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"Integrals," Chapter 7 of the Class 12 NCERT Mathematics textbook, covers the fundamental idea of integration, which is the inverse process of differentiation. This chapter discusses various integration strategies and how to use them. The substitution method, which is essential for reducing the complexity of complex integration problems by simplifying the integrand, is the subject of Exercise 7.5.
Students practice using the substitution method to evaluate integrals in Exercise 7.5. To simplify the integrand into a more manageable form, this method entails altering the variable of integration. When integrals involve composite functions and direct integration is difficult, it is especially helpful.
This exercise's solutions walk you through the process of choosing a suitable replacement and adjusting the integral in accordance with it. Simplifying the expression through substitution, differentiating the substitution to replace dx, and modifying the limits in the case of a definite integral are important strategies.
This section offers thorough, sequential solutions to every issue in Exercise 7.5, ensuring that students comprehend the rational steps required to use the substitution method correctly. By clearly outlining each step and showing how substitutions change the integral, these solutions aid in technique mastery.
Among the frequently used substitutions covered here are u=ax+b, u=sinx, and u=lnx. The explanation of each substitution includes instances of integrals in the situations where they work best, which helps students identify patterns and select the most appropriate course of action.
Various formulas used in the exercise are,
Rational Fraction | Respective Partial Fraction |
|---|---|
(px+q)/(x-a)(x-b), a !=b | A/(x-a) + B/(x-b) |
(px+q)/(x-a)2 | A/(x-a) + B/(x-a)2 |
(px2+qx+r)/(x-a)(x-b)(x-c) | A/(x-a) + B/(x-b) + C/(x-c) |
(px2+qx+r)/(x-a)2(x-b) | A/(x-a) + B/(x-a)2 + C/(x-b) |
(px2+qx+r)/(x-a)(x2+bx+c) | A/(x-a) + (Bx+C)/(x2 + bx + c) |
Answer:
Let x/(x+1)(x+2) = A/ (x+1) + B/(x+2)
=> x = A(x+2)+B(x+1)
Equating the coefficient of x and constant term, we obtain
A + B = 1
2A + B = 0
On solving we obtain
A = -1 nd B = 2
So x/(x+1)(x+2) = -1/ (x+1) + 2/(x+2)
=> ∫x/ (x+1)(x+2) dx
= ∫ -1/ (x+1) + 2/(x+2) dx
= -log|x+1| + 2log|x+2| + C
= log(x+2)2 - log|x+1| + C
= log (x+2)2/ (x+1) + C
Answer:
Let 1/ (x+3)(x-3) = A/ (x+3) + B/(x-3)
1 = A(x-3) + (x+2)
Equating the coefficient of x and constant term, we obtain
A + B = 0
-3A + 3B=1
On solving we obtain
A = -1/6 and B = 1/6
So, 1/ (x+3)(x-3) = -1/6(x+3) + 1/6(x-3)
=> ∫ 1/ (x2-9) dx
= ∫ { -1/6(x+3) + 1/6(x-3)} dx
= -1/6 log| x+3| + 1/6 log|x-3| + C
= 1/6 log| (x-3)/(x+3) | + c
Answer:
Let (3x-1)/(x-1)(x-2)(x-3) = A/(x-1) + B/ (x-2) + C/(x-3)
3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)...(i)
Substituting x = 1, 2 and 3 respectively in equation (i), we obtain
A = 1, B = -5 and C = 4
(3x-1)/(x-1)(x-2)(x-3)
= 1/(x-1) - 5/ (x-2) + 4/(x-3)
=> ∫ (3x-1)/(x-1)(x-2)(x-3) dx
= ∫{1/(x-1) - 5/(x-2) + 4/(x-3) }dx
= log| x-1| - 5log|x-2| +4 log|x-3| + C
Answer:
Let x/(x-1)(x-2)(x-3) = A/(x-1) + B/ (x-2) + C/(x-3)
x= A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)...(i)
Substituting x = 1, 2 and 3 respectively in equation (i), we obtain
A = 1/2 , B = -2 and C = 3/2
So, x/(x-1)(x-2)(x-3) = 1 /2 (x-1) -2/ (x-2) + 3/2(x-3)
=> ∫x/(x-1)(x-2)(x-3) dx
= ∫ {1 /2 (x-1) -2/ (x-2) + 3/2(x-3) }dx
= 1/2 log |x-1| -2 log|x-2| + 3/2 log|x-3| +C
Answer:
Let 2x/ (x2+3x+2) = A/(x+1) + B/(x+2)
2x = A(x+2) + B(x+1)...(i)
Substituting x = -1 and -2 in equation (i), we obtain
A = -2 and B = 4
So,2x/ (x2+3x+2) = -2/(x+1) + 4/(x+2)
=> ∫ 2x/ (x2+3x+2) dx
= ∫ { 4/(x+1) - 2/(x+1)} dx
= 4log|x+2| - 2log|x+1| + C
Answer:
It can seen that the given integrand is not proper fraction.
