CBSE Class 12 Mathematics Integrals MCQs Set 08

Answer: (b) \( \pi/2\sqrt{2} \)

Explanation:
1. Put t = sin x, dt = cos x dx; limits 0 to 1/√2.
2. \( \int_{0}^{1/\sqrt{2}} \frac{dt}{\sqrt{1 - 2t^2}} = \frac{1}{\sqrt{2}}[\sin^{-1}(\sqrt{2}\,t)]_{0}^{1/\sqrt{2}} \).
3. \( = \frac{1}{\sqrt{2}}\cdot\frac{\pi}{2} = \frac{\pi}{2\sqrt{2}} \).

Teacher's Note:
1. For \( \frac{1}{\sqrt{1 - k^2t^2}} \), the antiderivative is \( \frac{1}{k}\sin^{-1}(kt) \).
2. Note 1 − 2sin²x = cos 2x, which shows the integrand is defined on [0, π/4).

Answer: (c) \( \frac{1}{4} \log 3 \)

Explanation:
1. Put t = sin x − cos x: dt = (cos x + sin x)dx and t² = 1 − sin 2x, so 3 + sin 2x = 4 − t²; t: −1 → 0.
2. \( \int_{-1}^{0} \frac{dt}{4 - t^2} = \frac{1}{4}\left[\log\frac{2 + t}{2 - t}\right]_{-1}^{0} \).
3. \( = \frac{1}{4}\left(0 - \log\frac{1}{3}\right) = \frac{1}{4}\log 3 \).

Teacher's Note:
1. With sin x + cos x in the numerator, use t = sin x − cos x; with sin x − cos x, use t = sin x + cos x.
2. Use \( \int \frac{dt}{a^2 - t^2} = \frac{1}{2a}\log\frac{a + t}{a - t} \) with a = 2.

Answer: (d) \( \frac{1}{4} \log 5 \)

Explanation:
1. Put t = 1 + 4cos²x: dt = −8 cos x sin x dx = −4 sin 2x dx; t: 5 → 1.
2. \( I = \frac{1}{4}\int_{1}^{5} \frac{dt}{t} = \frac{1}{4}\log 5 \).

Teacher's Note:
1. The numerator sin 2x is a constant multiple of the derivative of the denominator, giving a logarithm.
2. The minus sign from dt is absorbed by reversing the limits 5 → 1.

Answer: (a) \( \pi/\sqrt{3} \)

Explanation:
1. The integrand has period π and is symmetric about π/2, so \( I = 2\int_{0}^{\pi/2} \frac{dx}{1 + 2\sin^2 x} \).
2. Divide by cos²x: \( 2\int_{0}^{\pi/2} \frac{\sec^2 x\,dx}{1 + 3\tan^2 x} \); t = tan x: \( 2\int_{0}^{\infty} \frac{dt}{1 + 3t^2} \).
3. \( = 2\cdot\frac{1}{\sqrt{3}}\cdot\frac{\pi}{2} = \frac{\pi}{\sqrt{3}} \).

Teacher's Note:
1. Use ∫₀^(2a) f = 2∫₀^a f when f(2a − x) = f(x); this avoids crossing x = π/2, where tan x is undefined.
2. After dividing by cos²x, the denominator is sec²x + 2tan²x = 1 + 3tan²x.

Answer: (d) 0

Explanation:
1. Let \( I = \int_{\sin\theta}^{\mathrm{cosec}\,\theta} f(x)\,dx \) and put x = 1/t, dx = −dt/t²; the limits swap to cosec θ and sin θ.
2. \( I = \int_{\mathrm{cosec}\,\theta}^{\sin\theta} f\left(\frac{1}{t}\right)\left(-\frac{dt}{t^2}\right) = \int_{\sin\theta}^{\mathrm{cosec}\,\theta} \frac{f(1/t)}{t^2}dt \).
3. Given \( f(1/t) = -t^2 f(t) \), this becomes \( -\int_{\sin\theta}^{\mathrm{cosec}\,\theta} f(t)\,dt = -I \).
4. So 2I = 0 and I = 0.

