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# Integrals Class 12 Mathematics Chapter 7 Important Questions

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Integrals Class 12 Mathematics Chapter 7 Important Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 7 Integrals Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 7 Maths Extra Questions

## Integrals Class 12 Mathematics Extra Questions

Chapter 7 Integrals

1. {tex}\int {\frac{1}{{{e^x} + 1}}dx} {/tex} is equal to

1. log {tex}(1 + {e^{ – 2x}})+C{/tex}
2. {tex}\log ({e^{ -2x}} -2 x)+C{/tex}
3. – log(1 + e-x) + C
4. {tex}\log ({e^{ 3x}} + x)+C{/tex}
2. The function{tex}f(x) = \int\limits_0^x {\log (t + \sqrt {1 + {t^2})} dt} {/tex} is

1. an odd function
2. an even function
3. Neither odd nor Even
4. a periodic function
3. {tex}\int\limits_0^{\pi /2} {\log \left| {\cos x} \right|} {/tex}dx is equal to

1. {tex} – \frac{\pi }{2}\log 2{/tex}
2. {tex}\pi \;log{\text{ }}2{/tex}
3. {tex}\frac{\pi }{2}\;\log 3{/tex}
4. {tex}-\frac{\pi }{3}\;\log 3{/tex}
4. {tex}\int\limits_a^b {\frac{{\log x}}{x}dx} {/tex} is equal to

1. {tex}\frac{log\;(b-a)}{b-a}{/tex}
2. log (a+b).log (b–a)
3. {tex}\log(ab).log{(\frac{b}{a}}){/tex}
4. {tex}\frac{1}{2}\log(ab).log \left( {\frac{b}{a}} \right){/tex}
5. If {tex}\int {f(x)} {/tex} dx = f (x), then

1. f(x) = ax
2. f (x) = x
3. f(x) = 0
4. {tex}f(x)=e^x{/tex}
6. The function A(x) denotes the ________ function and is given by A (x) = {tex}\int\limits_a^x {f(x)} dx{/tex}.
7. The indefinite integral of {tex}2x^{\frac{1}{2}}{/tex} is ________.
8. The indefinite integral of 2x2 + 3 is ________.
9. Show that {tex}\int {\frac{{2x + 3}}{{{x^2} + 3x}}dx = \log \left| {{x^2} + 3x} \right| + C} {/tex}

10. Evaluate {tex}\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 – x ^ { 2 } } } d x.{/tex}

11. Evaluate {tex}\int \sin ^ { 3 } x d x.{/tex}

12. Evaluate the definite integral {tex}\int\limits_0^{\frac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx} {/tex}.

13. Integrate the following function {tex}\frac{1}{{\sqrt {7 – 6x – {x^2}} }}{/tex}.

14. Integrate the function (x2 + 1) log x

15. {tex}\int_0^{\pi /4} {\frac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} {/tex}.

