Hello Students! In this post, you will find the complete NCERT Solutions for Maths Class 12 Exercise 7.3.
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NCERT Solutions for Maths Class 12 Exercise 7.1
NCERT Solutions for Maths Class 12 Exercise 7.2
NCERT Solutions for Maths Class 12 Exercise 7.4
NCERT Solutions for Maths Class 12 Exercise 7.5
NCERT Solutions for Maths Class 12 Exercise 7.6
NCERT Solutions for Maths Class 12 Exercise 7.7
NCERT Solutions for Maths Class 12 Exercise 7.8
NCERT Solutions for Maths Class 12 Exercise 7.9
NCERT Solutions for Maths Class 12 Exercise 7.10
NCERT Solutions for Maths Class 12 Exercise 7.3
Find the
integrals of the functions in Exercises 1 to 22.
Maths Class
12 Ex 7.3 Question 1.
sin² (2x + 5)
Solution:
∫ sin² (2x + 5)
dx = ½ ∫ [1 – cos 2(2x + 5)] dx
= ½ ∫ [1 – cos (4x + 10)] dx
=
sin 3x cos 4x
Solution:
∫ sin 3x cos 4x = ½ ∫ [sin (3x + 4x) + cos (3x – 4x)] dx
= ½ ∫ [sin 7x + sin (–x)] dx
= ½ ∫ sin 7x dx – ½ ∫ sin x dx
=
cos 2x cos 4x cos 6x dx
Solution:
∫ cos 2x cos 4x
cos 6x dx
= ½ ∫ (2 cos
2x cos 4x) cos 6x dx
= ½ ∫ (cos 6x
+ cos 2x) cos 6x dx
= ¼ ∫ (2 cos2
6x + 2 cos 2x cos 6x) dx
Maths Class
12 Ex 7.3 Question 4.
sin3 (2x + 1) dx
Solution:
∫ sin3 (2x
+ 1) dx = ¼ ∫ [3 sin (2x + 1) – sin 3(2x
+ 1)] dx
Maths Class
12 Ex 7.3 Question 5.
sin3 x cos3 x
Solution:
Maths Class 12 Ex 7.3 Question 6.
sin x sin 2x sin 3x
Solution:
∫ sin x sin 2x
sin 3x dx = ½ ∫ (2 sin x sin 2x) sin
3x dx
= ½ ∫ (cos x – cos 3x) sin 3x dx
= ¼ ∫ (2
sin 3x cos x – 2 sin 3x cos 3x) dx
= ¼ ∫ (sin 4x + sin 2x – sin 6x)dx
=
sin 4x sin 8x
Solution:
∫ sin 4x sin 8x dx
= ½ ∫ 2 sin 4x sin 8x dx
Solution:
sin4 x dx
Solution:
Maths Class 12 Ex 7.3 Question 11.
cos4 2x
Solution:
Maths Class 12 Ex 7.3 Question 12.
Solution:
Solution:
Solution:
tan3 2x sec 2x
Solution:
Let I = ∫ tan3 2x sec 2x = ∫ (sec2
2x – 1) sec 2x tan 2x dx
= ∫ tan2 2x sec 2x tan 2x dx
Putting sec 2x = t,
then, 2 sec 2x tan 2x dx = dt
Maths Class
12 Ex 7.3 Question 16.
tan4 x
Solution:
Let I = ∫ tan4 x
dx = ∫ tan2 x tan2 x dx
= ∫ tan2 x (sec²
x – 1) dx
= ∫ tan2 x sec² x dx –
∫ (sec² x – 1) dx
= ∫ tan2 x sec² x dx –
∫ (sec² x + ∫ 1 dx … (i)
Putting tan x =
t, then, sec² x dx = dt
From equation (i),
we get
I = ∫ t2 dt
– tan x + x + c
= t3/3 – tan x + x + c
= 1/3 tan3 x – tan x + x + c
Maths Class
12 Ex 7.3 Question 17.
Solution:
Solution:
Solution:
Solution:
sin-1 (cos x)
Solution:
Maths Class 12 Ex 7.3 Question 22.
Solution:
Choose the correct answers in Exercises 23 and 24.
Maths Class
12 Ex 7.3 Question 23.
(A) tan x + cot x + c
(B) tan x + cosec x + c
(C) –tan x + cot x + c
(D) tan x + sec x + c
Solution:
Maths Class 12 Ex 7.3 Question 24.
(A) –cot (exx) + c
(B) tan (xex) + c
(C) tan (ex) + c
(D) cot (ex) + c
Solution:
Related Links:
NCERT Solutions for Maths Class 12 Exercise 7.1
NCERT Solutions for Maths Class 12 Exercise 7.2
NCERT Solutions for Maths Class 12 Exercise 7.4
NCERT Solutions for Maths Class 12 Exercise 7.5
NCERT Solutions for Maths Class 12 Exercise 7.6
NCERT Solutions for Maths Class 12 Exercise 7.7
NCERT Solutions for Maths Class 12 Exercise 7.8