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Chapter 7 Class 12 Integrals
Serial order wise

The given integral ∫ f(x) dx can be transformed into another form by changing the independent variable x to t by substituting x=g(t)

Β 

Consider   I = ∫ f(x) dx

Put Β  Β x = g(t) so that dx/dt = gβ€²(t)

We writeΒ  dx = gβ€² (t)dt

ThusΒ  I = ∫ f(x) dx= ∫ f(g(t))Β  gβ€²(t) dt

Β 

This change of variable formula is one of the important tools available to us in the name of integration by substitution.

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Question 1

∫ sin⁑ (tan -1 x)/(1 + x 2 ) dx is equal to:

(a) -sin⁑(tan -1 ⁑x + C 

(b) -cos⁑(tan -1 ⁑x + C

(c) tan⁑x + C 

(d) None of these

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Question 2

∫ tan⁑x  dx is equal to:

(a) sec⁑x + C    

(b) cot⁑x + C  

(c) log⁑|x| + C    

(d) None of these

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Question 3

∫ 2x/1 + x 2   dx is equal to:

(a) 1 + x 2 + C Β 

(b) log⁑ |1 + x 2 | + C

(c) log ⁑|2/1 + x 2 | + C 

(d) None of these

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Question 4

∫ sin⁑ (ax+b)  cos⁑ (ax+b)  dx is equal to:

(a) cos 2 Β  x (ax + b) + C Β 

(b) sin 2 ⁑(ax + b)

(c) (-1)/4a  cos⁑2  (ax + b) + C 

(d) None of these

This question is inspired from Ex 7.2, 5 - Chapter 7 Class 12 - Integrals

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Question 5

∫ 1/x + x log⁑x  dx is equal to:

(a) |1 + log⁑x | + C 

(b) log⁑|1 + log⁑x| + C

(c) log⁑x + C 

(d) None of these

This question is inspired from Ex 7.2, 3 - Chapter 7 Class 12 - Integrals

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Transcript

Question 3 The given integral ∫1▒〖𝑓(π‘₯)γ€— 𝑑π‘₯ can be transformed into another form by changing the independent variable π‘₯ to 𝑑 by substituting π‘₯=𝑔(𝑑) Consider I = ∫1▒〖𝑓(π‘₯)γ€— 𝑑π‘₯ Put π‘₯ = 𝑔(𝑑) so that 𝑑π‘₯/𝑑𝑑=𝑔′(𝑑) We write 𝑑π‘₯=𝑔^β€² (𝑑)𝑑𝑑 Thus I = ∫1▒〖𝑓(π‘₯)γ€— 𝑑π‘₯= ∫1▒〖𝑓(𝑔(𝑑)) 𝑔^β€² (𝑑) 𝑑𝑑〗 This change of variable formula is one of the important tools available to us in the name of integration by substitution. For example: ∫1β–’γ€–2π‘₯ sin⁑〖(π‘₯^2+1)γ€— 𝑑π‘₯γ€— Put π‘₯^2+1=𝑑 2π‘₯𝑑π‘₯=𝑑𝑑 Thus, ∫1β–’γ€–sin⁑〖(𝑑) 𝑑𝑑〗=βˆ’cos⁑(𝑑)+𝐢〗 = βˆ’cos⁑〖(π‘₯^2+1)+𝐢〗 Based on the above information, answer any four of the following questions. Question 1 ∫1β–’γ€–sin⁑〖(tan^(βˆ’1)⁑π‘₯)γ€—/(1 + π‘₯^2 ) 𝑑π‘₯γ€— is equal to: (a) βˆ’sin⁑(tan^(βˆ’1)⁑π‘₯ )+𝐢 (b) βˆ’cos⁑(tan^(βˆ’1)⁑π‘₯ )+𝐢 (c) tan⁑π‘₯+𝐢 (d) None of these ∫1β–’γ€–sin⁑〖(tan^(βˆ’1)⁑π‘₯)γ€—/(1 + π‘₯^2 ) 𝑑π‘₯γ€— Let γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑𝒙=𝒕 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝟏/(𝟏 + 𝒙^𝟐 )= 𝒅𝒕/𝒅𝒙 𝑑π‘₯/(1 + π‘₯^2 )=𝑑𝑑 Now, Integrating the function ∫1β–’γ€–π’”π’Šπ’β‘γ€–(〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑𝒙)γ€—/(𝟏 + 𝒙^𝟐 ) 𝒅𝒙〗 Putting tan^(βˆ’1)⁑π‘₯=𝑑 & 𝑑π‘₯/(1 + π‘₯^2 )=𝑑𝑑 = ∫1▒〖𝐬𝐒𝐧⁑𝒕 . 𝒅𝒕〗 = βˆ’cos⁑𝑑+𝐢 Putting back t = tan^(βˆ’1)⁑π‘₯ = βˆ’πœπ¨π¬β‘γ€–(𝒕𝒂𝒏^(βˆ’πŸ) 𝒙)γ€—+π‘ͺ So, the correct answer is (b) Question 2 ∫1β–’γ€–tan⁑π‘₯ 𝑑π‘₯γ€— is equal to: (a) sec⁑π‘₯+𝐢 (b) cot⁑π‘₯+𝐢 (c) log⁑〖|π‘₯|γ€—+𝐢 (d) None of these ∫1β–’γ€–tan⁑π‘₯ 𝑑π‘₯γ€—=∫1β–’γ€–sin⁑π‘₯/cos⁑π‘₯ 𝑑π‘₯γ€— Let 𝒄𝒐𝒔⁑𝒙=𝒕 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ βˆ’π’”π’Šπ’ 𝒙= 𝒅𝒕/𝒅𝒙 sin x 𝑑π‘₯=βˆ’π‘‘π‘‘ Now, ∫1β–’(π’”π’Šπ’ 𝒙)/(𝒄𝒐𝒔 𝒙) 𝒅𝒙 Putting π‘π‘œπ‘ β‘π‘₯=𝑑 & sin⁑π‘₯ 𝑑π‘₯=𝑑𝑑 = ∫1β–’γ€–πŸ/𝒕 . (βˆ’π’…π’•)γ€— = βˆ’βˆ«1▒𝒅𝒕/𝒕 = βˆ’log⁑〖|𝑑|γ€—+𝐢 Putting back t = cos⁑π‘₯ = βˆ’π₯π¨π’ˆ |𝒄𝒐𝒔 𝒙|+π‘ͺ = lo𝑔 |π‘π‘œπ‘  