CBSE Class 12 Mathematics Integrals MCQs Set 06

Answer: (a) 1

Explanation:
1. \( \int \log x\,dx = x\log x - x \).
2. \( [x\log x - x]_{1}^{e} = (e - e) - (0 - 1) = 1 \).

Teacher's Note:
1. Use log e = 1 and log 1 = 0 when evaluating at the limits.
2. ∫ log x dx = x log x − x comes from integrating by parts with 1 as the second function.

Answer: (c) \( \frac{\pi}{2} \)

Explanation:
1. \( x - x^2 = \frac{1}{4} - \left(x - \frac{1}{2}\right)^2 \), so \( \int \frac{dx}{\sqrt{x - x^2}} = \sin^{-1}(2x - 1) \).
2. \( [\sin^{-1}(2x - 1)]_{0}^{1/2} = \sin^{-1} 0 - \sin^{-1}(-1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \).

Teacher's Note:
1. Completing the square converts √(x − x²) into the standard form √(a² − u²).
2. The integrand is unbounded at x = 0 but the integral is finite, as the arcsine antiderivative shows.

Answer: (b) 104

Explanation:
1. \( \int (3x - 2)^2\,dx = \frac{(3x - 2)^3}{9} \).
2. \( \left[\frac{(3x - 2)^3}{9}\right]_{2}^{4} = \frac{1000 - 64}{9} = \frac{936}{9} = 104 \).

Teacher's Note:
1. For a power of a linear expression, divide by the new power and by the coefficient of x.
2. Expanding to 9x² − 12x + 4 gives the same answer and is a good check.

Answer: (a) 20/3

Explanation:
1. \( \int \left(x^{1/2} + x^{-1/2}\right)dx = \frac{2}{3}x^{3/2} + 2x^{1/2} \).
2. At 4: \( \frac{16}{3} + 4 = \frac{28}{3} \); at 1: \( \frac{2}{3} + 2 = \frac{8}{3} \).
3. Value \( = \frac{28}{3} - \frac{8}{3} = \frac{20}{3} \).

Teacher's Note:
1. Write roots as fractional powers before integrating.
2. √4 = 2 and 4^(3/2) = 8; evaluating powers carefully avoids most errors here.

Answer: (c) f(2) − f(1)

Explanation:
1. Since f′ = φ, f is an antiderivative of φ.
2. By the second fundamental theorem, \( \int_{1}^{2} \phi(x)\,dx = f(2) - f(1) \).

Teacher's Note:
1. The definite integral is the antiderivative at the upper limit minus its value at the lower limit.
2. Options with φ(2) − φ(1) confuse the integrand with its antiderivative.

Answer: (b) \( \frac{\pi a^2}{4} \)

Explanation:
1. Put x = a sin θ: \( \int_{0}^{\pi/2} a\cos\theta\cdot a\cos\theta\,d\theta = a^2\int_{0}^{\pi/2} \cos^2\theta\,d\theta \).
2. \( = a^2\cdot\frac{\pi}{4} = \frac{\pi a^2}{4} \).

Teacher's Note:
1. Geometrically, this is the area of a quarter circle of radius a, i.e. πa²/4.
2. The answer must have the dimension of a², which rules out πa/4 and πa/2.

Answer: (c) 2

Explanation:
1. \( 1 + \sin 2x = (\sin x + \cos x)^2 \) and sin x + cos x ≥ 0 on [0, π/2].
2. \( \int_{0}^{\pi/2} (\sin x + \cos x)\,dx = [\sin x - \cos x]_{0}^{\pi/2} = 1 - (-1) = 2 \).

Teacher's Note:
1. √(1 + sin 2x) = |sin x + cos x|; the modulus can be dropped only where the bracket is non-negative.
2. Both ∫₀^(π/2) sin x and ∫₀^(π/2) cos x equal 1, so the sum is 2.

Answer: (c) 1/2

Explanation:
1. \( 1 + \cos 2x = 2\cos^2 x \), so the integrand is \( \frac{1}{2}\sec^2 x \).
2. \( \frac{1}{2}[\tan x]_{0}^{\pi/4} = \frac{1}{2}(1 - 0) = \frac{1}{2} \).

Teacher's Note:
1. Use 1 + cos 2x = 2cos²x and 1 − cos 2x = 2sin²x to simplify such denominators.
2. The integrand is positive, so the negative options are ruled out immediately.

