Math Assignment Class XII Ch -07 | Integrals

Extra questions of chapter 07 Integrals class XII with answers and hints to the difficult questions, strictly according to the CBSE Board syllabus. Important and useful math.

MATHEMATICS ASSIGNMENT OF EXTRA QUESTION

STRICTLY ACCORDING TO THE PREVIOUS CBSE BOARD SAMPLE QUESTION PAPERS

FROM 2018 TO 2022

Find the following integrals

Question 1

$\mathbf{Find\: \: \int_{-2}^{2}(x^{3}+1)dx}$

Answer: 4

Question 2

$\mathbf{Find\: \: \int (cos^{2}2x-sin^{2}2x)dx}$

Answer $\mathbf{\frac{sin4x}{4}+C}$

Question 3

$\mathbf{Find\: \:I= \int xe^{(1+x^{2})}dx}$

Answer $\mathbf{\frac{1}{2}e^{(1+x^{2})}+C}$

Question 4

$\mathbf{Find\: \: \int \frac{3+3cosx}{x+sinx}dx}$

Answer $\mathbf{3log|x+sinx|+C}$

Question 5

$\mathbf{Evaluate\: \: I=\int_{-\pi /2}^{\pi /2}x^{2}sinxdx}$

Answer: 0

Solution Hint : x²sinx is an odd function so this integral equal to zero

Question 6

$\mathbf{Find\: \: I=\int e^{x}(1-cotx+cosec^{2}x)dx}$

Answer: I = e^x(1 - cot x) + C

Solution Hint

f(x) = 1 - cot x ⇒ f ' (x) = cosec²x

$\mathbf{Using\: \int e^{x}\left [ f(x)+f'(x) \right ]dx=e^{x}(1-cosx)+C}$

I = e^x(1 - cot x)+C

Question 7

$\mathbf{Find\: \: I=\int \frac{dx}{\sqrt{9-25x^{2}}}}$

Answer $\mathbf{\frac{1}{5}\: sin^{-1}\left ( \frac{5x}{3} \right )+C}$

Question 8

$\mathbf{Find\: \: I=\int \frac{1}{cos^{2}x(1-tanx)^{2}}dx}$

Answer $\mathbf{\frac{1}{1-tanx}+c}$

Solution Hint

$\mathbf{I=\int \frac{sec^{2}x}{(1-tanx)^{2}}dx}$

Now putting 1 - tan x = t

Question 9

$\mathbf{Evauate\: \: I=\int_{0}^{1}x(1-x)^{n}dx}$

Answer $\mathbf{\frac{1}{(n+1)(n+2)}}$

Solution Hint

$\mathbf{Using\: \: \int_{0}^{1}f(x)dx=\int_{0}^{1}f(a-x)dx}$

Question 10

Evaluate: $\mathbf{\int \frac{dx}{\sqrt{3-2x-x^{2}}}}$

Answer $\mathbf{sin^{-1}\left ( \frac{x+1}{2} \right )+C}$

Solution Hint

Taking '-' common from the denominator and then applying method of completing the square.

Question 11

Evaluate: $\mathbf{\int_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{tanx}}}$

Answer: π / 12

Solution Hint:

$\mathbf{I=\int_{\pi/6 }^{\pi/3}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx}$ ......(1)

$\mathbf{Using\; \: \int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx}$

$\mathbf{I=\int_{\pi/6 }^{\pi/3}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt{sinx}}dx}$ ..... (2)

Adding equation (1) and equation (2) we get

$\mathbf{2I = \int_{\pi /6}^{\pi /3}dx=\frac{\pi }{6}\; \: \: \Rightarrow I=\frac{\pi }{12}}$

Question 12

Evaluate: $\mathbf{\int_{0}^{4}|x-1|dx}$

Answer : 5

Solution Hint

x - 1 = 0 when x = 1, so given integral becomes

$\mathbf{I=\int_{0}^{1}-(x-1)dx+\int_{1}^{4}(x-1)dx}$

Integrate and putting the limit we get I = 5

Question 13

$\mathbf{Evaluate\: \: I=\int_{1}^{3}|x^{2}-2x|dx}$

Answer: 2

Question 14

$\mathbf{Evaluate\: \: I=\int \frac{Logx}{(1+logx)^{2}}dx}$

Answer $\mathbf{I=\frac{x}{1+logx}+C}$

Solution Hint

$\mathbf{I=\int \frac{Logx+1-1}{(1+logx)^{2}}dx}$

$\mathbf{I=\int \frac{Logx+1}{(1+logx)^{2}}-\int \frac{1}{\left ( 1+logx \right )^{2}}dx}$

