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NCERT Solutions Class 12 Maths Chapter 7 - Integrals

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1 	 / 	 2 1
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 E x e r c i s e 	 7 . 3
F i n d 	 t h e 	 i n t e g r a l s 	 o f 	 t h e 	 f o l l o w i n g 	 f u n c t i o n s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 9 .
1 . 	
A n s . 	
= 	
U s i n g 	
= 	 	
= 	 	 	 	 	 	
U s i n g
= 	 	 A n s .
2 . 	
A n s . 	 	 	 = 	 	
= 	 	 	 	 	 	 	 	 U s i n g 	 2 s i n B 	 c o s A = s i n ( A + B ) - s i n ( A - B )
Page 2


1 	 / 	 2 1
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 E x e r c i s e 	 7 . 3
F i n d 	 t h e 	 i n t e g r a l s 	 o f 	 t h e 	 f o l l o w i n g 	 f u n c t i o n s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 9 .
1 . 	
A n s . 	
= 	
U s i n g 	
= 	 	
= 	 	 	 	 	 	
U s i n g
= 	 	 A n s .
2 . 	
A n s . 	 	 	 = 	 	
= 	 	 	 	 	 	 	 	 U s i n g 	 2 s i n B 	 c o s A = s i n ( A + B ) - s i n ( A - B )
2 	 / 	 2 1
= 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
= 	 	
= 	 	
= 	 	 A n s .
3 . 	 	
A n s . 	 	
= 	
= 	 	 	 	 	 	
	 U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
= 	 	
= 	 	 	 	 	 	 	 	
= 	 	 	 	
U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
a n d 	 	 	 	 	 	 	 	
= 	 	
Page 3


1 	 / 	 2 1
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 E x e r c i s e 	 7 . 3
F i n d 	 t h e 	 i n t e g r a l s 	 o f 	 t h e 	 f o l l o w i n g 	 f u n c t i o n s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 9 .
1 . 	
A n s . 	
= 	
U s i n g 	
= 	 	
= 	 	 	 	 	 	
U s i n g
= 	 	 A n s .
2 . 	
A n s . 	 	 	 = 	 	
= 	 	 	 	 	 	 	 	 U s i n g 	 2 s i n B 	 c o s A = s i n ( A + B ) - s i n ( A - B )
2 	 / 	 2 1
= 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
= 	 	
= 	 	
= 	 	 A n s .
3 . 	 	
A n s . 	 	
= 	
= 	 	 	 	 	 	
	 U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
= 	 	
= 	 	 	 	 	 	 	 	
= 	 	 	 	
U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
a n d 	 	 	 	 	 	 	 	
= 	 	
3 	 / 	 2 1
= 	 	 	 	 	 	 	 A n s .
4 . 	
A n s . 	 	
= 	 	
	
= 	 	
= 	 	
= 	 	
= 	
A n o t h e r 	 M e t h o d 	 	 	
	 	 	 	 	
U s i n g 	
	 	 	 	 l e t 	 	
	 	 t h e r e f o r e 	 	
Page 4


1 	 / 	 2 1
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 E x e r c i s e 	 7 . 3
F i n d 	 t h e 	 i n t e g r a l s 	 o f 	 t h e 	 f o l l o w i n g 	 f u n c t i o n s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 9 .
1 . 	
A n s . 	
= 	
U s i n g 	
= 	 	
= 	 	 	 	 	 	
U s i n g
= 	 	 A n s .
2 . 	
A n s . 	 	 	 = 	 	
= 	 	 	 	 	 	 	 	 U s i n g 	 2 s i n B 	 c o s A = s i n ( A + B ) - s i n ( A - B )
2 	 / 	 2 1
= 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
= 	 	
= 	 	
= 	 	 A n s .
3 . 	 	
A n s . 	 	
= 	
= 	 	 	 	 	 	
	 U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
= 	 	
= 	 	 	 	 	 	 	 	
= 	 	 	 	
U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
a n d 	 	 	 	 	 	 	 	
= 	 	
3 	 / 	 2 1
= 	 	 	 	 	 	 	 A n s .
4 . 	
A n s . 	 	
= 	 	
	
