Chapter 1: CET
Part A: Questions 1 to 4 - Student-Friendly Solutions
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This document gives the question, correct option, and step-by-step explanation in exam-ready format. Focus on the standard integration formulas and substitution/integration-by-parts methods. |
Quick Formula Revision
1. ∫(1/x) dx = log x + C
2. ∫sec²x dx = tan x + C
3. 1 - cos 2x = 2sin²x
4. ∫cosec²x dx = -cot x + C
5. ∫tan⁻¹x dx = x tan⁻¹x - ½ log(1 + x²) + C
Question 1
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Evaluate: ∫(1/x + sec²x) dx = ________ |
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(A) log x - tan x + C |
(B) log x + tan x + C |
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(C) -1/x² + sec x + C |
(D) log x + sec x tan x + C |
Correct Answer: (B) log x + tan x + C
Explanation
Split the integral term by term: ∫(1/x + sec²x) dx = ∫(1/x) dx + ∫sec²x dx.
Use the standard formula: ∫(1/x) dx = log x + C.
Use another standard formula: ∫sec²x dx = tan x + C.
Therefore, ∫(1/x + sec²x) dx = log x + tan x + C.
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Exam Tip: When the integrand is a sum, integrate each term separately. Do not differentiate sec x tan x here; that belongs to derivative formulas. |
Question 2
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Evaluate: ∫ 1/(1 - cos 2x) dx = ________ |
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(A) cot x/2 + C |
(B) (cosec x · cot x)/2 + C |
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(C) -(cosec x · cot x)/2 + C |
(D) -cot x/2 + C |
Correct Answer: (D) -cot x/2 + C
Explanation
Use the trigonometric identity: 1 - cos 2x = 2sin²x.
So, 1/(1 - cos 2x) = 1/(2sin²x) = ½ cosec²x.
Now integrate: ∫1/(1 - cos 2x) dx = ½∫cosec²x dx.
Since ∫cosec²x dx = -cot x + C, the answer is -cot x/2 + C.
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Exam Tip: For questions involving 1 - cos 2x or 1 + cos 2x, first convert using double-angle identities. |
Question 3
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Evaluate: ∫(2x² + 4x + 5)(x + 1) dx = ________ |
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(A) (2x² + 4x + 5)²/8 + C |
(B) (x + 1)/8 + C |
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(C) (2x² + 4x + 5)/8 + C |
(D) (2x² + 4x + 5)²/2 + C |
Correct Answer: (A) (2x² + 4x + 5)²/8 + C
Explanation
Observe the inside expression: u = 2x² + 4x + 5.
Differentiate it: du/dx = 4x + 4 = 4(x + 1). Hence, (x + 1)dx = du/4.
Substitute in the integral: ∫(2x² + 4x + 5)(x + 1) dx = ∫u · (du/4).
This becomes ¼∫u du = u²/8 + C.
Replace u by the original expression: (2x² + 4x + 5)²/8 + C.
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Exam Tip: Look for a function multiplied by something proportional to its derivative. This is a common substitution pattern. |
Question 4
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Evaluate: ∫tan⁻¹x dx = ________ |
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(A) x tan⁻¹x + ½ log(1 + x²) + C |
(B) 1/(1 + x²) + C |
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(C) x tan⁻¹x - ½ log(1 + x²) + C |
(D) tan⁻¹x + tan⁻¹x + C |
Correct Answer: (C) x tan⁻¹x - ½ log(1 + x²) + C
Explanation
Use integration by parts: ∫u dv = uv - ∫v du.
Take u = tan⁻¹x and dv = dx. Then du = 1/(1 + x²) dx and v = x.
So, ∫tan⁻¹x dx = x tan⁻¹x - ∫x/(1 + x²) dx.
For the remaining integral, take t = 1 + x². Then dt = 2x dx, so x dx = dt/2.
Hence, ∫x/(1 + x²) dx = ½∫dt/t = ½ log(1 + x²).
Therefore, ∫tan⁻¹x dx = x tan⁻¹x - ½ log(1 + x²) + C.
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Exam Tip: For inverse trigonometric functions alone, integration by parts is usually the fastest method. |
Final Answer Key
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Question |
Correct Option |
Answer |
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1 |
B |
log x + tan x + C |
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2 |
D |
-cot x/2 + C |
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3 |
A |
(2x² + 4x + 5)²/8 + C |
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4 |
C |
x tan⁻¹x - ½ log(1 + x²) + C |