Teacher's Note:
1. Limits that are reciprocals (sin θ and cosec θ) suggest the substitution x = 1/t.
2. A concrete check: f(x) = 1/x² − 1 satisfies the relation, and its integral from k to 1/k is 0.

Answer: (d) none of these

Explanation:
1. Divide inside the root by cos⁴x: \( \frac{1}{\sqrt{\cos x\sin^3 x}} = \frac{\sec^2 x}{\sqrt{\tan^3 x}} \).
2. Put t = tan x: \( \int_{1/\sqrt{3}}^{1} t^{-3/2}\,dt = \left[-\frac{2}{\sqrt{t}}\right]_{1/\sqrt{3}}^{1} = -2 + 2\cdot 3^{1/4} \).
3. \( 2\cdot 3^{1/4} - 2 \approx 0.63 \), which is not 1, 1/3 or −2: none of these.

Teacher's Note:
1. When the powers of sin x and cos x add up to an even negative number (here −4 under the root gives −2), use t = tan x.
2. The integrand is positive, so −2 is impossible; the exact value 2(3^(1/4) − 1) settles the rest.

Answer: (c) −1

Explanation:
1. f′(a) = tan(π/3) = √3 and f′(b) = tan(π/4) = 1.
2. \( \int_{a}^{b} f'(x)f''(x)\,dx = \left[\frac{(f'(x))^2}{2}\right]_{a}^{b} = \frac{1 - 3}{2} = -1 \).

Teacher's Note:
1. The slope of the tangent equals tan of the angle it makes with the x-axis.
2. ∫ g·g′ dx = g²/2, here with g = f′.

Answer: (c) \( \frac{[f(b)]^2 - [f(a)]^2}{2} \)

Explanation:
1. Since g = f′, \( \int_{a}^{b} f(x)g(x)\,dx = \int_{a}^{b} f(x)f'(x)\,dx \).
2. \( = \left[\frac{(f(x))^2}{2}\right]_{a}^{b} = \frac{[f(b)]^2 - [f(a)]^2}{2} \).

Teacher's Note:
1. The integral of a function times its derivative is half the square of the function.
2. f(b) − f(a) would be ∫ g dx, not ∫ f g dx; read the integrand carefully.

Answer: (b) \( \frac{1}{15} \log_e 6 \)

Explanation:
1. Put t = tan(x/2): \( 9\cos x + 12\sin x = \frac{9 - 9t^2 + 24t}{1 + t^2} \), dx = 2dt/(1 + t²); limits 0 to 1.
2. \( \int_{0}^{1} \frac{2\,dt}{9 + 24t - 9t^2} = \int_{0}^{1} \frac{2\,dt}{25 - (3t - 4)^2} \); with u = 3t − 4 (u: −4 → −1): \( \frac{2}{3}\int_{-4}^{-1} \frac{du}{25 - u^2} \).
3. \( = \frac{2}{3}\cdot\frac{1}{10}\left[\log\frac{5 + u}{5 - u}\right]_{-4}^{-1} = \frac{1}{15}\left(\log\frac{2}{3} - \log\frac{1}{9}\right) = \frac{1}{15}\log_e 6 \).

Teacher's Note:
1. Alternatively write 9 cos x + 12 sin x = 15 sin(x + α); the factor 15 = √(9² + 12²) appears in the answer.
2. Integrals in calculus use natural logarithms, so log₁₀ in option (a) is wrong even though the number 6 matches.

Answer: (b) \( \frac{1}{2} I \)

Explanation:
1. Put t = x², dt = 2x dx; limits 0 to 1.
2. \( \int_{0}^{1} \frac{x e^{x^2}}{x^2 + 1}dx = \frac{1}{2}\int_{0}^{1} \frac{e^t}{t + 1}dt = \frac{1}{2}I \).