= 	 	
= 	 	
= 	 	
= 	
A n o t h e r 	 M e t h o d 	 	 	
	 	 	 	 	
U s i n g 	
	 	 	 	 l e t 	 	
	 	 t h e r e f o r e 	 	
4 	 / 	 2 1
	 	 	 	 t h u s 	
	 	 	 	 A n s
5 . 	
A n s . 	 	
= 	 	
= 	
= 	
= 	 	
= 	 	
= 	 	
= 	 	
= 	
Page 5


1 	 / 	 2 1
N C E R T 	 s o l u t i o n
C h a p t e r 	 - 	 7
I n t e g r a l s 	 - 	 E x e r c i s e 	 7 . 3
F i n d 	 t h e 	 i n t e g r a l s 	 o f 	 t h e 	 f o l l o w i n g 	 f u n c t i o n s 	 i n 	 E x e r c i s e s 	 1 	 t o 	 9 .
1 . 	
A n s . 	
= 	
U s i n g 	
= 	 	
= 	 	 	 	 	 	
U s i n g
= 	 	 A n s .
2 . 	
A n s . 	 	 	 = 	 	
= 	 	 	 	 	 	 	 	 U s i n g 	 2 s i n B 	 c o s A = s i n ( A + B ) - s i n ( A - B )
2 	 / 	 2 1
= 	 	 	 	 	 	 	 	 	 	 	 	 	 	 	
= 	 	
= 	 	
= 	 	 A n s .
3 . 	 	
A n s . 	 	
= 	
= 	 	 	 	 	 	
	 U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
= 	 	
= 	 	 	 	 	 	 	 	
= 	 	 	 	
U s i n g 	 2 c o s A 	 c o s B = c o s ( A + B ) + c o s ( A - B )
a n d 	 	 	 	 	 	 	 	
= 	 	
3 	 / 	 2 1
= 	 	 	 	 	 	 	 A n s .
4 . 	
A n s . 	 	
= 	 	
	
= 	 	
= 	 	
= 	 	
= 	
A n o t h e r 	 M e t h o d 	 	 	
	 	 	 	 	
U s i n g 	
	 	 	 	 l e t 	 	
	 	 t h e r e f o r e 	 	
4 	 / 	 2 1
	 	 	 	 t h u s 	
	 	 	 	 A n s
5 . 	
A n s . 	 	
= 	 	
= 	
= 	
= 	 	
= 	 	
= 	 	
= 	 	
= 	
5 	 / 	 2 1
A n o t h e r 	 m e t h o d 	 	 	 l e t 	 	 	 I = 	 	 	 	
l e t 	 c o s x = t
- s i n x 	 d x = d t
s i n x 	 	 d x 	 = 	 - d t
	 	 a n s .
	
6 . 	 	
A n s . 	 	
= 	 	
= 	 	 	 	 	 	 	 	 	 	 U s i n g 	 2 s i n 	 A 	 s i n 	 B = c o s 	 ( A - B ) 	 - 	 c o s 	 ( 	 A + B )
	 	
= 	 	 	
u s i n g 	 2 c o s A 	 s i n B = s i n ( A + B ) - s i n ( A - B )
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FAQs on NCERT Solutions Class 12 Maths Chapter 7 - Integrals

1. What are integrals?
Ans. Integrals are mathematical tools used to calculate the area under a curve or to find the accumulation of quantities over a given interval. They are an important concept in calculus and have applications in various fields such as physics, engineering, and economics.
2. What is the significance of integrals?
Ans. Integrals have several significant applications. They can be used to find the area between curves, calculate volumes of irregular shapes, determine the average value of a function, solve differential equations, and analyze rates of change. They are fundamental in understanding the concept of calculus and its real-world applications.
3. How do I solve integrals?
Ans. To solve integrals, you need to determine the antiderivative of the function to be integrated. This involves finding a function whose derivative is equal to the given function. Once you have the antiderivative, you can evaluate the integral by plugging in the limits of integration and subtracting the value at the lower limit from the value at the upper limit.
4. What are definite integrals?
Ans. Definite integrals are a type of integral that have specific limits of integration. They represent the accumulation of quantities over a given interval. The value of a definite integral gives the net signed area between the curve and the x-axis within the specified limits.
5. Are there any techniques to simplify integral calculations?
Ans. Yes, there are several techniques to simplify integral calculations. Some common methods include integration by substitution, integration by parts, partial fraction decomposition, and trigonometric substitutions. These techniques help to transform complex integrals into simpler forms, making them easier to solve.
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