Therefore, on diving (1-x2) by x(1-2x), we obtain
(1-x2)/{x(1-2x)} = 1/2 + 1/2 { (2-x)/x(1-2x)}
Let (2-x)/x(1-2x) = A/x B/(1-2x)
=> (2-x) = A(1-2x) + BX...(i)
Substituting x = 0 and 1/2 in equation (i), we obtain
A = 2 and B = 3
So, (2-x)/x(1-2x) = 2/x+ 3/(1-2x)
Substituting in equation (1), we obtain
(1-x2)/{x(1-2x)} = 1/2 + 1/2 { 2/x+ 3/(1-2x)}
=> ∫ (1-x2)/{x(1-2x)} dx
= ∫{ 1/2 + 1/2 { 2/x+ 3/(1-2x)}}dx
= x/2 + log|x| + 3/(2(-2) log|1-2x| + C
= x/2 + log|x| - 3/2 log|1-2x| + C
Answer:
Let x/(x2+1)(x-1) = (Ax+B)/(x2+1) + C/(x-1)
x = (Ax+B)(x-1) + C(x2+1)
x = Ax2 - Ax + Bx-B+Cx2 + C
Equating the coefficient of x2,x and constant term, we obtain
A + C = 0
-A + B = 1
-B + C = 0
On solving these equation, we obtain
A = -1/2 , B = 1/2 and C = 1/2
From equation (1), we obtain
So, x/(x2+1)(x-1) = { (-1/2 x + 1/2)/(x2+1)} + (1/2)/(x-1)
=> ∫x/(x2+1)(x-1) = -1/2 ∫x/(x2+1) dx + 1/2 ∫1/(x2+1) dx + 1/2 ∫ 1/ (x-1) dx
=1/4 ∫ 2x/(x2+1) dx + 1/2 tan-1x + 1/2 log| x-1 | + C
Consider ∫ 2x/(x2+1) dx , let (x2+1) = t => 2x dx = dt
=> ∫ 2x/(x2+1) dx = ∫dt/t = log|t| = log | x2+1|
∫ x/(x2+1)(x-1) dx
= -1/4 log |x2+1| + 1/2 tan-1x + 1/2 log |x-1| + C
= 1/2 log |x-1| -1/4 log |x2+1| + 1/2 tan-1x + C
Answer:
x/(x-1)2(x+2) = A/(x-1) + B/(x-1)2 + C/(x+2)
x = A(x-1)(x+2) + B(x+2) + C(x-1)2
Substituting x = 1, we obtain
B = 1/3
Equating the coefficient of x2 and constant term, we obtain
A + C = 0
-2A + 2B + C = 0
On solving, we obtain
A = 2/9 and C = -2/9
x/(x-1)2(x+2) = 2 /9(x-1) + 1/3(x-1)2 - 2 /9(x+2)
=> ∫x/(x-1)2(x+2) dx
= 2/9∫1/(x-1) + 1/3 ∫1/(x-1)2 dx -2/9 ∫ 1/(x+2) dx
= 2/9 log|x-1| + 1/3 {-1/(x-1)} - 2/9 log |x+2| + C
= 2/9 log |(x-1)/(x+2) | - 1/ 3(x-1) + C
Answer:
Let (3x+5)/(x3-x2-x+1) = A/(x-1) + B/(x-1)2 + C/(x+1)
3x + 5 = A(x-1)(x+1) + B(x+1) + C(x-1)2
3x + 5 = A(x2-1) + B(x+1) + C(x2+1-2x)...(i)
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficient of x2 and x , we obtain
A+C = 0
B-2C = 3
On solving, we obtain
A = -1/2 and C = 1/2
So, (3x+5)/(x3-x2-x+1) = -1/2(x-1) + 4/(x-1)2 + 1/2(x+1)
=> ∫ (3x+5)/{(x-1)2(x+1)} dx
= -1/2∫1/(x-1) dx + 4∫1/(x-1)2 dx +1/2 ∫1/(x+1) dx
= -1/2 log | x-1| + 4 {-1/(x-1)} + 1/2 log | x+1 | + C