Answer: (b) \( I_2 \gt I_1 \)

Explanation:
1. For 1 ≤ x ≤ 2, \( \sqrt{1 + x^2} > \sqrt{x^2} = x \), so \( \frac{1}{\sqrt{1 + x^2}} < \frac{1}{x} \).
2. Integrating over [1, 2] preserves the inequality: \( I_1 < I_2 \), i.e. \( I_2 > I_1 \).

Teacher's Note:
1. To compare two integrals over the same interval, compare the integrands pointwise.
2. Numerically, \( I_1 = \log\frac{2 + \sqrt{5}}{1 + \sqrt{2}} \approx 0.56 \) while \( I_2 = \log 2 \approx 0.69 \).

Answer: (c) 19

Explanation:
1. \( \int_{-2}^{4} f = \int_{-2}^{-1} + \int_{-1}^{0} + \int_{0}^{1} + \int_{1}^{2} + \int_{2}^{3} + \int_{3}^{4} \).
2. Each piece from n to n + 1 equals n², with n = −2, −1, 0, 1, 2, 3.
3. Sum = 4 + 1 + 0 + 1 + 4 + 9 = 19.

Teacher's Note:
1. Break the interval at integers so that each piece matches the given formula.
2. The last piece is from 3 to 4, so n goes only up to 3, not 4; this is the usual trap.

Answer: (c) 116/15

Explanation:
1. Put t = 1 + x (x = t − 1): \( \int_{1}^{4} (t - 1)\sqrt{t}\,dt = \int_{1}^{4} (t^{3/2} - t^{1/2})\,dt \).
2. \( = \left[\frac{2}{5}t^{5/2} - \frac{2}{3}t^{3/2}\right]_{1}^{4} = \left(\frac{64}{5} - \frac{16}{3}\right) - \left(\frac{2}{5} - \frac{2}{3}\right) \).
3. \( = \frac{112}{15} + \frac{4}{15} = \frac{116}{15} \).

Teacher's Note:
1. Substitute for the expression under the root and write x in terms of the new variable.
2. Change the limits too: x = 0, 3 give t = 1, 4.

Answer: (b) \( \frac{2\sqrt{3}}{3} + \frac{1}{2} \log 3 \)

Explanation:
1. Write x + 2 = (x + 1) + 1 and \( x^2 + 2x - 3 = (x + 1)^2 - 4 \).
2. Antiderivative: \( \sqrt{x^2 + 2x - 3} + \log\left|x + 1 + \sqrt{x^2 + 2x - 3}\right| \).
3. At \( x = \frac{4\sqrt{3}}{3} - 1 \): x + 1 = 4/√3 and the root is \( \sqrt{\frac{16}{3} - 4} = \frac{2}{\sqrt{3}} \), giving \( \frac{2}{\sqrt{3}} + \log\frac{6}{\sqrt{3}} = \frac{2\sqrt{3}}{3} + \log 2\sqrt{3} \).
4. At x = 1: 0 + log 2. Value \( = \frac{2\sqrt{3}}{3} + \log\sqrt{3} = \frac{2\sqrt{3}}{3} + \frac{1}{2}\log 3 \).

Teacher's Note:
1. For (linear)/√(quadratic), split the numerator into a multiple of the derivative of the quadratic plus a constant.
2. Use \( \int \frac{du}{\sqrt{u^2 - a^2}} = \log\left|u + \sqrt{u^2 - a^2}\right| \) for the constant part.

Answer: (c) 1/3

Explanation:
1. Minors of the first row: \( M_{11} = 3 - 4\sin^2 x \), \( M_{12} = (3 + 4\sin x) - 4\sin x(1 + \sin x) = 3 - 4\sin^2 x \), \( M_{13} = (3 + 4\sin x)\sin x - 3(1 + \sin x) = 4\sin^2 x - 3 \).
2. Expand along R₁: \( f(x) = (\sin x + \sin 2x + \sin 3x)(3 - 4\sin^2 x) - \sin 2x(3 - 4\sin^2 x) + \sin 3x(4\sin^2 x - 3) \).
3. The sin 2x and sin 3x terms cancel, leaving \( f(x) = \sin x(3 - 4\sin^2 x) = 3\sin x - 4\sin^3 x = \sin 3x \).
4. \( \int_{0}^{\pi/2} \sin 3x\,dx = \left[-\frac{\cos 3x}{3}\right]_{0}^{\pi/2} = 0 + \frac{1}{3} = \frac{1}{3} \).

