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CBSE NCERT Solutions for Class 12 Maths Chapter 07
Back of Chapter Questions
Exercise 7.
1. Find an anti-derivative (or integral) of the function sin 2𝑥 by the method of inspection.
[2 Marks]
Solution:
Anti-derivative of sin 2𝑥 is a function of 𝑥 whose derivative is sin 2𝑥.
As we know that,
𝑑 𝑑𝑥 (cos 2𝑥) = −2 sin 2𝑥^ [1 Mark]
sin 2𝑥 = −
(cos 2𝑥)
∴ sin 2𝑥 =
𝑑 𝑑𝑥 (−^
1 2 cos 2𝑥)
Hence, the anti – derivative of sin 2𝑥 is −
1 2 cos 2𝑥^ [1 Mark]
2. Find an anti-derivative (or integral) of the function cos 3𝑥 by the method of inspection.
[2 Marks]
Solution:
Anti-derivative of cos 3𝑥 is a function of 𝑥 whose derivative is cos 3𝑥.
As we know that,
𝑑 𝑑𝑥 (sin 3𝑥) = 3 cos 3𝑥^ [1 Mark]
⇒ cos 3𝑥 =
(sin 3𝑥)
∴ cos 3𝑥 = (^) 𝑑𝑥𝑑 (^13 sin 3𝑥)
Hence, the anti-derivative of cos 3𝑥 is 13 sin 3𝑥 [1 Mark]
3. Find an anti-derivative (or integral) of the function 𝑒2𝑥^ by the method of inspection. [2 Marks]
Solution:
Anti-derivative of 𝑒2𝑥^ is the function of 𝑥 whose derivative is 𝑒2𝑥.
As we know that,
𝑑 𝑑𝑥 (𝑒
2𝑥) = 2𝑒2𝑥 (^) [1 Mark]
⇒ 𝑒2𝑥^ =
∴ 𝑒2𝑥^ =
1 2
𝑑 𝑑𝑥 (𝑒
Hence, the anti-derivative of 𝑒2𝑥^ is 12 𝑒2𝑥^ [1 Mark]
4. Find an anti-derivative (or integral) of the function (𝑎𝑥 + 𝑏)^2 by the method of inspection.
[2 Marks]
Solution:
Anti-derivative of (𝑎𝑥 + 𝑏)^2 is the function of 𝑥 whose derivative is (𝑎𝑥 + 𝑏)^2.
As we know that,
𝑑 𝑑𝑥 (𝑎𝑥 + 𝑏)
(^3) = 3𝑎(𝑎𝑥 + 𝑏) (^2) [1 Mark]
⇒ (𝑎𝑥 + 𝑏)^2 =
(𝑎𝑥 + 𝑏)^3
∴ (𝑎𝑥 + 𝑏)^2 = 𝑑𝑥𝑑 ( 3𝑎^1 (𝑎𝑥 + 𝑏)^3 )
Hence, the anti-derivative of (𝑎𝑥 + 𝑏)^2 is (^) 3𝑎^1 (𝑎𝑥 + 𝑏)^3 [1 Mark]