Teacher's Note:
1. Choose the substitution that makes the new integral identical to the given one.
2. The factor x dx = ½ dt supplies the ½.

Answer: (a) \( \int_{0}^{1} f(x)\,dx \)

Explanation:
1. Put t = 1 − x, dx = −dt; x = 0 → t = 1, x = 1 → t = 0.
2. \( \int_{0}^{1} f(1 - x)\,dx = \int_{1}^{0} f(t)(-dt) = \int_{0}^{1} f(t)\,dt = \int_{0}^{1} f(x)\,dx \).

Teacher's Note:
1. This is the property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx with a = 1.
2. The property is the basis of most 'replace x by a − x and add' tricks.

Answer: (d) 0

Explanation:
1. f(x) = sin²⁷x is odd. In the second sum put x = −u: \( \int_{-2n}^{2n + 1} f(x)\,dx = \int_{-2n - 1}^{2n} f(-u)\,du = -\int_{-2n - 1}^{2n} f(u)\,du \).
2. So each term of the second sum is the negative of the matching term of the first sum.
3. The two sums cancel, and the total is 0.

Teacher's Note:
1. For odd f, reflecting the interval [p, q] to [−q, −p] changes the sign of the integral.
2. No integral needs to be evaluated; recognising oddness is enough.

Answer: (c) a + b + c = 0

Explanation:
1. \( \int_{0}^{3} g = \int_{0}^{1} g + \int_{1}^{3} g \), so the condition gives \( \int_{0}^{1} (3ax^2 + 2bx + c)\,dx = 0 \).
2. \( [ax^3 + bx^2 + cx]_{0}^{1} = a + b + c = 0 \).

Teacher's Note:
1. When two integrals with a common part are equal, the difference over the remaining interval is zero.
2. 3ax² + 2bx + c is the derivative of ax³ + bx² + cx, which makes the evaluation immediate.

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

Explanation:
1. sin 2πx ≥ 0 on [0, ½] and ≤ 0 on [½, 1]; the two parts have equal areas.
2. \( \int_{0}^{1} |\sin 2\pi x|\,dx = 2\int_{0}^{1/2} \sin 2\pi x\,dx = 2\left[-\frac{\cos 2\pi x}{2\pi}\right]_{0}^{1/2} \).
3. \( = 2\cdot\frac{1 + 1}{2\pi} = \frac{2}{\pi} \).

Teacher's Note:
1. Split a modulus integral at the zeros of the expression inside.
2. The integral of a modulus can never be negative or zero (unless the function is zero), ruling out (a) and (b).

Answer: (d) 1

Explanation:
1. |x − 1| = 1 − x on [0, 1] and x − 1 on [1, 2].
2. \( \int_{0}^{1} (1 - x)\,dx + \int_{1}^{2} (x - 1)\,dx = \frac{1}{2} + \frac{1}{2} = 1 \).

Teacher's Note:
1. Geometrically, the graph forms two triangles of base 1 and height 1, each of area ½.
2. Split at the point where the expression inside the modulus is zero.

Answer: (a) 5/2

Explanation:
1. |x| = −x on [−1, 0] and x on [0, 2].
2. \( \int_{-1}^{0} (-x)\,dx + \int_{0}^{2} x\,dx = \frac{1}{2} + 2 = \frac{5}{2} \).

Teacher's Note:
1. The two triangles have areas ½·1·1 and ½·2·2.
2. Without the modulus the answer would be 3/2, which is a distractor.

Answer: (b) 5/2

Explanation:
1. |2 − x| = 2 − x on [0, 2] and x − 2 on [2, 3].
2. \( \int_{0}^{2} (2 - x)\,dx + \int_{2}^{3} (x - 2)\,dx = 2 + \frac{1}{2} = \frac{5}{2} \).