Teacher's Note:
1. Expand the determinant to a single function of x before integrating; here two of the three terms cancel.
2. 3 sin x − 4 sin³x = sin 3x is the triple-angle formula that makes the integral immediate.

Answer: (a) \( p = -\frac{5}{6}, q = 1, r = \frac{1}{3} \)

Explanation:
1. Partial fractions: \( \frac{x^2 + 1}{(2x + 1)(x - 1)(x + 1)} = -\frac{5/3}{2x + 1} + \frac{1/3}{x - 1} + \frac{1}{x + 1} \).
2. Integrate: \( -\frac{5}{6}\log(2x + 1) + \frac{1}{3}\log(x - 1) + \log(x + 1) \) from 2 to 3.
3. \( = -\frac{5}{6}\log\frac{7}{5} + \frac{1}{3}\log 2 + \log\frac{4}{3} \).
4. So \( p = -\frac{5}{6}, q = 1, r = \frac{1}{3} \).

Teacher's Note:
1. Find each partial-fraction coefficient by the cover-up rule: put x = −1/2, 1, −1 in turn.
2. For \( \frac{1}{2x + 1} \) the integral is \( \frac{1}{2}\log(2x + 1) \); the factor ½ is easy to miss.

Answer: (a) 2

Explanation:
1. \( \int_{1}^{a} \left(\frac{3}{2}\sqrt{x} + 1 - \frac{1}{\sqrt{x}}\right)dx = [x^{3/2} + x - 2\sqrt{x}]_{1}^{a} = a\sqrt{a} + a - 2\sqrt{a} \).
2. Dividing by √a: the condition is a + √a − 2 < 4, i.e. \( (\sqrt{a} + 3)(\sqrt{a} - 2) < 0 \), so √a < 2, i.e. 0 < a < 4.
3. Among the options only a = 2 lies in (0, 4).

Teacher's Note:
1. Treat √a as a new unknown; the inequality becomes a quadratic in √a.
2. a = 4 gives exactly 4, which fails the strict inequality.

Answer: (b) 1

Explanation:
1. \( \cos 3x + 1 = 4\cos^3 x - 3\cos x + 1 = (\cos x + 1)(2\cos x - 1)^2 \).
2. So the integrand is \( (\cos x + 1)(2\cos x - 1) = 2\cos^2 x + \cos x - 1 = \cos 2x + \cos x \).
3. \( \int_{0}^{\pi/2} (\cos 2x + \cos x)\,dx = \left[\frac{\sin 2x}{2} + \sin x\right]_{0}^{\pi/2} = 0 + 1 = 1 \).

Teacher's Note:
1. Factorising 4c³ − 3c + 1 using the root c = −1 (and the repeated root c = ½) cancels the denominator.
2. After cancellation, rewrite 2cos²x − 1 as cos 2x for a one-step integration.

Answer: (b) \( 2\log_2 (2/a) \)

Explanation:
1. log₂a is a constant. \( \int_{0}^{2} (t - \log_2 a)\,dt = \left[\frac{t^2}{2}\right]_{0}^{2} - 2\log_2 a = 2 - 2\log_2 a \).
2. \( = 2(\log_2 2 - \log_2 a) = 2\log_2\frac{2}{a} \).

Teacher's Note:
1. Any expression not containing the variable of integration (here t) behaves as a constant.
2. Write 1 as log₂2 to combine the logs into a single term.

Answer: (b) \( \frac{32\sqrt{2}}{\pi^3} \)

Explanation:
1. \( \frac{d}{dx}\left[x^3\sin\frac{1}{x}\right] = 3x^2\sin\frac{1}{x} + x^3\cos\frac{1}{x}\cdot\left(-\frac{1}{x^2}\right) = 3x^2\sin\frac{1}{x} - x\cos\frac{1}{x} \), the integrand.
2. Value \( = \left[x^3\sin\frac{1}{x}\right]_{0}^{4/\pi} = \frac{64}{\pi^3}\sin\frac{\pi}{4} - 0 \) (the term → 0 as x → 0).
3. \( = \frac{64}{\pi^3}\cdot\frac{\sqrt{2}}{2} = \frac{32\sqrt{2}}{\pi^3} \).

Teacher's Note:
1. Look for an integrand that is the derivative of a product; the factor 3x² next to x hints at x³.
2. x³ sin(1/x) → 0 as x → 0 because sin(1/x) is bounded.