5. Find an anti-derivative (or integral) of the function sin 2𝑥 – 4𝑒3𝑥^ by the method of inspection.
[2 Marks]
8. Find the integral (^) ∫(𝑎𝑥^2 + 𝑏𝑥 + 𝑐)𝑑𝑥 [2 Marks]
Solution:
The given integral is (^) ∫(𝑎𝑥^2 + 𝑏𝑥 + 𝑐)𝑑𝑥
= 𝑎 ∫ 𝑥^2 𝑑𝑥 + 𝑏 ∫ 𝑥𝑑𝑥 + 𝑐 ∫ 1. 𝑑𝑥
= 𝑎 (𝑥
3 3 ) + 𝑏 (
𝑥^2 2 ) + 𝑐𝑥 + 𝐶^ [∵ ∫ 𝑥
𝑛+1 + 𝐶 𝑎𝑛𝑑 ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶]
[1 Mark]
𝑎𝑥^3 3 +^
𝑏𝑥^2 2 + 𝑐𝑥 + 𝐶
Hence, the integral ∫(𝑎𝑥^2 + 𝑏𝑥 + 𝑐)𝑑𝑥 =
𝑎𝑥^3 3 +^
𝑏𝑥^2 2 + 𝑐𝑥 + 𝐶
[1 Mark]
9. Find the integral (^) ∫(2𝑥^2 + 𝑒𝑥) 𝑑𝑥. [2 Marks]
Solution:
The given integral is ∫(2𝑥^2 + 𝑒𝑥) 𝑑𝑥
= 2 ∫ 𝑥^2 𝑑𝑥 + ∫ 𝑒𝑥𝑑𝑥
𝑥^3 3 ) + 𝑒
𝑥 + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥𝑛+
𝑛+1 + 𝐶 𝑎𝑛𝑑^ ∫ 𝑒
𝑎 + 𝐶]
[1 Mark]
= 23 𝑥^3 + 𝑒𝑥^ + 𝐶
Hence, the integral (^) ∫(2𝑥^2 + 𝑒𝑥) 𝑑𝑥 = 23 𝑥^3 + 𝑒𝑥^ + 𝐶.
[1 Mark]
10. Find the integral ∫ (√𝑥 −
1 √𝑥)
2 𝑑𝑥 [2 Marks]
Solution:
The given integral is ∫ (√𝑥 −
1 √𝑥)
2 𝑑𝑥
= ∫ 𝑥𝑑𝑥 + ∫ 𝑥^1 𝑑𝑥 − 2 ∫ 1. 𝑑𝑥
[ 𝟏𝟐 Mark]
𝑥^2 2 + log|𝑥| − 2𝑥 + 𝐶^ [∵ ∫ 𝑥
𝑛+1 + 𝐶, ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝐶 𝑎𝑛𝑑 ∫^
1 𝑥 𝑑𝑥 = log|𝑥| + 𝐶 ]
[1 Mark]
Hence, the integral ∫ (√𝑥 −
1 √𝑥)
2 𝑑𝑥 =
𝑥^2 2 + log|𝑥| − 2𝑥 + 𝐶 [ 𝟏𝟐 Mark]
11. Find the integral (^) ∫
𝑥^3 +5𝑥^2 − 𝑥^2 𝑑𝑥^ [2 Marks]
Solution:
The given integral is (^) ∫
𝑥^3 +5𝑥^2 − 𝑥^2 𝑑𝑥
= ∫(𝑥 + 5 − 4𝑥−2)𝑑𝑥
[
𝟏 𝟐 Mark]
= 𝑥
2 2 + 5𝑥 − 4 (
𝑥− −1 ) + 𝐶^ [∵ ∫ 𝑥
𝑛+1 + 𝐶]
[1 Mark]
𝑥^2
Hence, the integral (^) ∫ 𝑥
(^3) +5𝑥 (^2) − 𝑥^2 𝑑𝑥 =^
𝑥^2 2 + 5𝑥 +^
4 𝑥 + 𝐶 [ 𝟏𝟐 Mark]
Solution:
The given integral is (^) ∫(1 − 𝑥)√𝑥𝑑𝑥
3 (^2) ) 𝑑𝑥
[ 𝟏𝟐 Mark]
3 2 3 2
5 2 5 2
+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶]^ [1 Mark]