Teacher's Note:
1. Areas of the two triangles are ½·2·2 = 2 and ½·1·1 = ½.
2. A modulus integral is always positive, so −3/2 is impossible.

Answer: (d) 12

Explanation:
1. On [1, 5], |1 − x| = x − 1, so \( \int_{1}^{5} |1 - x|\,dx = \int_{1}^{5} (x - 1)\,dx = 8 \).
2. \( \int_{1}^{5} |x - 3|\,dx = \int_{1}^{3} (3 - x)\,dx + \int_{3}^{5} (x - 3)\,dx = 2 + 2 = 4 \).
3. Total = 8 + 4 = 12.

Teacher's Note:
1. Integrate each modulus term separately, splitting only where that term changes sign.
2. Triangle areas give the answers instantly: ½·4·4 = 8 and two triangles of area 2.

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

Explanation:
1. log x < 0 on [1/e, 1) and ≥ 0 on [1, e]. Using \( \int \log x\,dx = x\log x - x \):
2. \( \int_{1/e}^{1} (-\log x)\,dx = -\left[(0 - 1) - \left(-\frac{1}{e} - \frac{1}{e}\right)\right] = 1 - \frac{2}{e} \).
3. \( \int_{1}^{e} \log x\,dx = (e - e) - (0 - 1) = 1 \). Total \( = 2 - \frac{2}{e} = 2\left(1 - \frac{1}{e}\right) \).

Teacher's Note:
1. Split at x = 1, where log x changes sign.
2. At x = 1/e, x log x = (1/e)(−1) = −1/e; signs here are where most slips happen.

Answer: (b) \( 2 - \sqrt{2} \)

Explanation:
1. sin(x − π/4) ≤ 0 on [0, π/4] and ≥ 0 on [π/4, π/2]; by symmetry the two parts are equal.
2. \( 2\int_{\pi/4}^{\pi/2} \sin\left(x - \frac{\pi}{4}\right)dx = 2\left[-\cos\left(x - \frac{\pi}{4}\right)\right]_{\pi/4}^{\pi/2} \).
3. \( = 2\left(-\frac{1}{\sqrt{2}} + 1\right) = 2 - \sqrt{2} \).

Teacher's Note:
1. The interval [0, π/2] is symmetric about π/4, where the sine changes sign; this halves the work.
2. 2 − √2 ≈ 0.59 is positive, as a modulus integral must be.

Answer: (b) \( 4/3\sqrt{3} \)

Explanation:
1. 3x² − 1 = 0 at x = 1/√3; it is negative before and positive after.
2. \( \int_{0}^{1/\sqrt{3}} (1 - 3x^2)\,dx = \left[x - x^3\right]_{0}^{1/\sqrt{3}} = \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} = \frac{2}{3\sqrt{3}} \).
3. \( \int_{1/\sqrt{3}}^{1} (3x^2 - 1)\,dx = [x^3 - x]_{1/\sqrt{3}}^{1} = 0 + \frac{2}{3\sqrt{3}} \). Total \( = \frac{4}{3\sqrt{3}} \).

Teacher's Note:
1. Find the zero of the expression inside the modulus within the interval and split there.
2. Without the modulus the integral would be 0, which is a deliberate distractor.

Answer: (b) \( 2(\sqrt{2} - 1) \)

Explanation:
1. sin x − cos x ≤ 0 on [0, π/4] and ≥ 0 on [π/4, π/2]; by symmetry both parts are equal.
2. \( 2\int_{0}^{\pi/4} (\cos x - \sin x)\,dx = 2[\sin x + \cos x]_{0}^{\pi/4} = 2(\sqrt{2} - 1) \).

Teacher's Note:
1. Replacing x by π/2 − x swaps sin and cos, so the two halves of the interval contribute equally.
2. Without the modulus the value would be 0.

Answer: (a) 1

Explanation:
1. |1 − x| = 1 − x on [0, 1] and x − 1 on [1, 2].
2. \( \int_{0}^{1} (1 - x)\,dx + \int_{1}^{2} (x - 1)\,dx = \frac{1}{2} + \frac{1}{2} = 1 \).

