝐀
2
𝐀 = 2 sin
𝐀
2
− cos
𝐀
2
cos
4
𝐀−sin
4
𝐀
1+cos 4 𝐀
Solution: Let 𝐀 =
cos
4
𝐀−sin
4
𝐀
1+cos 4 𝐀
cos
2
𝐀−sin
2
𝐀 cos
2
𝐀+sin
2
𝐀
1+cos 4 𝐀
cos
2
𝐀−sin
2
𝐀 cos
2
𝐀+sin
2
𝐀
2cos
2
2 𝐀
cos
2
𝐀−sin
2
𝐀 cos
2
𝐀+sin
2
𝐀
2 cos 2 𝐀
1
2
cos
2
𝐀−sin
2
𝐀 sin
2
𝐀+cos
2
𝐀
cos 2 𝐀
1
2
cos
2
𝐀−sin
2
𝐀
cos 2 𝐀
1
2
cos 2 𝐀
cos 2 𝐀
1
2
1
2
- tan
2
𝐀 − cot
2
Solution: Let 𝐀 = tan
2
− cot
2
𝐀 = tan
2
𝐀 𦐀𝐀 − cot
2
cos 4 𝐀 = cos
2
2 𝐀 − sin
2
cos 4 𝐀 = cos
2
2 𝐀 − 1 − cos
2
cos 4 𝐀 = cos
2
2 𝐀 − 1 + cos
2
cos 4 𝐀 = 2cos
2
1 + cos 4 𝐀 = 2cos
2
sin
2
𝐀 + cos
2
And cos 2 𝐀 = cos
2
𝐀 − sin
2
𝐀 = sec
2
𝐀 − 1 𦐀𝐀 − cosec
2
𝐀 = sec
2
𝐀 𦐀𝐀 − 1 𦐀𝐀− cosec
2
𝐀 = tan 𝐀 − 𝐀 − − cot 𝐀 + 𝐀 + 𝐀
𝐀 = tan 𝐀 + cot 𝐀 + 𝐀
cos 2 𝐀
sin
2
𝐀 cos
2
𝐀
Solution: Let 𝐀 =
cos 2 𝐀
sin
2
𝐀 cos
2
𝐀
cos
2
𝐀−sin
2
𝐀
sin
2
𝐀 cos
2
𝐀
cos
2
𝐀
sin
2
𝐀 cos
2
𝐀
sin
2
𝐀
sin
2
𝐀 cos
2
𝐀
1
sin
2
𝐀
1
cos
2
𝐀
𝐀 = cosec
2
𝐀 − sec
2
𝐀 = cosec
2
𝐀 𦐀𝐀 − sec
2
𝐀 =− cot 𝐀 − tan 𝐀 + 𝐀
𝐀 sin
2
𝐀+𝐀 cos
2
𝐀
sin
2
𝐀 cos
2
𝐀
Solution: Let 𝐀 =
𝐀 sin
2
𝐀+𝐀 cos
2
𝐀
sin
2
𝐀 cos
2
𝐀
𝐀 sin
2
𝐀
sin
2
𝐀 cos
2
𝐀
𝐀 cos
2
𝐀
sin
2
𝐀 cos
2
𝐀
𝐀
cos
2
𝐀
𝐀
sin
2
𝐀
𝐀 = 𝐀 sec
2
𝐀 + 𝐀 cosec
2
𝐀 = 𝐀 sec
2
𝐀 𦐀𝐀 + 𝐀 cosec
2
𝐀 = 𝐀 tan 𝐀 − 𝐀 cot 𝐀 + 𝐀
1 −cos 2 𝐀
1+cos 2 𝐀
Solution: Let 𝐀 =
1 −cos 2 𝐀
1+cos 2 𝐀
1 −cos 2 𝐀
1+cos 2 𝐀
2 sin
2
𝐀
2 cos
2
𝐀
sin
2
𝐀
cos
2
𝐀
𝐀 = tan
2
𝐀 = sec
2
𝐀 = sec
2
𝐀 = tan 𝐀 – 𝐀 + 𝐀
sec
2
𝐀
cosec
2
𝐀
Solution: Let 𝐀 =
sec
2
𝐀
cosec
2
𝐀
1
cos
2
𝐀
1
sin
2
𝐀
1
cos
2
𝐀
sin
2
𝐀
1
sin
2
𝐀
cos
2
𝐀
𝐀 = tan
2
cos 2 𝐀 =
cos
2
sin
2
cos 2 𝐀 = 1 − sin
2
𝐀 − sin
2
cos 2 𝐀 = 1 − 2 sin
2
2 sin
2
𝐀 = 1 − cos 2 𝐀
1 − cos 2 𝐀 = 2 sin
2
Similarly,
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = cos
2
𝐀 − 1 − cos
2
cos 2 𝐀 = cos
2
𝐀 − 1 + cos
2
cos 2 𝐀 = 2 cos
2
1 + cos 2 𝐀 = 2 cos
2
𝐀 = sec
2
𝐀 = sec
2
𝐀 = tan 𝐀 − 𝐀 + 𝐀
sec 2 𝐀 − 1
sec 2 𝐀 +
Solution: Let 𝐀 =
sec 2 𝐀 − 1
sec 2 𝐀 +
1
cos 2 𝐀
− 1
1
cos 2 𝐀
1 −cos 2 𝐀
cos 2 𝐀
1+cos 2 𝐀
cos 2 𝐀
1 −cos 2 𝐀
cos 2 𝐀
cos 2 𝐀
1+cos 2 𝐀
1 −cos 2 𝐀
1+cos 2 𝐀
2 sin
2
𝐀
2 cos
2
𝐀
sin
2
𝐀
cos
2
𝐀
𝐀 = tan
2
𝐀 = sec
2
𝐀 = sec
2
𝐀 = tan 𝐀 – 𝐀 + 𝐀