16. Evaluate {tex}\int \frac { \sqrt { 1 – \sin x } } { 1 + \cos x } e ^ { \frac { – x } { 2 } } d x.{/tex}

17. Evaluate {tex}\int\limits_0^1 {x\log \left( {1 + 2x} \right)dx} {/tex}

18. Evaluate {tex}\int _ { 1 } ^ { 3 } \left( 2 x ^ { 2 } + 5 x \right) d x{/tex} as a limit of a sum.

Chapter 7 Integrals

Solution

1. –log(1 + e-x) + C, Explanation: {tex}\int {\frac{{{e^{ – x}}}}{{1 + {e^{ – x}}}}} dx = – \int {\frac{{ – {e^{ – x}}}}{{1 + {e^{ – x}}}}} dx = – \log (1 + {e^{ – x}})+C{/tex}
1. an even function, Explanation: t = -u
f(-x) = {tex}\int\limits_0^x {\log ( – u + \sqrt {1 + {u^2}} } )( – du){/tex}
= {tex} – \int\limits_0^\pi {\log \left( {\frac{{1 + u + {u^2}}}{{\sqrt {1 + {u^2} + u} }}} \right)} ( – du){/tex}
= {tex}\int\limits_0^x {\log (u + \sqrt {1 + {u^2}} } )du{/tex} = f(x)
{tex}\Rightarrow{/tex} f(-x) = f(x) {tex}\Rightarrow{/tex} f is an even function
1. {tex} – \frac{\pi }{2}\log 2{/tex}Explanation: {tex}\int\limits_0^{\pi /2} {\log \left| {\cos x} \right|} dx{/tex}
{tex}= \int\limits_0^{\pi /2} {\log \left| {\cos (\frac{\pi }{2} – x)} \right|} dx {/tex}
{tex} = \int\limits_0^{\pi /2} {\log \left| {\sin x} \right|dx} {/tex}
{tex} = – \frac{\pi }{2}\;\log 2{/tex}(standard result)
1. {tex}\frac{1}{2}\log(ab).log \left( {\frac{b}{a}} \right){/tex}Explanation: {tex}= {\int\limits_a^b {{{(log\;x)}^{^1}}} ^{}}(\frac{1}{x})dx{/tex}
{tex} \\(let\; log\;x=t\;then\;\frac1xdx=dt) \\ {/tex}
{tex}= \left[ {\frac{{{{(log\;x)}^2}}}{2}} \right]_a^b \\ {/tex}
{tex}= \frac{1}{2}\left[ {{{(log\;b)}^2} – {{(log\;a)}^2}} \right] {/tex}
{tex}\\ = \frac{1}{2}(log\;b + log\;a)(log\;b – log\;a){/tex}
{tex}= \frac{1}{2}\log (ab)\log \frac{b}{a} {/tex}
1. {tex}f(x)=e^x{/tex}, mExplanation: {tex}{\frac{d}{{dx}}} (f(x)) = f(x){/tex}
It implies that the function remains same after integrating or differentiating it. So the function must be ex
1. area
2. {tex}\frac{4}{3} x^{\frac{3}{2}}+c{/tex}
3. {tex}=\frac{4}{5} x^{\frac{5}{4}}+c{/tex}
4. Let {tex}I = \int {\frac{{2x + 3}}{{{x^2} + 3x}}dx} {/tex}
Put x2 + 3x = t
{tex}\Rightarrow{/tex} (2x + 3)dx = dt
{tex}\therefore I = \int {\frac{1}{t}dt = \log \left| t \right| + C} {/tex}
= log |(x2 + 3x)| + C
5. According to the question , {tex}I = \int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 1 – x ^ { 2 } } } d x{/tex}
= {tex}\left[ \sin ^ { – 1 } x \right] _ { 0 } ^ { 1 } \quad \left[ \because \int_b^a \frac { 1 } { \sqrt { 1 – x ^ { 2 } } } d x = \sin ^ { – 1 } (a) – \sin ^ { – 1 } (b) \right]{/tex}
{tex}= sin^{-1}(1)-sin^{-1}(0){/tex}
{tex}= \sin ^ { – 1 } \left( \sin \frac { \pi } { 2 } \right) – \sin ^ { – 1 } ( \sin 0 ) {/tex}
{tex}= \frac { \pi } { 2 } – 0 = \frac { \pi } { 2 }{/tex}