= 1/2log| (x+1)/ (x-1)| - 4/(x-1) + C
Answer:
(2x-3)/(x2-1)(2x+3) = (2x-3)/(x+1)(x-1)(2x+3)
Let (2x-3)/(x+1)(x-1)(2x+3) = A/(x+1) + B/(x-1) + C/(2x+3)
=> (2x-3) = A(x-1)(2x+3) + B(2x2+5x+3) + C(x2-1)
=> (2x-3) = A(2x2+x-3) + B(2x2+5x+3) + C(x2-1)
=>(2x-3) = (2A+2B+C)x2 + (A+5B)x+(-3A+3B-C)
Equating the coefficient of x2 and x , we obtain
B = -1/10 , A = 5/2 and C = -24/5
So, (2x-3)/(x+1)(x-1)(2x+3) = 5/2(x+1) + -1 /10 (x-1) - 24/5 (2x+3)
=> ∫(2x-3)/(x2-1)(2x+3) = 5/2∫1/(x+1) dx - -/10 ∫1/(x-1) dx - 24/5 ∫1/(2x+3) dx
= 5/2 log | x+1| - 1/10 log | x-1 | - 24/5*2 log | 2x+3 | + C
= 5/2 log | x+1| - 1/10 log | x-1 | - 12/5 log | 2x+3 | + C
Answer:
5x/(x+1)(x2-4) = 5x/(x+1)(x+2)(x-2)
Let 5x/(x+1)(x+2)(x-2) = A/(x+1) + B/(x+2) + C/(x-2)
5x = A(x+2)(x-2) + B(x+1)(x-2) + C(x+1)(x+2)...(i)
Substituting x = -1,-2 and 2 in equation (1), we obtain
A = 5/2, B = -5/2 and C = 5/6
So, 5x/(x+1)(x+2)(x-2) = 5 /3(x+1) - 5/2(x+2) + 5 /6 (x-2)
=> ∫ 5x/(x+1)(x+2)(x-2) dx = ∫ 5 /3(x+1) dx - ∫5/2(x+2) dx + ∫ 5 /6 (x-2) dx
= 5/3 log|x+1| -5.2 log|x+2| + 5/6 log |x-2| + C
Answer:
It can seen that the given integrand is not proper fraction.
Therefore, on diving (x3+x+1) by (x2-1), we obtain
(x3+x+1)/(x2-1) = x + (2x+1)/(x2-1)
Let (2x+1)/(x2-1) = A/(x+1) + B/(x-1)
2x+1 = A(x-1) + B(x+1)...(i)
Substituting x = 1,-1 in equation (1), we obtain
A = 1/2 and B = 3/2
So,(x3+x+1)/(x2-1) = x + 1/2(x+1) + 3/2(x-1)
=> ∫ ,(x3+x+1)/(x2-1) dx = ∫x dx + 1/2 ∫1/(x+1) dx + 3/2 ∫ 1/ (x-1) dx
= x2 /2 + 1/2 log |x+1| + 3/2 log|x-1| + C
Answer:
Let 2/(1-x)(1+x2) = A/(1-x) + (Bx+c)/(1+x2)
2 = A(1+x) + (Bx+C)(1-x)
2 = A+Ax2+Bx-Bx2+C-Cx
Equating the coefficient of x2,x and constant term, we obtain
A-B = 0
B-C = 0
A+C = 2
On solving these equation, we obtain
A = 1, B = 1 and C = 1
So, 2/ (1-x)(1+x2) = 1/(1-x) + (x+1)/(1+x2)
=> ∫2/ (1-x)(1+x2) dx
= ∫1/(1-x) dx + ∫x/(1+x2) dx + ∫ 1/ (1+x2) dx
= - ∫1/(x-1) dx + 1/2∫2x/(1+x2) dx + ∫ 1/ (1+x2) dx
= -log|x-1 | + 1/2 log|1+x2| + tan-1x + C
Answer:
Let (3x-1)/(x+2)2 = A /(x+2) + B/(x+2)2
=> 3x - 1 = A(x+2) + B
Equating the coefficient of x and constant term, we obtain
A = 3
2A +B = -1 => B = -7