$\mathbf{I=\int \frac{1}{(1+logx)}-\int \frac{1}{\left ( 1+logx \right )^{2}}dx}$

Now integrating the first integral by parts we get

$\mathbf{I=\frac{1}{1+logx}\int 1dx-\int \left ( \frac{d}{dx}\left ( \frac{1}{1+logx} \right )\int 1dx \right )dx-\int \frac{1}{(1+logx)^{2}}dx}$

$\mathbf{I=\frac{x}{1+logx}+\int \left ( \frac{1}{\left ( 1+logx \right )^{2}}\times \frac{1}{x}\times x \right )dx -\int \frac{1}{(1+logx)^{2}}dx}$

$\mathbf{I=\frac{x}{1+logx}+\int \left ( \frac{1}{\left ( 1+logx \right )^{2}} \right )dx -\int \frac{1}{(1+logx)^{2}}dx+C}$

$\mathbf{I=\frac{x}{1+logx}+C}$

Question 15

$\mathbf{Evaluate\: \: I=\int \frac{sin2x}{\sqrt{9-cos^{4}x}}dx}$ ...... (1)

Answer $\mathbf{ I=-sin^{-1}\left ( \frac{cos^{2}x}{3} \right )+c}$

Solution Hint

Putting cos²x = t ⇒ - 2 cosx sinx dx = dt ⇒ Sin2x dx = - dt

Putting these values in equation (1) we get

$\mathbf{ I=\int \frac{-dt}{\sqrt{9-t^{2}}}dx}$

$\mathbf{ I=\int \frac{-dt}{\sqrt{3^{2}-t^{2}}}dx=-sin^{-1}\left ( \frac{t}{3} \right )+c}$

$\mathbf{ I=-sin^{-1}\left ( \frac{cos^{2}x}{3} \right )+c}$

Question 16

$\mathbf{Evaluate\: \: I=\int_{-1}^{2}|x^{3}-3x^{2}+2x|dx}$

Answer : I = 11/4

Solution Hint: Factorise the given function we get

$\mathbf{ I=\int_{-1}^{2}|x(x-1)(x-2)|dx}$

x(x - 1)(x - 2) = 0 ⇒ x = 0, 1, 2

Given integral can be written as

$\mathbf{ I=\int_{-1}^{0}|x^{3}-3x^{2}+2x|dx+\int_{0}^{1}|x^{3}-3x^{2}+2x|dx+\int_{1}^{2}|x^{3}-3x^{2}+2x|dx}$

$\mathbf{ I=-\int_{-1}^{0}\left ( x^{3}-3x^{2}+2x \right )dx+\int_{0}^{1}\left ( x^{3}-3x^{2}+2x \right )dx-\int_{1}^{2}\left ( x^{3}-3x^{2}+2x \right )dx}$

Find the integrals and putting the limits we get

$\mathbf{I=\frac{9}{4}+\frac{1}{4}+\frac{1}{4}=\frac{11}{4}}$

Question 17

$\mathbf{Find\: \: I=\int \frac{x^{2}+1}{(x^{2}+2)(x^{2}+3)}dx}$

Answer $\mathbf{-\frac{1}{\sqrt{2}}tan^{-1}\left ( \frac{x}{\sqrt{2}} \right )+\frac{2}{\sqrt{3}}tan^{-1}\left ( \frac{x}{\sqrt{3}} \right )+C}$

Solution Hint:

Putting x² = t then use partial fraction and then integrate

Question 18

$\mathbf{Find: \: \:I= \int \frac{(x^{2}+sin^{2}x)sec^{2}x}{1+x^{2}}dx}$

Answer: tan x - tan^-1x + C

Solution Hint:

Adding and subtracting '1' in the numerator we get

$\mathbf{I= \int \frac{(x^{2}+1+sin^{2}x-1)sec^{2}x}{1+x^{2}}dx}$

Separating the denominator we get

$\mathbf{I= \int \left ( 1-\frac{cos^{2}x}{1+x^{2}} \right ) sec^{2}xdx}$

$\mathbf{I= \int \left ( sec^{2}x-\frac{1}{1+x^{2}} \right ) dx}$

I = tan x - tan^-1x + C

Question 19

$\mathbf{Find\: \: I=\int \frac{e^{x}(x-3)}{(x-1)^{3}}dx}$

Answer $\mathbf{I=\frac{e^{x}}{(x-1)^{2}}+C}$

Solution Hint

$\mathbf{I=\int \left [ \frac{e^{x}(x-1)-2}{(x-1)^{3}} \right ]dx}$

Separating the denominator we get

$\mathbf{I=\int \left [ \frac{e^{x}}{(x-1)^{2}}-\frac{2}{(x-1)^{3}} \right ]dx}$

$\mathbf{I=\frac{e^{x}}{(x-1)^{2}}+C}$

Question 20

$\mathbf{Find\: \: I=\int \frac{(x^{4}-x)^{1/4}}{x^{5}}dx}$

Answer $\mathbf{\Rightarrow I=\frac{4}{15}\left ( 1-\frac{1}{x^{3}} \right )^{5/4}+C}$

Solution Hint

Taking x⁴ common from the numerator we get

$\mathbf{I=\int x\left ( 1-\frac{x}{x^{4}} \right )^{1/4}\times \frac{1}{x^{5}}dx}$

$\mathbf{I=\int \left ( 1-\frac{1}{x^{3}} \right )^{1/4}\times \frac{1}{x^{4}}dx}$ ............. (1)

$\mathbf{Putting\: \: \left ( 1-\frac{1}{x^{3}} \right )^{1/4}=t}$

$\mathbf{\Rightarrow \left ( 1-\frac{1}{x^{3}} \right )=t^{4}}$

Differentiating both side we get

$\mathbf{\frac{dx}{x^{4}}=\frac{4}{3}t^{3}dt}$

Putting all these values in eqn. (1) we get

$\mathbf{I=\frac{4}{3}\int t^{4}dt=\frac{4}{3}\times \frac{t^{5}}{5}+C}$

$\mathbf{\Rightarrow I=\frac{4}{15}\left ( 1-\frac{1}{x^{3}} \right )^{5/4}+C}$

Question 21

$\mathbf{Evaluate\: \: I=\int_{-1}^{1}\frac{x+|x|+1}{x^{2}+2|x|+1}dx}$

Answer: 2log2

Solution Hint

$\mathbf{ I=\int_{-1}^{1}\frac{x}{x^{2}+2|x|+1}dx+\int_{-1}^{1}\frac{|x|+1}{x^{2}+2|x|+1}dx}$

I = I₁ + I₂

Since I₁is an odd function so I₁ = 0

$\mathbf{ I=I_{2}=\int_{-1}^{1}\frac{|x|+1}{x^{2}+2|x|+1}dx}$

Since is an even function so

$\mathbf{ I=I_{2}=2\int_{0}^{1}\frac{x+1}{x^{2}+2x+1}dx}$

$\mathbf{ \Rightarrow I=2\int_{0}^{1}\frac{x+1}{(x+1)^{2}}dx=2\int_{0}^{1}\frac{1}{(x+1)}dx}$

$\mathbf{ \Rightarrow I=\left [ 2log|x+1| \right ]_{0}^{1}}$

⇒ I = 2log2

Question 22

$\mathbf{Evaluate\: \: I=\int \frac{x^{4}+1}{x(x^{2}+1)^{2}}dx}$

Answer $\mathbf{I=log\: x+\frac{2}{x^{2}+1}+C}$

Solution Hint:

Multiply and divide by x we get

$\mathbf{I=\int \frac{\left ( x^{4}+1 \right )x}{x^{2}(x^{2}+1)^{2}}dx}$

Now putting x² = t ⇒ 2xdx = dt ⇒ xdx = dt/2 we get

$\mathbf{I=\frac{1}{2}\int \frac{t^{2}+1}{t(t+1)^{2}}dt}$

Using partial fraction to solve this integral

$\mathbf{ \frac{t^{2}+1}{t(t+1)^{2}}=\frac{A}{t}+\frac{B}{t+1}+\frac{C}{(t+1)^{2}}}$

Solving these fractions we get, A = 1, B = 0, C = - 2 and the given integral becomes

$\mathbf{I=\frac{1}{2}\int \left ( \frac{1}{t}+\frac{0}{t+1}+\frac{-2}{(t+1)^{2}} \right )dx}$

$\mathbf{I=\frac{1}{2}\left ( logt+\frac{4}{t+1} \right )+C}$

$\mathbf{I=\frac{1}{2}log\: t+\frac{2}{t+1}+C}$

$\mathbf{I=\frac{1}{2}log\: x^{2}+\frac{2}{x^{2}+1}+C}$

$\mathbf{I=log\: x+\frac{2}{x^{2}+1}+C}$

Question 23

$\mathbf{Evaluate\: \:I= \int e^{x}\left ( \frac{\sqrt{1+sin2x}}{1+cos2x} \right )dx}$