Teacher's Note:
1. |1 − x| = |x − 1|; the graph is a V with vertex at x = 1.
2. Two triangles of base 1 and height 1 give a total area of 1.

Answer: (c) 20

Explanation:
1. Put t = x + 2: \( \int_{-2}^{6} |t|\,dt = \int_{-2}^{0} (-t)\,dt + \int_{0}^{6} t\,dt \).
2. \( = 2 + 18 = 20 \).

Teacher's Note:
1. Shifting the variable moves the vertex of the V-graph to the origin.
2. Triangle areas: ½·2·2 = 2 and ½·6·6 = 18.

Answer: (d) 4

Explanation:
1. |sin x| = sin x on [0, π] and −sin x on [π, 2π].
2. \( \int_{0}^{\pi} \sin x\,dx - \int_{\pi}^{2\pi} \sin x\,dx = 2 - (-2) = 4 \).

Teacher's Note:
1. |sin x| has period π, so ∫₀^(2π) |sin x| dx = 2∫₀^π sin x dx.
2. ∫₀^π sin x dx = 2 is a value worth remembering.

Answer: (a) 1

Explanation:
1. On [−1, 0), [x] = −1 so f(x) = x + 1; on [0, 1), [x] = 0 so f(x) = x.
2. \( \int_{-1}^{0} (x + 1)\,dx + \int_{0}^{1} x\,dx = \frac{1}{2} + \frac{1}{2} = 1 \).

Teacher's Note:
1. x − [x] is the fractional part {x}, periodic with period 1, and ∫ over any unit interval is ½.
2. Split at integers, where [x] jumps.

Answer: (a) 31

Explanation:
1. [√x + 2] = [√x] + 2. [√x] = 0 on [0, 1), 1 on [1, 4), 2 on [4, 9).
2. \( \int_{0}^{9} [\sqrt{x}]\,dx = 0\cdot 1 + 1\cdot 3 + 2\cdot 5 = 13 \).
3. \( \int_{0}^{9} 2\,dx = 18 \), so the total is 13 + 18 = 31.

Teacher's Note:
1. [x + n] = [x] + n for an integer n; take the integer out first.
2. [√x] changes value where √x is an integer, i.e. at the perfect squares 1, 4, 9.

Answer: (b) 7/3

Explanation:
1. [x] = 0 on [0, 1) and 1 on [1, 2).
2. \( \int_{0}^{2} x^2[x]\,dx = \int_{1}^{2} x^2\,dx = \frac{8 - 1}{3} = \frac{7}{3} \).

Teacher's Note:
1. The piece where [x] = 0 contributes nothing.
2. Split at the integers inside the interval before integrating.

Answer: (c) 1/8

Explanation:
1. Split at ½: \( I = \int_{0}^{1/2} x\left(\frac{1}{2} - x\right)dx + \int_{1/2}^{1} x\left(x - \frac{1}{2}\right)dx \).
2. First: \( \left[\frac{x^2}{4} - \frac{x^3}{3}\right]_{0}^{1/2} = \frac{1}{16} - \frac{1}{24} = \frac{1}{48} \).
3. Second: \( \left[\frac{x^3}{3} - \frac{x^2}{4}\right]_{1/2}^{1} = \frac{1}{12} + \frac{1}{48} = \frac{5}{48} \). So \( I = \frac{6}{48} = \frac{1}{8} \).

Teacher's Note:
1. Split at the zero of the expression inside the modulus, and change its sign on the negative side.
2. Fractions with denominators 16, 24, 48 are easiest handled by converting all to 48ths.

Answer: (c) \( \sqrt{2} - 1 \)

Explanation:
1. [x²] = 0 for 0 ≤ x < 1 and 1 for 1 ≤ x < √2.
2. \( \int_{0}^{\sqrt{2}} [x^2]\,dx = \int_{1}^{\sqrt{2}} 1\,dx = \sqrt{2} - 1 \).