π‘₯|^(βˆ’1)+𝐢 = lo𝑔 1/(|π‘π‘œπ‘  π‘₯|)+𝐢 = π₯π¨π’ˆ |𝒔𝒆𝒄 𝒙|+π‘ͺ So, the correct answer is (d) Question 3 ∫1β–’γ€–2π‘₯/(1 + π‘₯^2 ) 𝑑π‘₯γ€— is equal to: (a) 1+π‘₯^2+𝐢 (b) log⁑〖|1+π‘₯^2 |+𝐢〗 (c) log⁑|2/(1 + π‘₯^2 )|+𝐢 (d) None of these ∫1β–’γ€–2π‘₯/(1 + π‘₯^2 ) 𝑑π‘₯γ€— Let 𝟏+𝒙^𝟐=𝒕 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ πŸπ’™= 𝒅𝒕/𝒅𝒙 2π‘₯ 𝑑π‘₯=𝑑𝑑 Now, Integrating the function ∫1β–’γ€–πŸπ’™/(𝟏 + 𝒙^𝟐 ) 𝒅𝒙〗 Putting 1+π‘₯^2=𝑑 & 2π‘₯𝑑π‘₯=𝑑𝑑 = ∫1β–’γ€–πŸ/𝒕 . 𝒅𝒕〗 = log⁑〖|𝑑|γ€—+𝐢 Putting back t = 1+π‘₯^2 = π₯π’π’ˆ|𝟏+𝒙^𝟐 |+π‘ͺ So, the correct answer is (b) Question 4 ∫1β–’γ€–sin⁑〖(π‘Žπ‘₯+𝑏)γ€— cos⁑〖(π‘Žπ‘₯+𝑏)γ€— 𝑑π‘₯γ€— is equal to: (a) cos^2 π‘₯ (π‘Žπ‘₯+𝑏)+𝐢 (b) sin^2⁑〖(π‘Žπ‘₯+𝑏)γ€— (c) (βˆ’1)/4π‘Ž cos⁑2 (π‘Žπ‘₯+𝑏)+𝐢 (d) None of these Taking the given function π’”π’Šπ’β‘(𝒂𝒙 + 𝒃) 𝒄𝒐𝒔⁑(𝒂𝒙 + 𝒃) = 1/2 sin⁑(2(π‘Žπ‘₯+𝑏)) = 𝟏/𝟐 π’”π’Šπ’β‘(πŸπ’‚π’™+πŸπ’ƒ) Let (πŸπ’‚π’™+πŸπ’ƒ)=𝒕 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 2π‘Ž+0=𝑑𝑑/𝑑π‘₯ 2π‘Ž=𝑑𝑑/𝑑π‘₯ 2π‘Ž.𝑑π‘₯=𝑑𝑑 𝒅𝒙=𝒅𝒕/πŸπ’‚ Integrating the function ∫1β–’γ€–" " sin⁑(π‘Žπ‘₯+𝑏) cos⁑(π‘Žπ‘₯+𝑏)" " γ€—.𝑑π‘₯ = 1/2 ∫1β–’γ€–sin⁑(2π‘Žπ‘₯+2𝑏)γ€— .𝑑π‘₯ Putting 𝑑=2π‘Žπ‘₯+2𝑏 & 𝑑π‘₯=𝑑𝑑/2π‘Ž = 𝟏/𝟐 ∫1β–’γ€–π’”π’Šπ’ 𝒕〗 .𝒅𝒕/πŸπ’‚ = 1/4π‘Ž ∫1β–’γ€–sin⁑(𝑑)γ€— .𝑑𝑑 = 1/4π‘Ž [βˆ’cos⁑𝑑+𝐢1] = βˆ’ 1/4π‘Ž . cos⁑𝑑 + 𝐢1/4π‘Ž = βˆ’ 𝟏/πŸ’π’‚ . 𝒄𝒐𝒔⁑𝒕 + π‘ͺ Putting back 𝑑=2π‘Žπ‘₯+2𝑏 = βˆ’ 1/4π‘Ž . cos⁑〖(2π‘Žπ‘₯+2𝑏)γ€— + 𝐢 = βˆ’ 𝟏/πŸ’π’‚ . π’„π’π’”β‘γ€–πŸ(𝒂𝒙+𝒃)γ€—+π‘ͺ So, the correct answer is (c) Question 5 ∫1β–’γ€–1/(π‘₯ + π‘₯ log⁑π‘₯ ) 𝑑π‘₯γ€— is equal to: (a) |1+log⁑π‘₯ |+𝐢 (b) log⁑〖|1+log⁑π‘₯ |+𝐢〗 (c) log⁑π‘₯+𝐢 (d) None of these ∫1β–’γ€–1/(π‘₯ + π‘₯ log⁑π‘₯ ) 𝑑π‘₯γ€— = ∫1β–’γ€–1/(π‘₯(1 + log⁑π‘₯)) 𝑑π‘₯γ€— Let 1 + log 𝒙= 𝒕 Differentiate 𝑀.π‘Ÿ.𝑑.π‘₯ 0 + 𝑑𝑑/𝑑π‘₯= 1/π‘₯ 0 + 𝑑𝑑/𝑑π‘₯= 1/π‘₯ 𝒅𝒕.𝒙=𝒅𝒙 Now, ∫1β–’1/(π‘₯ + π‘₯ log⁑π‘₯ ) .𝑑π‘₯ =∫1β–’1/(π‘₯ (1 + log⁑π‘₯ ) ) .𝑑π‘₯" " Putting 1+log⁑π‘₯ & 𝑑π‘₯=π‘₯ 𝑑𝑑 = ∫1β–’1/(π‘₯(𝑑)) 𝑑𝑑.π‘₯ = ∫1β–’1/𝑑 𝑑𝑑 = π‘™π‘œπ‘”|𝑑|+𝐢 Putting back 𝑑 =1+π‘™π‘œπ‘”β‘π‘₯ = π’π’π’ˆ|𝟏+π₯𝐨𝐠⁑𝒙 |+π‘ͺ So, the correct answer is (b)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.