2 3 𝑥
3 (^2) − 2 5 𝑥
5 (^2) + 𝐶 [1 Mark]
Hence, the integral (^) ∫(1 − 𝑥)√𝑥𝑑𝑥 = 23 𝑥
3 (^2) − 2 5 𝑥
5 (^2) + 𝐶
[
𝟏 𝟐 Mark]
15. Find the integral (^) ∫ √𝑥(3𝑥^2 + 2𝑥 + 3)𝑑𝑥 [2 Marks]
Solution:
The given integral is (^) ∫ √𝑥(3𝑥^2 + 2𝑥 + 3)𝑑𝑥
5 (^2) + 2𝑥
3 (^2) + 3𝑥
1 (^2) ) 𝑑𝑥
[ 𝟏𝟐 Mark]
7 2 7 2
5 2 5 2
3(𝑥
3 (^2) ) 3 2
+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶]^ [1 Mark]
6 7 𝑥
7 (^2) + 45 𝑥
5 (^2) + 2𝑥
3 (^2) + 𝐶
Hence, the integral (^) ∫ √𝑥(3𝑥^2 + 2𝑥 + 3)𝑑𝑥 = 67 𝑥
7 (^2) + 4 5 𝑥
5 (^2) + 2𝑥
3 (^2) + 𝐶
[ 𝟏𝟐 Mark]
16. Find the integral ∫(2𝑥 − 3 cos 𝑥 + 𝑒𝑥)𝑑𝑥 [4 Marks]
Solution:
The given integral is ∫(2𝑥 − 3 cos 𝑥 + 𝑒𝑥)𝑑𝑥
= 2 ∫ 𝑥𝑑𝑥 − 3 ∫ cos 𝑥 𝑑𝑥 + ∫ 𝑒𝑥𝑑𝑥 [
𝟏 𝟐 Mark]
=
2𝑥^2 2 − 3(sin 𝑥) + 𝑒
𝑥 (^) + 𝐶 [3 Marks]
[∵ ∫ 𝑥𝑛𝑥𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶 , ∫ 𝑒
𝑎 + 𝐶 and ∫ 𝑐𝑜𝑠𝑥 𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝐶]
= 𝑥^2 − 3 sin 𝑥 + 𝑒𝑥^ + 𝐶
Hence, the integral (^) ∫(2𝑥 − 3 cos 𝑥 + 𝑒𝑥)𝑑𝑥 = 𝑥^2 − 3 sin 𝑥 + 𝑒𝑥^ + 𝐶 [ 𝟏𝟐 Mark]
17. Find the integral ∫(2𝑥^2 − 3 sin 𝑥 + 5√𝑥)𝑑𝑥 [4 Marks]
Solution:
The given integral is (^) ∫(2𝑥^2 − 3 sin 𝑥 + 5√𝑥)𝑑𝑥
= 2 ∫ 𝑥^2 𝑑𝑥 − 3 ∫ sin 𝑥𝑑𝑥 + 5 ∫ 𝑥
1 (^2) 𝑑𝑥 [
𝟏 𝟐 Mark]
3 3 − 3(− cos 𝑥) + 5 (
𝑥
3 2 3 2
[∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶,^ and ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶]^ [3 Marks]
= 23 𝑥^3 + 3 cos 𝑥 + 103 𝑥
3 (^2) + 𝐶
Hence, the integral ∫(2𝑥^2 − 3 sin 𝑥 + 5√𝑥)𝑑𝑥 =
2 3 𝑥
(^3) + 3 cos 𝑥 + 10 3 𝑥
3 (^2) + 𝐶
[
𝟏 𝟐 Mark]
18. Find the integral ∫ sec 𝑥(sec 𝑥 + tan 𝑥)𝑑𝑥 [2 Marks]
[ 𝟏𝟐 Mark]
20. Find the integral ∫