Teacher's Note:
1. [x²] changes value where x² is an integer, i.e. at x = 1, √2, √3, ….
2. The answer must be positive and less than √2, which leaves only √2 − 1.

Answer: (b) 4

Explanation:
1. \( \int_{0}^{2\pi} \sin x\,dx = 0 \).
2. \( \int_{0}^{2\pi} |\sin x|\,dx = 2\int_{0}^{\pi} \sin x\,dx = 4 \).
3. Total = 0 + 4 = 4.

Teacher's Note:
1. sin x + |sin x| equals 2 sin x on [0, π] and 0 on [π, 2π], giving 2 × 2 = 4 directly.
2. Split sums of terms and treat each separately.

Answer: (c) \( 2\sqrt{2} \)

Explanation:
1. sin x + cos x = √2 sin(x + π/4), which is ≥ 0 on [0, 3π/4] and ≤ 0 on [3π/4, π].
2. \( \int_{0}^{3\pi/4} (\sin x + \cos x)\,dx = [\sin x - \cos x]_{0}^{3\pi/4} = \sqrt{2} + 1 \).
3. \( \int_{3\pi/4}^{\pi} -(\sin x + \cos x)\,dx = -[\sin x - \cos x]_{3\pi/4}^{\pi} = -(1 - \sqrt{2}) = \sqrt{2} - 1 \).
4. Total \( = 2\sqrt{2} \).

Teacher's Note:
1. Writing a sin x + b cos x as R sin(x + α) shows where it changes sign.
2. Shortcut: \( \int_{0}^{\pi} \sqrt{2}\,|\sin(x + \pi/4)|\,dx = \sqrt{2}\int_{0}^{\pi} |\sin u|\,du = 2\sqrt{2} \) (|sin| has period π).

Answer: (a) \( \frac{41}{2} \)

Explanation:
1. On [k, k + 1), [x] = k, and \( \int_{k}^{k + 1} kx\,dx = \frac{k(2k + 1)}{2} \).
2. For k = −2, −1, 0, 1, 2, 3 these are 3, ½, 0, 3/2, 5, 21/2.
3. Sum \( = 3 + \frac{1}{2} + 0 + \frac{3}{2} + 5 + \frac{21}{2} = \frac{41}{2} \).

Teacher's Note:
1. A general formula for each unit piece saves repeating the same integration six times.
2. On negative pieces, both [x] and x are negative, so their products give positive contributions.

Answer: (d) \( \frac{28}{3} \)

Explanation:
1. |1 − x²| = 1 − x² on [−1, 1] and x² − 1 elsewhere.
2. \( \int_{-2}^{2} |1 - x^2|\,dx = 2\left[\int_{0}^{1} (1 - x^2)\,dx + \int_{1}^{2} (x^2 - 1)\,dx\right] = 2\left[\frac{2}{3} + \frac{4}{3}\right] = 4 \).
3. \( \int_{2}^{3} (x^2 - 1)\,dx = \frac{27 - 8}{3} - 1 = \frac{16}{3} \). Total \( = 4 + \frac{16}{3} = \frac{28}{3} \).

Teacher's Note:
1. Use evenness over the symmetric part [−2, 2] and handle the leftover [2, 3] separately.
2. Split at both zeros x = ±1 of 1 − x².

Answer: (c) \( \frac{1}{3}(a^3 + b^3) \)

Explanation:
1. x|x| = −x² for x < 0 and x² for x ≥ 0.
2. \( \int_{a}^{0} (-x^2)\,dx + \int_{0}^{b} x^2\,dx = -\left(0 - \frac{a^3}{3}\right) + \frac{b^3}{3} = \frac{a^3 + b^3}{3} \).