6. Let {tex} I = \int \sin ^ { 3 } x d x = \int \frac { 3 \sin x – \sin 3 x } { 4 } d x{/tex} {tex}\left[ \because \sin 3 x = 3 \sin x – 4 \sin ^ { 3 } x \right]{/tex}
{tex}= \frac { 1 } { 4 } \int 3 \sin x d x – \frac { 1 } { 4 } \int \sin 3 x d x{/tex}
{tex}= \frac { 1 } { 4 } \left( – 3 \cos x + \frac { \cos 3 x } { 3 } \right) + C{/tex}
7. {tex}\int\limits_0^{\frac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx} {/tex}
{tex}= 2\int\limits_0^{\frac{\pi }{4}} {{{\sec }^2}xdx + \int\limits_0^{\frac{\pi }{4}} {{x^3}dx + 2\int\limits_0^{\frac{\pi }{4}} {1dx} } } {/tex}
{tex}= 2\left( {\tan x} \right)_0^{\frac{\pi }{4}} + \left( {\frac{{{x^4}}}{4}} \right)_0^{\frac{\pi }{4}} + 2\left( x \right)_0^{\frac{\pi }{4}}{/tex}
{tex}= 2\left( {\tan \frac{\pi }{4} – \tan {0^o}} \right) + \frac{{{{\left( {\frac{\pi }{4}} \right)}^4}}}{4} – 0 {/tex} {tex}+ 2\left( {\frac{\pi }{4} – 0} \right){/tex}
{tex}= 2\left( {1 – 0} \right) + \frac{{\left( {\frac{{{\pi ^4}}}{{256}}} \right)}}{4} + \frac{{2\pi }}{4}{/tex}
{tex}= 2 + \frac{{{\pi ^4}}}{{1024}} + \frac{\pi }{2}{/tex}
{tex}= \frac{{{\pi ^4}}}{{1024}} + \frac{\pi }{2} + 2{/tex}
8. {tex}\int {\frac{1}{{\sqrt {7 – 6x – {x^2}} }}dx} {/tex}
{tex} = \int {\frac{1}{{\sqrt { – {x^2} – 6x + 7} }}dx} {/tex}
{tex}= \int {\frac{1}{{\sqrt { – \left( {{x^2} + 6x – 7} \right)} }}dx} {/tex}
{tex}= \int {\frac{1}{{\sqrt { – \left( {{x^2} + 6x + 9 – 9 – 7} \right)} }}dx} {/tex}
{tex}= \int {\frac{1}{{\sqrt { – \left\{ {{{\left( {x + 3} \right)}^2} – 16} \right\}} }}dx} {/tex}
{tex}= \int {\frac{1}{{\sqrt {{{\left( 16 \right)}} – {{\left( {x + 3} \right)}^2}} }}dx} {/tex}
{tex}= \int {\frac{1}{{\sqrt {{{\left( 4 \right)}^2} – {{\left( {x + 3} \right)}^2}} }}dx} {/tex}
{tex}= {\sin ^{ – 1}}\left( {\frac{{x + 3}}{4}} \right) + c{/tex}
{tex}\left[ {\because \int {\frac{1}{{{a^2} – {x^2}}}dx = {{\sin }^{ – 1}}\frac{x}{a}} } \right]{/tex}
9. {tex}\int {\left( {{x^2} + 1} \right)\log xdx} {/tex}
{tex} = \int {\left( {\log x} \right)\left( {{x^2} + 1} \right)dx} {/tex}
[Applying product rule] {tex}= \log x\left( {\frac{{{x^3}}}{3} + x} \right) – \int {\frac{1}{x}\left( {\frac{{{x^3}}}{3} + x} \right)dx}{/tex}
{tex}= \left( {\frac{{{x^3}}}{3} + x} \right)\log x – \int {\left( {\frac{{{x^2}}}{3} + 1} \right)dx}{/tex}
{tex}= \left( {\frac{{{x^3}}}{3} + x} \right)\log x – \frac{1}{3}\int {{x^2}dx – \int {1dx} } {/tex}
{tex}= \left( {\frac{{{x^3}}}{3} + x} \right)\log x – \frac{1}{3}\frac{{{x^3}}}{3} – x + c{/tex}
{tex}= \left( {\frac{{{x^3}}}{3} + x} \right)\log x – \frac{{{x^3}}}{9} – x + c{/tex}
10. {tex}I = \int_0^{\pi /4} {\frac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} {/tex}
Dividing Numerator and Denominator by {tex}{\cos ^4}x{/tex}
{tex}= \int_0^{\pi /4} {\frac{{\frac{{\sin x.\cos x}}{{{{\cos }^4}x}}}}{{\frac{{{{\cos }^4}x}}{{{{\cos }^4}x}} + \frac{{{{\sin }^4}x}}{{{{\cos }^4}x}}}}} dx{/tex}
