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JEE Exam  >  JEE Tests  >  Mathematics (Maths) for JEE Main & Advanced  >  Test: Integrals- 2 - JEE MCQ

Test: Integrals- 2 - JEE MCQ


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25 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Integrals- 2

Test: Integrals- 2 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Integrals- 2 questions and answers have been prepared according to the JEE exam syllabus.The Test: Integrals- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Integrals- 2 below.
Solutions of Test: Integrals- 2 questions in English are available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & Test: Integrals- 2 solutions in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Integrals- 2 | 25 questions in 25 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
Test: Integrals- 2 - Question 1
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One value of ∫ f'(x) dx is 

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As (∫f'(x)dx) = f(x) + C, therefore, one value of (∫ f'(x) dx) is f(x)

Test: Integrals- 2 - Question 2
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∫ log x dx is equal to 

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Test: Integrals- 2 - Question 3
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Test: Integrals- 2 - Question 6
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if ∫ g(x) dx = g(x), then  is equal to 
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Test: Integrals- 2 - Question 7
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If f (x) be a function such that 

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∫ log xdx = xlog x - x = x(log x -1)

 = x(log x - log e) = 

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Test: Integrals- 2 - Question 9
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Dividing numerator and denominator by cos2x and substituting tanx = t , we get :

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Test: Integrals- 2 - Question 11
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∫ sec2 x cosec2 x dx is equal to 
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Test: Integrals- 2 - Question 15
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Note that sin9x is an odd function, therefore, 

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Test: Integrals- 2 - Question 17
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∫ log(1-x) dx - ∫ log x dx = 0

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Test: Integrals- 2 - Question 21
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∫ (tan x + cot x) dx is equal to 
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Correct Answer :- c

Explanation : ∫(tanx + cotx)dx

= ∫tanxdx + ∫cotxdx

= ∫sinx/cosx dx + ∫cosx/sinx dx

I1 = sinx/cosx

Put t = cosx

dt = -sinx dx

I2 = cosx/sinx

Put t = sinx

dt = cosx dx

So, we get

=> ∫-dt/t + ∫dt/t

=> - ln t + ln t + c

=> -ln cosx + ln sinx + c

=> log (sinx/cosx) + c

=> log(tanx)

Test: Integrals- 2 - Question 22
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 ∫ log(log x) + (log x)-1) dx is equal to 
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