Answer: (b) |b| − |a|

Explanation:
1. For x < 0, |x| = −x, so \( \frac{x}{|x|} = -1 \) on [a, b].
2. \( \int_{a}^{b} (-1)\,dx = -(b - a) = a - b \).
3. Since a, b < 0: a = −|a| and b = −|b|, so a − b = |b| − |a|.

Teacher's Note:
1. x/|x| is the signum function: −1 for negative x and +1 for positive x.
2. Express the answer with moduli carefully: for a negative number, a = −|a|.

Answer: (c) \( -\pi/3 \)

Explanation:
1. \( \int_{\alpha}^{2\alpha} \cos 2x\,dx = \frac{\sin 4\alpha - \sin 2\alpha}{2} = \cos 3\alpha\sin\alpha \).
2. So \( \sin\alpha(1 + \cos 3\alpha) = 0 \). In (−π, 0), sin α ≠ 0, so cos 3α = −1.
3. 3α ∈ (−3π, 0) gives 3α = −π, i.e. \( \alpha = -\frac{\pi}{3} \).

Teacher's Note:
1. Use sin C − sin D = 2 cos((C + D)/2) sin((C − D)/2) to factor the result.
2. Check the interval: 3α = −3π would give α = −π, which is excluded from the open interval.

Answer: (a) 3

Explanation:
1. \( \int_{0}^{36} \frac{dx}{2x + 9} = \frac{1}{2}[\log(2x + 9)]_{0}^{36} = \frac{1}{2}\log\frac{81}{9} \).
2. \( = \frac{1}{2}\log 9 = \log 3 \), so k = 3.

Teacher's Note:
1. For 1/(ax + b), the antiderivative is (1/a) log(ax + b); keep the factor 1/a.
2. Forgetting the ½ would give k = 9, which is one of the distractors.

Answer: (c) \( \pi/4 \)

Explanation:
1. Numerator \( = \left[\frac{t^2}{2}\right]_{\pi/2}^{x} = \frac{x^2}{2} - \frac{\pi^2}{8} \), which → 0; the denominator also → 0.
2. By L'Hospital's rule, the limit \( = \lim_{x \to \pi/2} \frac{x}{2\cos(2x - \pi)} = \frac{\pi/2}{2\cdot 1} = \frac{\pi}{4} \).

Teacher's Note:
1. The derivative of \( \int_{a}^{x} f(t)\,dt \) with respect to x is f(x), which makes L'Hospital's rule quick here.
2. Confirm the 0/0 form before differentiating top and bottom.

Answer: (a) 2

Explanation:
1. By parts: \( \int_{0}^{1} t f''(t)\,dt = [t f'(t)]_{0}^{1} - \int_{0}^{1} f'(t)\,dt = f'(1) - [f(1) - f(0)] \).
2. \( = 1 - f(1) + 1 = 2 - f(1) = 0 \), so f(1) = 2.

Teacher's Note:
1. When an integrand contains f″ multiplied by a polynomial, integrate by parts to bring in f′ and f.
2. The term t f′(t) vanishes at t = 0 because of the factor t.

Answer: (b) \( \frac{\pi}{4} \)

Explanation:
1. Put t = cos x, dt = −sin x dx; x = 0 → t = 1, x = π/2 → t = 0.
2. \( \int_{1}^{0} \frac{-dt}{1 + t^2} = \int_{0}^{1} \frac{dt}{1 + t^2} = [\tan^{-1} t]_{0}^{1} = \frac{\pi}{4} \).

Teacher's Note:
1. sin x dx in the numerator signals the substitution t = cos x.
2. The minus sign from dt is absorbed by swapping the limits.

Answer: (d) \( \frac{1}{3}(1 - \cos a^3) \)

Explanation:
1. Put t = x³, dt = 3x² dx; limits 0 to a³.
2. \( \frac{1}{3}\int_{0}^{a^3} \sin t\,dt = \frac{1}{3}[-\cos t]_{0}^{a^3} = \frac{1}{3}(1 - \cos a^3) \).

Teacher's Note:
1. When the derivative of the inner function (3x² for x³) appears as a factor, substitute the inner function.
2. Remember cos 0 = 1, which gives the 1 inside the bracket.

