allery∫ 1/(1 + x²) dx =
- x²/2
- ln(1 + x²)
- arctan x + C
- tan x
Show answer and solution
Correct answer: C. arctan x + C
∫ dx/(1 + x²) = arctan x + C (= tan⁻¹ x + C). Standard result.
Used in trig substitution x = tan θ.
JEE Advanced Unit 6B: Definite Integration & Riemann Sums — practice questions — page 2
Explore 42 free Unit 6B: Definite Integration & Riemann Sums questions in JEE Advanced Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 31–42 of 42.
∫ 1/(1 + x²) dx =Read the answer and worked solution →∫ from 0 to π sin²x dx =Read the answer and worked solution →∫ x sin x dx (integration by parts):Read the answer and worked solution →∫ sec²x dx =Read the answer and worked solution →Area under y = x² from x = 0 to x = 2:Read the answer and worked solution →∫ from 0 to ∞ e^(-x) dx =Read the answer and worked solution →∫ x ln x dx (integration by parts: u = ln x, dv = x dx):Read the answer and worked solution →∫ from 1 to e (1/x) dx =Read the answer and worked solution →Reduction formula example: ∫ sinⁿ x dx (for n ≥ 2) =Read the answer and worked solution →∫ from 0 to 2 (x² - 2x) dx =Read the answer and worked solution →For ∫ dx / sqrt(a² - x²), let x = a sin θ. Result:Read the answer and worked solution →∫ from 0 to π/2 sin x cos x dx =Read the answer and worked solution →
Worked practice questions
∫ from 0 to π sin²x dx =
- π/2 (use sin²x = (1 - cos 2x)/2)
- 2
- π/4
- π
Show answer and solution
Correct answer: A. π/2 (use sin²x = (1 - cos 2x)/2)
sin²x = (1 - cos 2x)/2. ∫₀^π sin²x dx = (1/2) ∫₀^π (1 - cos 2x) dx = (1/2)[π - 0] = π/2.
∫ x sin x dx (integration by parts):
- sin x - x cos x + C
- -x cos x + sin x + C
- x cos x
- cos x
Show answer and solution
Correct answer: B. -x cos x + sin x + C
u = x, du = dx; dv = sin x dx, v = -cos x.
We have ∫ x sin x dx = -x cos x + ∫ cos x.
So dx = -x cos x + sin x + C.
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