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)

 

Consider   I = ∫ f(x) dx

Put    x = g(t) so that dx/dt = g′(t)

We write  dx = g′ (t)dt

Thus  I = ∫ f(x) dx= ∫ f(g(t))  g′(t) dt

 

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

Slide21.JPG

Slide22.JPG

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

Slide23.JPG Slide24.JPG

Question 2

∫ tan⁡x  dx is equal to:

(a) sec⁡x + C    

(b) cot⁡x + C  

(c) log⁡|x| + C    

(d) None of these

Slide25.JPG Slide26.JPG

Question 3

∫ 2x/1 + x 2   dx is equal to:

(a) 1 + x 2 + C  

(b) log⁡ |1 + x 2 | + C

(c) log ⁡|2/1 + x 2 | + C 

(d) None of these

Slide27.JPG Slide28.JPG

Question 4

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

(a) cos 2   x (ax + b) + C  

(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

Slide29.JPG Slide30.JPG Slide31.JPG

 

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

Slide33.JPG Slide34.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.