Answer: (a) \( \frac{\pi}{2} \)

Explanation:
1. Divide numerator and denominator by cos⁴x: \( \int_{0}^{\pi/4} \frac{\sec^4 x\,dx}{1 - \tan^2 x + \tan^4 x} \).
2. With t = tan x, sec²x dx = dt and sec²x = 1 + t²: \( \int_{0}^{1} \frac{1 + t^2}{t^4 - t^2 + 1}dt = \int_{0}^{1} \frac{1 + 1/t^2}{(t - 1/t)^2 + 1}dt \).
3. Put u = t − 1/t (u: −∞ → 0): \( \int_{-\infty}^{0} \frac{du}{u^2 + 1} = 0 - \left(-\frac{\pi}{2}\right) = \frac{\pi}{2} \).

Teacher's Note:
1. For integrands in even powers of sin and cos, divide by cos⁴x (or cos²x) and substitute t = tan x.
2. Dividing by t² and writing the denominator in terms of t − 1/t is the standard trick for (1 + t²)/(t⁴ + kt² + 1).

Answer: (b) 2

Explanation:
1. \( \frac{\sqrt{\tan x}}{\sin x\cos x} = \frac{\sqrt{\tan x}\,\sec^2 x}{\tan x} = \frac{\sec^2 x}{\sqrt{\tan x}} \).
2. With t = tan x: \( \int_{0}^{1} t^{-1/2}\,dt = [2\sqrt{t}]_{0}^{1} = 2 \).

Teacher's Note:
1. Writing sin x cos x = tan x · cos²x brings in sec²x, which pairs with t = tan x.
2. The integrand blows up at x = 0, but the integral is still finite (t^(−1/2) is integrable at 0).

Answer: (d) \( \frac{\sqrt{e} - 1}{e} \)

Explanation:
1. Put t = −1/x, so \( dt = \frac{dx}{x^2} \); x = 1 → t = −1, x = 2 → t = −1/2.
2. \( \int_{-1}^{-1/2} e^t\,dt = e^{-1/2} - e^{-1} = \frac{1}{\sqrt{e}} - \frac{1}{e} = \frac{\sqrt{e} - 1}{e} \).

Teacher's Note:
1. 1/x² is the derivative of −1/x, which is exactly the exponent; substitute the exponent.
2. Bring both terms to the denominator e: \( \frac{1}{\sqrt{e}} = \frac{\sqrt{e}}{e} \).

Answer: (d) \( \frac{\pi^2}{32} \)

Explanation:
1. Put t = tan⁻¹x, \( dt = \frac{dx}{1 + x^2} \); limits 0 to π/4.
2. \( \int_{0}^{\pi/4} t\,dt = \frac{1}{2}\left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{32} \).

Teacher's Note:
1. f(x)·f′(x) integrates to f(x)²/2; here f = tan⁻¹x.
2. Squaring π/4 gives π²/16, then halve it; stopping at π²/16 is the common slip.

Answer: (b) 1

Explanation:
1. First integral \( = \frac{1}{2}\log\frac{1 + \tan^2 x}{1 + 1/e^2} \).
2. In the second, \( \frac{1}{t(1 + t^2)} = \frac{1}{t} - \frac{t}{1 + t^2} \), giving \( \left[\log t - \frac{1}{2}\log(1 + t^2)\right] = \frac{1}{2}\log\frac{t^2}{1 + t^2} \); from 1/e to cot x: \( \frac{1}{2}\log\frac{\cos^2 x}{1/(e^2 + 1)} \).
3. Sum \( = \frac{1}{2}\log\left(\sec^2 x\cos^2 x\cdot\frac{e^2}{e^2 + 1}\cdot(e^2 + 1)\right) = \frac{1}{2}\log e^2 = 1 \).

Teacher's Note:
1. The answer does not depend on x; you can check by putting x = π/4, where both upper limits equal 1.
2. Substituting t = 1/u in the second integral turns it into the first integrand with limits tan x to e, so the total is \( \int_{1/e}^{e} \frac{t\,dt}{1 + t^2} \) with the same value 1.

Answer: (d) 1

Explanation:
1. Put t = 1/x, \( dt = -\frac{dx}{x^2} \); x = 1/π → t = π, x = 2/π → t = π/2.
2. \( \int_{\pi}^{\pi/2} -\sin t\,dt = \int_{\pi/2}^{\pi} \sin t\,dt = [-\cos t]_{\pi/2}^{\pi} = 1 - 0 = 1 \).

Teacher's Note:
1. 1/x² dx is −d(1/x), so the substitution t = 1/x is natural.
2. Swapping limits removes the minus sign; keep track of it carefully.

