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### Math Assignment Class XII Ch -07 | Integrals

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# 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. assignment for the students of class XII*

**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

Question 1

**Answer: 4**

**Question 2**

**Answer **

**Question 3**

**Answer **

**Question 4**

**Answer **

**Question 5**

**Answer: 0**

**Solution Hint : x**

^{2}sinx is an odd function so this integral equal to zero**Question 6**

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

**Solution Hint**

**f(x) = 1 - cot x ⇒ f ' (x) = cosec ^{2}x **

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

**Question 7**

**Answer **

**Question 8**

**Answer **

**Solution Hint**

**Now putting 1 - tan x = t**

**Question 9**

**Answer **

**Solution Hint**

**Question 10**

**Evaluate: **

**Answer **

**Solution Hint**

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

**Question 11**

**Evaluate: **

**Answer: Ï€ / 12**

**Solution Hint:**

** ......(1) **

**..... (2)**

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

**Question 12**

**Evaluate: **

**Answer : 5**

**Solution Hint**

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

**Integrate and putting the limit we get I = 5**

**Question 13**

**Answer: 2**

**Question 14**

**Answer **

**Solution Hint**

** **

**Now integrating the first integral by parts we get**

**Question 15**

** ...... (1)**

**Answer **

**Solution Hint**

**Putting cos ^{2}x = t ⇒ **

**- 2 cosx sinx dx = dt**

**⇒**

**Sin2x dx = - dt**

**Putting these values in equation (1) we get**

**Question 16**

**Answer **

**Solution Hint: ****Factorise the given function we get **

**x(x - 1)(x - 2) = 0 ⇒ x = 0, 1, 2**

**Given integral can be written as**

**Find the integrals and putting the limits we get**

**Question 17**

**Answer **

**Solution Hint:**

**Putting x ^{2} = t then use partial fraction and then integrate**

**Question 18**

**Answer: tan x - tan**

^{-1}x + C**Solution Hint:**

**Adding and subtracting '1' in the numerator we get**

**Separating the denominator we get**

**I = tan x - tan**

^{-1}x + C**Question 19**

**Answer **

**Solution Hint**

**Separating the denominator we get**

**Question 20**

**Answer **

**Solution Hint**

**Taking x ^{4} common from the numerator we get**

** ............. (1)**

** **

**Differentiating both side we get**

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

**Question 21**

**Answer: 2log2**

**Solution Hint**

**I = I _{1} + I_{2} **

**Since I**

_{1 }is an odd function so I_{1}= 0

**Since is an even function so**

**⇒ I = 2log2**

**Question 22**

**Answer**

**Solution Hint:**

**Multiply and divide by x we get**

**Now putting x**

^{2}= t ⇒ 2xdx = dt ⇒ xdx = dt/2 we get

**Using partial fraction to solve this integral**

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

**Question 23**

**Answer:**

**Solution Hint**

**Question 24**

**Answer**

**Solution Hint**

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

**Now using partial fraction we get**

**Now putting t = sinx we get**

**Question 25**

**Answer**

**Solution Hint**

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

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

**Now given integral becomes**

**Divide numerator and denominator by Cos**

^{2}x we get

**Putting tan x = t ⇒ sec ^{2}x dx
= dt we get**

**Integrating this and putting the limit we get**

**Question 26**

**Ans: 1**

**Solution Hint**

**Question 27**

**Ans:**

**Question 28**

**Ans: sin ^{-1}(sin x - cos x)**

**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- t ^{2}**

**Making all substitution we get**

**Question 29**

**Answer**

**Question 30**

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### Comments

Keep going . Very good work.

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