sec
3
𝐀 + cosec
3
𝐀
sec 𝐀 cosec 𝐀
Solution: Let
sec
3
𝐀 + cosec
3
𝐀
sec 𝐀 cosec 𝐀
sec
3
𝐀
sec 𝐀 cosec 𝐀
cosec
3
𝐀
sec 𝐀 cosec 𝐀
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = 1 − sin
2
𝐀 − sin
2
cos 2 𝐀 = 1 − 2 sin
2
2 sin
2
𝐀 = 1 − cos 2 𝐀
cos 2 𝐀 = 2 sin
2
Similarly,
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = cos
2
𝐀 − 1 − cos
2
cos 2 𝐀 = cos
2
𝐀 − 1 + cos
2
cos 2 𝐀 = 2 cos
2
1 + cos 2 𝐀 = 2 cos
2
1
4
1
4
sin 2 𝐀 +
1
8
1
8
sin 4 𝐀
4
3
8
1
4
sin 2 𝐀 +
1
32
sin 4 𝐀 + 𝐀
cos
4
𝐀
sin
2
𝐀
Solution: Let 𝐀 =
cos
4
𝐀
sin
2
𝐀
cos
2
𝐀
2
sin
2
𝐀
1 −sin
2
𝐀
2
sin
2
𝐀
1 − 2 sin
2
𝐀+ sin
4
𝐀
sin
2
𝐀
1
sin
2
𝐀
2 sin
2
sin
2
𝐀
sin
4
𝐀
sin
2
𝐀
𝐀 = cosec
2
𝐀 − 2 + sin
2
𝐀 = cosec
2
𝐀 𦐀𝐀− 2 𦐀𝐀 + sin
2
𝐀 = cosec
2
1 − cos 2 𝐀
2
𝐀 = cosec
2
1
2
1 − cos 2 𝐀 𦐀𝐀
𝐀 = cosec
2
1
2
1
2
cos 2 𝐀𦐀𝐀
𝐀 =− cot
1
2
1
2
sin 2 𝐀
2
𝐀 =− cot 𝐀 +
1
2
1
4
sin 2 𝐀 + 𝐀
𝐀 =− cot 𝐀 −
3
2
1
4
sin 2 𝐀 + 𝐀
cos 2 𝐀−cos 2 𝐀
cos 𝐀−cos 𝐀
Solution: Let 𝐀 =
cos 2 𝐀−cos 2 𝐀
cos 𝐀−cos 𝐀
2 cos
2
𝐀− 1 − 2 cos
2
𝐀− 1
cos 𝐀−cos 𝐀
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = 1 − sin
2
𝐀 − sin
2
cos 2 𝐀 = 1 − 2 sin
2
2 sin
2
𝐀 = 1 − cos 2 𝐀
sin
2
1 − cos 2 𝐀
2
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = cos
2
cos
2
cos 2 𝐀 = cos
2
𝐀 − 1 + cos
2
cos 2 𝐀 = cos
2
𝐀 − sin
2
cos 2 𝐀 = cos
2
𝐀 − 1 − cos
2
cos 2 𝐀 = cos
2
𝐀 − 1 + cos
2
2 cos
2
𝐀− 1 − 2 cos
2
𝐀+ 1
cos 𝐀−cos 𝐀
2 cos
2
𝐀− 2 cos
2
𝐀
cos 𝐀−cos 𝐀
2 cos
2
𝐀
− cos
2
𝐀
cos 𝐀−cos 𝐀
cos
2
𝐀− cos
2
𝐀
cos 𝐀−cos 𝐀
cos 𝐀−cos 𝐀 cos 𝐀+cos 𝐀
cos 𝐀−cos 𝐀
𝐀 = 2 cos 𝐀 + cos 𝐀 𦐀𝐀
𝐀 = 2 cos 𝐀𦐀𝐀 + 2 cos 𝐀 𦐀𝐀
𝐀 = 2 sin 𝐀 + 2 cos 𝐀 𝐀 + 𝐀
𝐀 = 2 sin 𝐀 + 2𝐀 cos 𝐀 + 𝐀
sin 𝐀 cos 𝐀+cos
2
𝐀
1+sin 2 𝐀
Solution: Let 𝐀 =
sin 𝐀 cos 𝐀+cos
2
𝐀
1+sin 2 𝐀
sin 𝐀 cos
𝐀+ cos
2
𝐀
1+2 sin 𝐀 cos 𝐀
sin 𝐀 cos 𝐀+cos
2
𝐀
1+2 sin 𝐀 cos 𝐀
sin 𝐀 cos 𝐀+cos
2
𝐀
sin
2
𝐀+cos
2
𝐀+2 sin 𝐀 cos 𝐀
sin 𝐀 cos 𝐀+cos
2
𝐀
sin 𝐀+cos 𝐀
2
sin 𝐀 cos 𝐀+cos
2
𝐀
sin 𝐀+cos 𝐀
cos 𝐀 sin 𝐀+cos 𝐀
sin 𝐀+cos 𝐀
𝐀 = cos 𝐀𦐀𝐀
𝐀 = sin 𝐀 + 𝐀
cos 𝐀 = 𝐀𤀀𝐀𝐀𝐀𝐀𝐀𝐀
sin 2 𝐀 = sin 𝐀 cos 𝐀 + cos 𝐀 sin 𝐀
sin 2 𝐀 = 2 sin 𝐀 cos 𝐀
And,
sin
2
𝐀 + cos
2