Answer: (d) None of these

Explanation:
1. Put t = 3 + 4 sin x, dt = 4 cos x dx; limits 3 to \( 3 + 2\sqrt{3} \).
2. \( \frac{1}{4}\int_{3}^{3 + 2\sqrt{3}} \frac{dt}{t} = \frac{1}{4}\log\frac{3 + 2\sqrt{3}}{3} \).
3. The options have denominator 2 inside the log, which differs from 3, so the answer is none of these.

Teacher's Note:
1. When the numerator is a constant multiple of the derivative of the denominator, the integral is a logarithm.
2. Compute the new upper limit carefully: \( 4\sin\frac{\pi}{3} = 2\sqrt{3} \).

Answer: (b) \( \log\left(\frac{1 + e}{2e}\right) - \frac{1}{e} + 1 \)

Explanation:
1. Put t = eˣ, dx = dt/t: \( \int_{1}^{e} \frac{dt}{t^2(1 + t)} \).
2. \( \frac{1}{t^2(1 + t)} = \frac{1}{t^2} - \frac{1}{t} + \frac{1}{1 + t} \), so the integral \( = \left[-\frac{1}{t} - \log t + \log(1 + t)\right]_{1}^{e} \).
3. \( = \left(-\frac{1}{e} - 1 + \log(1 + e)\right) - (-1 + \log 2) = \log\frac{1 + e}{2e} - \frac{1}{e} + 1 \).

Teacher's Note:
1. After t = eˣ, check that dx = dt/t; the extra t in the denominator is easy to drop.
2. Combine logs at the end: \( \log(1 + e) - 1 - \log 2 = \log\frac{1 + e}{2e} \).

Answer: (b) \( 4 - \pi \)

Explanation:
1. Put t² = eˣ − 1, so 2t dt = eˣ dx and eˣ + 3 = t² + 4; limits 0 to 2.
2. \( \int_{0}^{2} \frac{2t^2}{t^2 + 4}dt = 2\int_{0}^{2} \left(1 - \frac{4}{t^2 + 4}\right)dt = 2\left[t - 2\tan^{-1}\frac{t}{2}\right]_{0}^{2} \).
3. \( = 2\left(2 - 2\cdot\frac{\pi}{4}\right) = 4 - \pi \).

Teacher's Note:
1. Substituting the whole square root as t removes the radical completely.
2. At x = log 5, eˣ = 5, so t = √4 = 2.

Answer: (a) \( I_1 = I_2 \)

Explanation:
1. In \( I_1 \) put x = eᵗ, dx = eᵗ dt; x = e → t = 1, x = e² → t = 2.
2. \( I_1 = \int_{1}^{2} \frac{e^t}{t}\,dt = I_2 \).

Teacher's Note:
1. A substitution that matches the integrand and limits shows two integrals are equal without evaluating them.
2. Neither integral has an elementary antiderivative, so comparison by substitution is the only route.

Answer: (b) \( \frac{1}{6} \)

Explanation:
1. Put t = cos x, dt = −sin x dx; limits 1 to √3/2.
2. \( \int_{\sqrt{3}/2}^{1} t^{-3}\,dt = \left[-\frac{1}{2t^2}\right]_{\sqrt{3}/2}^{1} = -\frac{1}{2} + \frac{1}{2}\cdot\frac{4}{3} = \frac{1}{6} \).

Teacher's Note:
1. Equivalently, \( \frac{\sin x}{\cos^3 x} = \tan x\sec^2 x \), whose integral is \( \frac{\tan^2 x}{2} \): \( \frac{1}{2}\cdot\frac{1}{3} = \frac{1}{6} \).
2. Choosing t = tan x often gives the shortest working for tan·sec² patterns.

Answer: (a) \( \frac{2}{\log 3}(3^{\sqrt{2}} - 1) \)

Explanation:
1. Put t = √x, \( dt = \frac{dx}{2\sqrt{x}} \); limits 0 to √2.
2. \( 2\int_{0}^{\sqrt{2}} 3^t\,dt = 2\left[\frac{3^t}{\log 3}\right]_{0}^{\sqrt{2}} = \frac{2}{\log 3}(3^{\sqrt{2}} - 1) \).

Teacher's Note:
1. The factor \( \frac{1}{\sqrt{x}} \) beside a function of √x signals t = √x.
2. Use \( \int a^t\,dt = \frac{a^t}{\log a} \); at t = 0 the value is 1/log 3, not 0.