2−3 sin 𝑥 cos^2 𝑥 𝑑𝑥^ [4 Marks]
Solution:
The given integral is (^) ∫ 2−3 sin 𝑥cos (^2) 𝑥 𝑑𝑥
2 cos^2 𝑥 −^
3 sin 𝑥 cos^2 𝑥) 𝑑𝑥
∫ 2 sec^2 𝑥𝑑𝑥 − 3 ∫ tan 𝑥 sec 𝑥𝑑𝑥^ [∵^
sin 𝑥 cos 𝑥 = tan 𝑥 𝑎𝑛𝑑^
1 cos 𝑥 = sec 𝑥] [1 Mark]
= 2 tan 𝑥 − 3 sec 𝑥 + 𝐶
[2 Marks]
[∵ ∫ sec^2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶 𝑎𝑛𝑑 ∫ tan 𝑥 sec 𝑥𝑑𝑥 = sec 𝑥 + 𝐶]
Hence, the integral ∫
2−3 sin 𝑥 cos^2 𝑥 𝑑𝑥 = 2 tan 𝑥 − 3 sec 𝑥 + 𝐶
[1 Mark]
21. The anti-derivative of (√𝑥 +
1 √𝑥)^ equals
(A)
1 3 𝑥
1 (^3) + 2𝑥
1 (^2) + 𝐶
(B) 23 𝑥
2 (^3) + 1 2 𝑥
(C) 23 𝑥
3 (^2) + 2𝑥
1 (^2) + 𝐶
(D)
3 2 𝑥
3 (^2) + 1 2 𝑥
1 (^2) + 𝐶 [2 Marks]
Solution:
1 (^2) 𝑑𝑥 + ∫ 𝑥−
1 (^2) 𝑑𝑥
3 2 3 2
1 2 1 2
+ 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶]
3 (^2) + 2𝑥
1 (^2) + 𝐶
The anti-derivative of (√𝑥 + (^) √𝑥^1 ) = 23 𝑥
3 (^2) + 2𝑥
1 (^2) + 𝐶 [2 Marks]
Hence, the correct answer is (C)
22. If
𝑑 𝑑𝑥 𝑓(𝑥) = 4𝑥
𝑥^4 such that^ 𝑓(2) = 0.^ Then^ 𝑓(𝑥)^ is
(A) 𝑥^4 + (^) 𝑥^13 − (^1298)
(B) 𝑥^3 + 𝑥^14 + 1298
(C) 𝑥^4 +
1 𝑥^3 +^
129 8
(D) 𝑥^3 +
1 𝑥^4 −^
129 8 [4 Marks]
Solution:
Given that, (^) 𝑑𝑥𝑑 𝑓(𝑥) = 4𝑥^3 − (^) 𝑥^34
Thus, anti-derivative of 4𝑥^3 −
3 𝑥^4 = 𝑓(𝑥)^ [1 Mark]
∴ 𝑓(𝑥) = ∫ 4𝑥^3 −
𝑥^4
⇒ 𝑓(𝑥) = 4 ∫ 𝑥^3 𝑑𝑥 − 3 ∫(𝑥−4)𝑑𝑥
𝑥^4 4 ) − 3 (
𝑥− −3 ) + 𝐶^ [∵ ∫ 𝑥
𝑛+1 + 𝐶]^ [1 Mark]
⇒ 𝑓(𝑥) = 𝑥^4 +
1 𝑥^3 + 𝐶
Also,
𝑓(2) = 0
∴ 𝑓(2) = (2)^4 + (^) (2)^13 + 𝐶 = 0
Hence, integration of the function
2𝑥 1+𝑥^2 = log(1 + 𝑥
[ 𝟏𝟐 Mark]
2. Integrate the function
(log 𝑥)^2 𝑥 [4 Marks]
Solution:
We need to integrate
(log 𝑥)^2 𝑥
Let log|𝑥| = 𝑡 [ 𝟏𝟐 Mark]
Differentiating with respect to 𝑡, we get 1 𝑥
𝑑𝑥 𝑑𝑡 = 1
∴ (^) 𝑥^1 𝑑𝑥 = 𝑑𝑡 [1 Mark]