{tex}= \int_0^{\pi /4} {\frac{{\tan x.{{\sec }^2}x}}{{1 + {{\tan }^4}x}}} dx{/tex}
{tex}= \int_0^{\pi /4} {\frac{{\tan x.{{\sec }^2}x}}{{1 + {{\left( {{{\tan }^2}x} \right)}^2}}}} dx{/tex}
put {tex}{\tan ^2}x = t{/tex}
{tex}2\tan x.{\sec ^2}xdx = dt{/tex}
when x=0, t=0 and when x={tex}\frac{π}{4}{/tex} t=1
{tex} \therefore I= \frac{1}{2}\int_0^1 {\frac{{dt}}{{1 + {t^2}}}}{/tex}
{tex}= \frac{1}{2}\left[ {{{\tan }^{ – 1}}t} \right]_0^1{/tex}
{tex} = \frac{1}{2}.\frac{\pi }{4} = \frac{\pi }{8}{/tex}
11. Given, {tex}I=\int \frac { \sqrt { 1 – \sin x } } { 1 + \cos x } \cdot e ^ { \frac { – x } { 2 } } d x{/tex}
Let {tex}\frac { – x } { 2 } = t \Rightarrow d x = – 2 d t{/tex}
{tex} I = \int \frac { \sqrt { 1 – \sin ( – 2 t ) } } { 1 + \cos ( – 2 t ) } e ^ { t } ( – 2 d t ){/tex} {tex}[\because x = -2t]{/tex}
{tex}= – 2 \int e ^ { t } \frac { \sqrt { 1 + \sin 2 t } } { 1 + \cos 2 t } d t{/tex}{tex}[ \because \sin ( – \theta ) = – \sin \theta \text { and } \cos ( – \theta ) = \cos \theta ]{/tex}
{tex}= – 2 \int e ^ { t } \frac { \sqrt { cos^2x+sin^2x+2sinx cosx } } { 1 + \cos 2 t } d t [\because cos^2x+sin^2x=1 , sin2x=2sinx cosx ]{/tex}
{tex}= – 2 \int e ^ { t } \left( \frac { \sqrt { ( \cos t + \sin t ) ^ { 2 } } } { 2 \cos ^ { 2 } t } \right) d t [\because (a+b)^2 =a^2+b^2+2ab]{/tex}
{tex}= – 2 \int e ^ { t } \left( \frac { \cos t + \sin t } { 2 \cos ^ { 2 } t } \right) d t{/tex}
{tex}= – \int e ^ { t } ( \sec t + \tan t \sec t ) d t{/tex}{tex}[\because{1\over cosx} = secx , {sinx\over cosx }= tanx]{/tex}
we know that , {tex}\int e ^ { t } \left[ f ( t ) + f ^ { \prime } ( t ) \right] d t = e^tf(t)+C{/tex}
Now, consider {tex}f(t) = sec\ t{/tex}
then {tex}f'(t) = sec\ t\ tan\ t{/tex}
{tex}\therefore I= e ^ { t} \sec t + C{/tex}
{tex}= – e ^ { – x / 2 } \sec \frac { x } { 2 } + C{/tex}{tex}\left[ \because t = \frac { – x } { 2 } \text { and } \sec ( – \theta ) = \sec \theta \right]{/tex}
{tex}I = – e ^ { – x / 2 } \sec \frac { x } { 2 } + C{/tex}
12. {tex}I = \int\limits_0^1 {x\log \left( {1 + 2x} \right)dx} {/tex}
{tex} = \left[ {\log \left( {1 + 2x} \right)\frac{{{x^2}}}{2}} \right]_0^1 – \int {\frac{1}{{1 + 2x}}.2.\frac{{{x^2}}}{2}dx} {/tex}
{tex} = \frac{1}{2}\left[ {{x^2}\log \left( {1 + 2x} \right)} \right]_0^1 – \int {\frac{{{x^2}}}{{1 + 2x}}dx} {/tex}
{tex}= \frac{1}{2}\left[ {\log 3 – 0} \right]_0^1 – \left[ {\int_0^1 {\left( {\frac{x}{2} – \frac{{\frac{x}{2}}}{{1 + 2x}}} \right)} dx} \right]{/tex}
{tex} = \frac{1}{2}\log 3 – \frac{1}{2}\int_0^1 x dx + \frac{1}{2}\int_0^1 {\frac{x}{{1 + 2x}}dx} {/tex}
{tex}= \frac{1}{2}\log 3 – \frac{1}{2}\left[ {\frac{{{x^2}}}{2}} \right]_0^1 + \frac{1}{2}\int_0^1 {\frac{{\frac{1}{2}\left( {2x + 1 – 1} \right)}}{{\left( {2x + 1} \right)}}} dx{/tex}