So, (3x-1)/(x+2)2 = 3/(x+2) - 7/(x+2)2
=> ∫(3x-1)/(x+2)2 dx = 3∫1/(x+2) dx - 7 ∫ x/(x+2)2
= 3log|x+2| -7{-1/(x+2)} +C
= 3log|x+2| + 7/(x+2) + C
Answer:
1/(x4-1) = 1/(x2-1)(x2+1) = 1/{(x+1)(x-1)(1+x2)}
Let 1/(x4-1) = 1/(x2-1)(x2+1) = A /(x+1) + B/(x-1) + (Cx+D)/(1+x2)
1 = A(x-1)(x2+1) + B(x+1)(x2+1) + (Cx+D)(x2-1)
1 = A(x3+x-x2-1)+B(x3+x+x2+1) + Cx3+Dx3-Cx-D
1 = (A+B+C)x3 + (-A+B+D)x2 + (A+B-C)x+(-A+B-D)
Equating the coefficient of x3,x2,x and constant term, we obtain
A+B+C = 0
-A+B+D = 0
A+B-C = 0
-A+B-D = 1
On solving these equation, we obtain
a = -1/4, b = 1/4, C = 0 and D = -1/2
∫1/(x4-1)dx = ∫-1/4(x+1)dx + ∫1/4(x-1)dx -∫1/2(1+x2)dx
=> ∫1/(x4-1)dx = -1/4 log|x-1| + 1/4 log |x--1| -1/2tan-1x + C
= 1/4 log |(x-1)/(x+1)| - 1/2 tan-1x + C
Answer:
1/x(xn+1)
Multiplying numerator and denominator by xn-1, we obtain
1/x(xn+1) = xn-1/{xn-1 x(xn+1)} = xn-1/xn{xn+1}
Let xn = t => xn-1 dx = dt
So, ∫1/{x(xn+1)} dx = ∫xn-1/{xn(xn+1} dx = 1/n ∫1/t(t+1) dt
Let 1 /t(t+1) = A/t + B/(t+1)
1 = A(1+t) + Bt...(i)
Substituting t=0,-1 in equation (i),we obtain
A = 1 and B = -1
So, 1/ t(t+1) = 1/t - 1/(1+t)
=> ∫1/x(xn+1) dx = 1/n ∫ {1/t - 1/(t+1)} dx
= 1/n[log|t| = log|t+1|] + C
= -1/n [ log|xn| - log|xn+1|] +C
= 1/n log |xn/xn+1| + C
Answer:
cosx/(1-sinx)(2-sinx)
Let sinx = t => cosx dx = dt
So, ∫cosx/(1-sinx)(2-sinx) dx = ∫dt/(1-t)(2-t)
Let 1/(1-t)(2-t) = A/(1-t) + B/(2-t)
1 = A(2-t)+B(1-t)...(i)
Substituting t=2 and then t=1 in equation (i),we obtain
A = 1 and B = -1
So, 1/(1-t)(2-t) = 1/(1-t) - 1/(2-t)
=> ∫cosx/(1-sinx()2-sinx) dx
= ∫ {1/(1-t) - 1/(2-t) }dt
= -log|1-t| + log|2-t|+ C
= log|(2-t)/(1-t)| + C
= log |(2-sinx)/(1-sinx)| + C
Answer:
(x2+1)(x2+2)/(x2+3)(x2+4) = 1 - (4x2+10)/(x2+3)(x2+4)
Let (4x2+10)/(x2+3)(x2+4) = (Ax+B)/(x2+3) + (Cx+D)/(x2+4)
4x2 + 10 = (Ax+B)(x2+4) + (Cx+D)(x2+3)
4x2 + 10 = Ax3 + 4Ax + Bx2 + 4B + Cx3 + 3Cx + Dx2 + 3D
4x2 + 10 = (A+C)x3 + (B+D)x2 + (4A+3C)x + (4B+3D)
Equating the coefficient of x3,x2,x and constant term, we obtain
A + C = 0
B + D = 4
4A + 3C = 0
4B + 3D = 10
On solving these equation, we obtain
A = 0, B = -2 , C= 0 and D = 6
So, (4x2+10)/(x2+3)(x2+4) = -2/(x2+3) + 6/(x2+4)