(log|𝑥|)^2 𝑥 𝑑𝑥 = ∫^
(𝑡)^2 𝑥 𝑑𝑥^ [
𝟏 𝟐 Mark]
= (^) ∫(𝑡)^2 𝑑𝑡
3 3 + 𝐶^ [∵ ∫ 𝑥
𝑛+1 + 𝐶]^ [1 Mark]
=
(log|𝑥|)^3 3 + 𝐶^ [substituting^ 𝑡] [
𝟏 𝟐 Mark]
Hence, integration of the function (log 𝑥)
2 𝑥 =^
(log|𝑥|)^3 3 + 𝐶. [
𝟏 𝟐 Mark]
3. Integrate the function (^) 𝑥+𝑥 log 𝑥^1 [4 Marks]
Solution:
We need to integrate (^) 𝑥+𝑥 log 𝑥^1
𝑥(1 + log 𝑥)
Let 1 + log 𝑥 = 𝑡 [
𝟏 𝟐 Mark]
Differentiating with respect to 𝑡, we get 1 𝑥
𝑑𝑥 𝑑𝑡 = 1
∴ (^) 𝑥^1 𝑑𝑥 = 𝑑𝑡 [1 Mark]
1 𝑥(1+log 𝑥) 𝑑𝑥^ = ∫^
1 𝑡 𝑑𝑡^ [∵ ∫^
1 𝑥 𝑑𝑥 = log 𝑥 + 𝐶] [
𝟏 𝟐 Mark]
= log|𝑡| + 𝐶 [
𝟏 𝟐 Mark]
= log|1 + log 𝑥| + 𝐶 [substituting 𝑡] [1 Mark]
Hence, integration of the function
1 𝑥+𝑥 log 𝑥 =^ = log|1 + log 𝑥| + 𝐶 [ 𝟏𝟐 Mark]
4. Integrate the function sin 𝑥 sin (cos 𝑥)^ [4 Marks]
Solution:
We need to integrate sin 𝑥 sin (cos 𝑥)
Let cos 𝑥 = 𝑡 [ 𝟏𝟐 Mark]
Differentiating with respect to 𝑡, we get
−sin 𝑥
𝑑𝑥 𝑑𝑡 = 1
∴ −sin 𝑥 𝑑𝑥 = 𝑑𝑡 [1 Mark]
⇒ ∫ sin 𝑥 sin(cos 𝑥)𝑑𝑥 = − ∫ sin 𝑡 𝑑𝑡 [ 𝟏𝟐 Mark]
= −[− cos 𝑡] + 𝐶 [∵ ∫ sin 𝑥 dx = − 𝑐𝑜𝑠 𝑥 + 𝐶] [1 Mark]
Solution:
We need to integrate √𝑎𝑥 + 𝑏
Let 𝑎𝑥 + 𝑏 = 𝑡 [
𝟏 𝟐 Mark]
Differentiating with respect to 𝑡, we get
𝑎 𝑑𝑥𝑑𝑡 = 1
⇒ 𝑎𝑑𝑥 = 𝑑𝑡
∴ 𝑑𝑥 =
1 𝑎 𝑑𝑡
[1 Mark]
1 (^2) 𝑑𝑥 = 1 𝑎 ∫ 𝑡
1 (^2) 𝑑𝑡
[ 𝟏𝟐 Mark]
= 𝑎^1 ( 𝑡
3 2 3 2
) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶]
[1 Mark]
= 3𝑎^2 (𝑎𝑥 + 𝑏)
3 (^2) + 𝐶 [substituting 𝑡]
[ 𝟏𝟐 Mark]
Hence, integration of the function (^) √𝑎𝑥 + 𝑏 = (^) 3𝑎^2 (𝑎𝑥 + 𝑏)
3 (^2) + 𝐶
[ 𝟏𝟐 Mark]
7. Integrate the function 𝑥√𝑥 + 2 [4 Marks]
Solution:
We need to integrate 𝑥√𝑥 + 2
Let 𝑥 + 2 = 𝑡 [ 𝟏𝟐 Mark]
Differentiating with respect to 𝑡, we get 𝑑𝑥 𝑑𝑡 = 1
[1 Mark]