Answer: $\mathbf{e^{x}\: secx+C}$

Solution Hint

$\mathbf{I= \int e^{x} \left [ \frac{\sqrt{\left ( sinx+cosx \right )^{2}}}{1+cos2x} \right ] dx}$

$\mathbf{I= \int e^{x} \left [ \frac{ sinx+cosx }{2cos^{2}x} \right ] dx}$

$\mathbf{I= \frac{1}{2}\int e^{x} \left [ \frac{ sinx }{cos^{2}x}+\frac{cosx}{cos^{2}x} \right ] dx}$

$\mathbf{I= \frac{1}{2}\int e^{x} \left [ secx\: tanx+secx \right ] dx}$

$\mathbf{I= \frac{1}{2}\int e^{x} \left [ secx+secx\: tanx \right ] dx}$

$\mathbf{Using\: \: \int e^{x}\left [ f(x)+f'(x) \right ]=e^{x}f(x)+C}$

$\mathbf{=\frac{1}{2}e^{x}\: secx+C}$

Question 24

$\mathbf{Find\: \:I= \int \frac{secx}{1+cosecx}dx}$

Answer

$\mathbf{I=\frac{1}{4}log(1+sinx)+\frac{1}{2(1+sinx)}-\frac{1}{4}log(1-sinx)+C}$

Solution Hint

$\mathbf{I= \int \frac{secx}{1+cosecx}dx}$

$\mathbf{I= \int \frac{sinx}{cosx(1+sinx)}dx}$

$\mathbf{I= \int \frac{sinx\: cosx}{cos^{2}x(1+sinx)}dx}$

$\mathbf{I= \int \frac{sinx\: cosx}{\left ( 1-sin^{2}x \right )(1+sinx)}dx}$

$\mathbf{I= \int \frac{sinx\: cosx}{\left ( 1+sinx \right )^{2}(1-sinx)}dx}$

Putting sin x = t ⇒ cosx dx = dt we get

$\mathbf{I=\int \frac{tdt}{(1+t)^{2}(1-t)}}$

Now using partial fraction we get

$\mathbf{I=\frac{1}{4}\int \frac{dt}{1+t}-\frac{1}{2}\int \frac{dt}{(1+t)^{2}}+\frac{1}{4}\int \frac{dt}{1-t}}$

$\mathbf{I=\frac{1}{4}log(1+t)+\frac{1}{2(1+t)}-\frac{1}{4}log(1-t)+C}$

Now putting t = sinx we get

$\mathbf{I=\frac{1}{4}log(1+sinx)+\frac{1}{2(1+sinx)}-\frac{1}{4}log(1-sinx)+C}$

Question 25

$\mathbf{Evaluate:\: \: I=\int_{-\pi /4}^{\pi /4}\left ( \frac{x+\frac{\pi }{4}}{2-cos2x} \right )dx}$

Answer $\mathbf{I=\frac{\sqrt{3}}{18}\pi ^{2}}$

Solution Hint

$\mathbf{I=\int_{-\pi /4}^{\pi /4}\left ( \frac{x}{2-cos2x} \right )dx+\frac{\pi }{4}\int_{-\pi /4}^{\pi /4}\left ( \frac{1}{2-cos2x} \right )dx}$

$\mathbf{Let\: f(x)= \frac{x}{2-cos2x}}$

$\mathbf{\Rightarrow f(-x)= \frac{-x}{2-cos2x} =-f(x)}$

⇒ f(x) = -f(x) ⇒ f(x) is an odd function so

$\mathbf{\int_{-\pi /4}^{\pi /4}\left ( \frac{x}{2-cos2x} \right )dx\: =0}$

$\mathbf{g(x)=\frac{\pi }{4}\left ( \frac{1}{2-cos2x} \right )}$

$\mathbf{g(-x)=\frac{\pi }{4}\left ( \frac{1}{2-cos2x} \right )=g(x)}$

⇒ g(-x) = g(x) ⇒ g(x) is an even function so

$\mathbf{\int_{-\pi /4}^{\pi /4}\left ( \frac{1}{2-cos2x} \right )dx=2\int_{0}^{\pi /4}\left ( \frac{1}{2-cos2x} \right )dx}$

Now given integral becomes

$\mathbf{I=0+\frac{\pi }{4}\times 2\int_{0}^{\pi /4}\left ( \frac{1}{2-cos2x} \right )dx}$