{tex}= \frac{1}{2}\log 3 – \frac{1}{2}\left[ {\frac{1}{2} – 0} \right] {/tex} {tex}+ \frac{1}{4}\int_0^1 {dx – \frac{1}{4}\int_0^1 {\frac{1}{{1 + 2x}}dx} } {/tex}
{tex}= \frac{1}{2}\log 3 – \frac{1}{4} + \frac{1}{4}\left[ x \right]_0^1 – \frac{1}{8}\left[ {\log \left| {\left( {1 + 2x} \right)} \right|} \right]_0^1{/tex}
{tex} = \frac{1}{2}\log 3 – \frac{1}{4} + \frac{1}{4} – \frac{1}{8}\left[ {\log 3 – \log 1} \right]{/tex}
{tex} = \frac{1}{2}\log 3 – \frac{1}{8}\log 3{/tex}
{tex}= \frac{3}{8}\log 3{/tex}
13. According to the question , {tex}I = \int _ { 1 } ^ { 3 } \left( 2 x ^ { 2 } + 5 x \right) d x{/tex}
On comparing the given integral with {tex} \int _ { a } ^ { b } f ( x ) d x,{/tex} we get
a = 1, b = 3 and f(x) =2x2 + 5x
We know that , {tex} \int_a^b f (x)dx = \mathop {\lim }\limits_{h \to 0} h[f(a) + f(a + h)+ f(a + 2h) + … + f(a + (n – 1)h) \ ]{/tex}…(i)
f(a) = f(1) = 2(1)2 + 5(1) = 2 + 5 = 7
f(a+h) = f(1 + h) = 2(1 + h)2 + 5(1 + h) = 2 + 2h2 + 4h + 5 + 5h = 2h2+ 9h + 7
f(a + 2h) = 2(1 + 2h)2 + 5(1 + 2h) = 2 + 8h2 + 8h + 5 +10h = 8h2 + 18h + 7
so on
f(a+(n-1)h) = f{1 + (n – 1)h} = 2{1 + (n – 1)h}2 + 5{1 + (n – 1)h}2 = 2 + 2(n – 1)2h2 + 4(n – 1)h + 5 + 5(n – 1)h = 2(n – 1)2h2 + 9(n – 1)h + 7
On putting all above values in (i), we get
{tex} {\int_1^3 {\left( {2{x^2} + 5x} \right)} dx = \mathop {\lim }\limits_{h \to 0} }{/tex}h[7 + (2h2 + 9h + 7) +(8h2 + 18h + 7)…….+2(n – 1)2h2 + 9(n – 1)h + 7] On rearranging terms , we get
{tex}= \mathop {\lim }\limits_{h \to 0} {/tex}h[7 + 7+ 7……..+ 7] + {tex} \mathop {\lim }\limits_{h \to 0} {/tex}h[2h2 + 8h2+………+2(n – 1)2h2] + {tex} \mathop {\lim }\limits_{h \to 0} {/tex}h[9h + 18h +……+9(n -1 )h] {tex}= \mathop {\lim }\limits_{h \to 0} {/tex}7nh + {tex} \mathop {\lim }\limits_{h \to 0} {/tex}2h3[ 12 + 22 + 32 +………+(n -1)2] +{tex} \mathop {\lim }\limits_{h \to 0} {/tex}9h2[ 1+ 2 +……..+(n – 1)] {tex} = \mathop {\lim }\limits_{h \to 0} 7(2) + \mathop {\lim }\limits_{h \to 0} \frac{{2{h^3} \cdot n(n – 1)(2n – 1)}}{6}{/tex}{tex} + \mathop {\lim }\limits_{h \to 0} \frac{{9{h^2} \cdot n(n – 1)}}{2}{/tex}{tex}[\because \sum n = \frac {n(n+1)}{2} , \sum n^2 = \frac {n(n+1)(2n+1)}{6}]{/tex}
{tex} = 14 + \mathop {\lim }\limits_{h \to 0} \frac{{nh(nh – h)(2nh – h)}}{3}{/tex}+ {tex}\mathop {\lim }\limits_{h \to 0} \frac{9}{2} \cdot nh(nh – h){/tex}
{tex}= 14 + \frac { 2 ( 2 – 0 ) ( 4 – 0 ) } { 3 } + \frac { 9 } { 2 } \cdot 2 ( 2 – 0 ){/tex}
{tex}= 14 + \frac { 16 } { 3 } + 18 {/tex}
{tex}= \frac { 42 + 16 + 54 } { 3 } {/tex}
{tex}= \frac { 112 } { 3 }{/tex} sq units.

## Chapter Wise Important Questions Class 12 Maths Part I and Part II

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