⇒ ∫ 𝑥√𝑥 + 2𝑑𝑥 = ∫(𝑡 − 2)√𝑡𝑑𝑡 [ 𝟏𝟐 Mark]
3 (^2) − 2𝑡
1 (^2) ) 𝑑𝑡
3 (^2) 𝑑𝑡 − 2 ∫ 𝑡
1 (^2) 𝑑𝑡
𝑡
5 2 5 2
𝑡
3 2 3 2
) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 =
𝑥𝑛+ 𝑛+1 + 𝐶]
[1 Mark]
5 (^2) −
3 (^2) + 𝐶
5 (^2) − 4 3 (𝑥 + 2)
3 (^2) + 𝐶 [substituting 𝑡]
[ 𝟏𝟐 Mark]
Hence, integration of the function, 𝑥√𝑥 + 2 = 25 (𝑥 + 2)
5 (^2) − 4 3 (𝑥 + 2)
3 (^2) + 𝐶
[ 𝟏𝟐 Mark]
8. Integrate the function 𝑥√1 + 2𝑥^2
[4 Marks]
Solution:
We need to integrate 𝑥√1 + 2𝑥^2
Let 1 + 2𝑥^2 = 𝑡 [
𝟏 𝟐 Mark]
Differentiating with respect to 𝑡, we get
4𝑥 𝑑𝑥𝑑𝑡 = 1
∴ 4𝑥𝑑𝑥 = 𝑑𝑡
[1 Mark]
= 43 (𝑥^2 + 𝑥 + 1)
3 (^2) + 𝐶 [substituting 𝑡]
[ 𝟏𝟐 Mark]
Hence, integration of the function, (4𝑥 + 2)√𝑥^2 + 𝑥 + 1 =
4 3 (𝑥
2 + 𝑥 + 1)^32 + 𝐶
[
𝟏 𝟐 Mark]
10. Integrate the function (^) 𝑥−√𝑥^1 [4 Marks]
Solution:
We need to integrate (^) 𝑥−√𝑥^1
Let (√𝑥 − 1) = 𝑡 [ 𝟏𝟐 Mark]
Differentiating with respect to 𝑡, we get 1 2√𝑥
𝑑𝑥 𝑑𝑡 = 1
∴ (^) 2√𝑥^1 𝑑𝑥 = 𝑑𝑡 [1 Mark]
2 𝑡 𝑑𝑡 [
𝟏 𝟐 Mark]
= 2 log|𝑡| + 𝐶 [∵ ∫
1 𝑥 𝑑𝑥 = log 𝑥 + 𝐶]^ [1 Mark]
= 2 log|√𝑥 − 1| + 𝐶 [substituting 𝑡] [
𝟏 𝟐 Mark]
Hence, integration of the function,
1 𝑥−√𝑥 = 2 log|√𝑥 − 1| + 𝐶 [
𝟏 𝟐 Mark]
11. Integrate the function
𝑥 √𝑥+4^ , 𝑥 > 0^ [4 Marks]
Solution:
We need to integrate
𝑥 √𝑥+
Let 𝑥 + 4 = 𝑡 [ 𝟏𝟐 Mark]
Differentiating with respect to 𝑡, we get 𝑑𝑥 𝑑𝑡 = 1
∴ 𝑑𝑥 = 𝑑𝑡 [1 Mark]
(𝑡−4) √𝑡^ 𝑑𝑡 [ 𝟏𝟐 Mark]
3 2 3 2
1 2 1 2
) + 𝐶 [∵ ∫ 𝑥𝑛𝑑𝑥 = 𝑥
𝑛+ 𝑛+1 + 𝐶]^ [1 Mark]
3 (^2) − 8(𝑡)
1 (^2) + 𝐶
1 (^2) − 8𝑡
1 (^2) + 𝐶
1 (^2) (𝑡 − 12) + 𝐶
1 (^2) (𝑥 + 4 − 12) + 𝐶 [substituting 𝑡]
[ 𝟏𝟐 Mark]
Hence, integration of the function
𝑥 √𝑥+4^ =^
2 3 √𝑥 + 4(𝑥 − 8) + 𝐶 [
𝟏 𝟐 Mark]