Answer: (d) \( \frac{\pi}{2ab} \)

Explanation:
1. Divide numerator and denominator by cos²x: \( \int_{0}^{\pi/2} \frac{\sec^2 x\,dx}{a^2 + b^2\tan^2 x} \).
2. Put t = tan x (t: 0 → ∞): \( \int_{0}^{\infty} \frac{dt}{a^2 + b^2 t^2} = \frac{1}{b^2}\cdot\frac{b}{a}\left[\tan^{-1}\frac{bt}{a}\right]_{0}^{\infty} \).
3. \( = \frac{1}{ab}\cdot\frac{\pi}{2} = \frac{\pi}{2ab} \).

Teacher's Note:
1. For a cos²x + b sin²x type denominators, divide by cos²x and use t = tan x.
2. Check with a = b = 1: the integral is ∫₀^(π/2) dx = π/2, which matches π/(2ab).

Answer: (d) None of these

Explanation:
1. Complete the square: \( zx - z^2 = \frac{x^2}{4} - \left(z - \frac{x}{2}\right)^2 \), so \( I_1 = e^{x^2/4}\int_{-x/2}^{x/2} e^{-u^2}\,du \) with u = z − x/2.
2. The integrand is even, so \( I_1 = 2e^{x^2/4}\int_{0}^{x/2} e^{-u^2}\,du \); put u = w/2: \( = e^{x^2/4}\int_{0}^{x} e^{-w^2/4}\,dw = e^{x^2/4} I_2 \).
3. Since the factor is \( e^{x^2/4} \), none of (a), (b), (c) is correct.

Teacher's Note:
1. Completing the square in the exponent is the standard way to handle e^(linear + quadratic).
2. An answer can be checked numerically: at x = 1.3, I₁/I₂ ≈ 1.53 = e^(0.4225), not eˣ or e^(x²/2).

Answer: (a) \( (\log x)^2 \)

Explanation:
1. log(x²) = 2 log x, so the integrand is \( \frac{2\log t}{t} \) (writing t for the variable of integration).
2. Put u = log t: \( \int_{0}^{\log x} 2u\,du = (\log x)^2 \).

Teacher's Note:
1. The variable inside the integral is a dummy; the upper limit x is a separate quantity.
2. f′/f-type and f·f′-type patterns: here log t and 1/t form f·f′.

Answer: (d) \( \tan^{-1} e - \frac{\pi}{4} \)

Explanation:
1. \( \frac{1}{e^x + e^{-x}} = \frac{e^x}{e^{2x} + 1} \); put t = eˣ (t: 1 → e).
2. \( \int_{1}^{e} \frac{dt}{1 + t^2} = \tan^{-1} e - \tan^{-1} 1 = \tan^{-1} e - \frac{\pi}{4} \).

Teacher's Note:
1. Multiplying numerator and denominator by eˣ converts the expression into the form \( \frac{f'}{1 + f^2} \).
2. This also equals \( \tan^{-1}\frac{e - 1}{e + 1} \), a positive number, so option (a), which is negative, is wrong.

Answer: (c) 0

Explanation:
1. Replace x by π/2 − x: sin x and cos x swap, so the integrand becomes \( (\cos x - \sin x)\log(\cos x + \sin x) \).
2. That is the negative of the original integrand, so I = −I.
3. Hence 2I = 0, i.e. I = 0.

Teacher's Note:
1. The property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx often shows an integral equals its own negative.
2. Alternatively, put t = sin x + cos x: dt = (cos x − sin x)dx and both limits give t = 1, so the value is 0.

Answer: (b) \( \log\left(\frac{9}{8}\right) \)

Explanation:
1. Put t = cos x, dt = −sin x dx (t: 1 → 0): \( \int_{0}^{1} \frac{t\,dt}{(t + 1)(t + 2)} \).
2. \( \frac{t}{(t + 1)(t + 2)} = \frac{2}{t + 2} - \frac{1}{t + 1} \).
3. \( [2\log(t + 2) - \log(t + 1)]_{0}^{1} = (2\log 3 - \log 2) - 2\log 2 = \log\frac{9}{8} \).

Teacher's Note:
1. Factorise cos²x + 3cos x + 2 = (cos x + 1)(cos x + 2) before using partial fractions.
2. The integrand is positive on (0, π/2), so log(8/9), which is negative, is impossible.