(x2+1)(x2+2)/(x2+3)(x2+4) = 1 - { -2/(x2+3) + 6/(x2+4) }
=> ∫(x2+1)(x2+2)/(x2+3)(x2+4) dx
= ∫ 1 + 2/(x2+3) - 6/(x2+4) } dx
= ∫ { 1 + 2/{x2+√(3)2 - 6/{x2+22} } dx
= x + 2{1/√3 tan-1x/√3} -6{1/2 tan-1x/2} + C
= x + 2/√3 tan-1x/√3 - 3 tan-1x/2 + C
Answer:
2x/(x2+1)(x3+3)
Let x2 = t => 2x dx = dt
∫ 2x/(x2+1)(x2+3) dx = ∫dt/(t+1)(t+3)...(i)
Let 1/(t+1)(t+3) = A/(t+1) + B?(t+3)
1 = A(t+3) + B(t+1)...(ii)
Substituting t = -3 and then t = -1 in equation (i), we obtain
A = 1/2 and B = -1/2
So, 1/(t+1)(t+3) = 1/2(t+1) - 1/2(t+3)
=>∫2x/(x2+1)(x2+3) dx = ∫{1/2(t+1) - 1/2(t+3)} dt
1/2 log| (t+1) |-1/2 log|t+3| + C
= 1/2 log |(t+1)/(t+3)| + C
= 1/2 log |(x2+1)/(x2+3)| + C
Answer:
1/x(x4-1)
Multiplying numerator and denominator by x3, we obtain
1/x(x4-1) = x3/ x4(x4-1)
∫ 1/x(x4-1) dx = ∫x 3/ x4(x4-1) dx
Let x4 = t => 4x3 dx = dt
∫ 1/x(x4-1) dx =1/4 ∫ dt/ t(t-1)
Let 1/t(t-1) = A/t + B/(t-1)
1 = A(t-1) + Bt...(i)
Substituting t= 0 and 1 in equation (i),we obtain
A = -1 and B = 1
=> ∫ 1/x(x4-1) dx = 1/4 ∫ { -1/t + 1/ (t-1) } dt
= 1/4 [-log|t| + log |t-1 | ] + C
= 1/4 log |(t-1)/t| + C
= 1/4 log | (x4-1)/x4 | + C
Answer:
1/(ex-1)
Let ex = t => ex dx = dt
=> ∫1/ (ex-1) dx = ∫1/(t-1) dt/t = ∫ 1/t(t-1) dt
Let 1/t(t-1) = A/t + B/(t-1)
1 = A(t-1) + Bt...(i)
Substituting t = 1 and t= 0 in equation (i), we obtain
A = -1 and B = 1
So, 1/t(t-1) = -1/t + 1/(t-1)
=> ∫ 1/t(t-1) dt
= log | (t-1)/t | + C
= log | (ex-1)/ex | + C
(A) log | (x-1)2/(x-2) | + C
(B) log | (x-2)2/(x-1) | + C
(C) log | {(x-1)/(x-2)}2 | + C
(D) log | (x-1)(x-2) | + C
Correct Answer is option (B)
(A) log|x| - 1/2 log(x2+1) + C
(B) log|x| + 1/2 log(x2+1) + C
(C) - log|x| + 1/2 log(x2+1) + C
(D) 1/2 log|x| + log(x2+1) + C
Correct Answer is option (A)
Exercise 7.5 focuses on definite integrals and their applications. It covers the evaluation of definite integrals using fundamental theorems of calculus, properties of definite integrals, and integration techniques. The exercise emphasizes understanding the relationship between definite integrals and areas under curves, as well as applying integration to solve real-world problems involving areas, volumes, and average values of functions.