$\mathbf{I=\frac{\pi }{2}\int_{0}^{\pi /4}\left ( \frac{1}{1+1-cos2x} \right )dx}$

$\mathbf{I=\frac{\pi }{2}\int_{0}^{\pi /4}\left ( \frac{1}{1+2sin^{2}2x} \right )dx}$

Divide numerator and denominator by Cos²x we get

$\mathbf{I=\frac{\pi }{2}\int_{0}^{\pi /4}\left ( \frac{sec^{2}x}{sec^{2}x+2tan^{2}x} \right )dx}$

$\mathbf{I=\frac{\pi }{2}\int_{0}^{\pi /4}\left ( \frac{sec^{2}x}{1+tan^{2}x+2tan^{2}x} \right )dx}$

$\mathbf{I=\frac{\pi }{2}\int_{0}^{\pi /4}\left ( \frac{sec^{2}x}{1+3tan^{2}x} \right )dx}$

Putting tan x = t ⇒ sec²x dx = dt we get

$\mathbf{I=\frac{\pi }{2}\int_{0}^{1}\left ( \frac{dt}{1+3t^{2}} \right )}$

Integrating this and putting the limit we get

$\mathbf{I=\frac{\sqrt{3}}{18}\pi ^{2}}$

Question 26

$\mathbf{Evaluate\: \: I=\int_{0}^{1}tan(sin^{-1}x)dx}$

Ans: 1

Solution Hint

$\mathbf{Using\: \: sin^{-1}x=tan^{-1}\left ( \frac{x}{\sqrt{1-x^{2}}} \right )}$

$\mathbf{I=\int_{0}^{1}tan\left [ tan^{-1}\left ( \frac{x}{\sqrt{1-x^{2}}} \right ) \right ]dx}$

$\mathbf{I=\int_{0}^{1}\frac{x}{\sqrt{1-x^{2}}}dx=-\left [ \sqrt{1-x^{2}} \right ]_{0}^{1}\textrm{}=1}$

Question 27

$\mathbf{Evaluate\: \: I=\int \frac{e^{x}}{x+1}[1+(x+1)log(x+1)]dx}$

Ans: $\mathbf{\frac{e^{x}}{x+1}+C}$

Question 28

$\mathbf{Evaluate\: \: I=\int_{\pi /6}^{\pi /3}\frac{sinx+cosx}{\sqrt{sin2x}}dx}$

$\mathbf{Answer:\;I=2sin^{-1}\left ( \frac{\sqrt{3}-1}{2} \right )}$

Solution Hint

Putting sin x - cos x = t ⇒ (sin x + cos x)dx = dt

Squaring sin x - cos x = t on both side we get: sin2x = 1- t²

Making all substitution we get

$\mathbf{\int \frac{dt}{\sqrt{1-t^{2}}}=sin^{-1}t=sin^{-1}(sinx-cosx)}$

$\mathbf{I=\int_{\pi /6}^{\pi /3}\frac{sinx+cosx}{\sqrt{sin2x}}dx=\left [ sin^{-1}(sinx-cosx) \right ]_{\pi /6}^{\pi /3}\textrm{}}$

$\mathbf{I=2sin^{-1}\left ( \frac{\sqrt{3}-1}{2} \right )}$

Question 29

$\mathbf{Evaluate:\: \: I=\int \frac{x}{(x-1)^{2}(x+2)}dx}$

Answer

$\mathbf{\frac{2}{9}log|x-1|-\frac{1}{3(x-1)}-\frac{2}{9}log|x+2|+C}$

Question 30

$\mathbf{ Evaluate\;\;I=\int_{0}^{3}x[x]dx}$

Solution :

$\mathbf{\int_{0}^{3}x[x]dx=\int_{0}^{1}x\times 0dx+\int_{1}^{2}x\times 1dx+\int_{2}^{3}x\times 2dx}$

$\mathbf{\int_{0}^{3}x[x]dx=1\int_{1}^{2}xdx+2\int_{2}^{3}xdx}$

$\mathbf{\int_{0}^{3}x[x]dx=\frac{13}{2}}$

Question 31

Find $\mathbf{\int x\sqrt{1+2x}dx}$

Answer: Putting 1 + 2x = t². then integrating w r t x we get

$\mathbf{\frac{(1+2x)^{5/2}}{10}-\frac{(1+2x)^{3/2}}{6}+C}$

Search This Blog

Featured Posts

Math Assignment Class XII Ch-8 | Applications of Integrations