Teacher's Note:
1. An antiderivative of x|x| valid for all x is x²|x|/3, which gives b³/3 − (a²|a|)/3 = (a³ + b³)/3 directly.
2. With a < 0, a³ is negative, which correctly reflects the negative area to the left of 0.

Answer: (a) 7

Explanation:
1. \( \int_{-1}^{3} |x - 2|\,dx = \int_{-1}^{2} (2 - x)\,dx + \int_{2}^{3} (x - 2)\,dx = \frac{9}{2} + \frac{1}{2} = 5 \).
2. \( \int_{-1}^{3} [x]\,dx = (-1) + 0 + 1 + 2 = 2 \) (one unit interval each).
3. Total = 5 + 2 = 7.

Teacher's Note:
1. Treat the modulus and the greatest-integer parts separately; each needs its own split points.
2. ∫ of [x] over [k, k + 1] is simply k.

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

Explanation:
1. \( \frac{1}{1 + \tan\theta} = \frac{\cos\theta}{\cos\theta + \sin\theta} \). Replacing θ by π/2 − θ gives \( I = \int_{0}^{\pi/2} \frac{\sin\theta}{\sin\theta + \cos\theta}d\theta \).
2. Adding the two forms: \( 2I = \int_{0}^{\pi/2} 1\,d\theta = \frac{\pi}{2} \).
3. So \( I = \frac{\pi}{4} \).

Teacher's Note:
1. The 'replace x by a − x and add' method works whenever f(x) + f(a − x) is simple.
2. Any integral of the form ∫₀^(π/2) g(sin)/(g(sin) + g(cos)) equals π/4.

Answer: (c) \( \frac{\pi}{12} \)

Explanation:
1. \( \frac{1}{1 + \sqrt{\cot x}} = \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \). Use \( \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx \) with a + b = π/2.
2. Then \( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}dx \); adding gives \( 2I = \int_{\pi/6}^{\pi/3} dx = \frac{\pi}{6} \).
3. So \( I = \frac{\pi}{12} \).

Teacher's Note:
1. When the limits add up to π/2, x → (a + b − x) swaps sin and cos.
2. The answer is half the length of the interval, (π/3 − π/6)/2 = π/12.

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

Explanation:
1. \( \frac{x\tan x}{\sec x + \cos x} = \frac{x\sin x}{1 + \cos^2 x} \). With x → π − x: \( I = \int_{0}^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x}dx \).
2. Adding: \( 2I = \pi\int_{0}^{\pi} \frac{\sin x\,dx}{1 + \cos^2 x} \); with t = cos x this is \( \pi\int_{-1}^{1} \frac{dt}{1 + t^2} = \pi\cdot\frac{\pi}{2} \).
3. So \( I = \frac{\pi^2}{4} \).

Teacher's Note:
1. For ∫₀^π x f(sin x) dx, use the result (π/2)∫₀^π f(sin x) dx; the x disappears.
2. Simplify the trigonometric expression first: multiply top and bottom by cos x.

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

Explanation:
1. Let I be the integral. Replacing x by π/2 − x gives \( I = \int_{0}^{\pi/2} \frac{2^{\cos x}}{2^{\cos x} + 2^{\sin x}}dx \).
2. Adding: \( 2I = \int_{0}^{\pi/2} 1\,dx = \frac{\pi}{2} \), so \( I = \frac{\pi}{4} \).

Teacher's Note:
1. Whenever the integrand is g(sin x)/(g(sin x) + g(cos x)) on [0, π/2], the answer is π/4.
2. There is no need to integrate 2^(sin x); the symmetry does all the work.

Answer: (d) \( \pi\left(\frac{\pi}{2} - 1\right) \)

Explanation:
1. \( \frac{x\tan x}{\sec x + \tan x} = \frac{x\sin x}{1 + \sin x} \). Using x → π − x: \( I = \frac{\pi}{2}\int_{0}^{\pi} \frac{\sin x}{1 + \sin x}dx \).
2. \( \frac{\sin x}{1 + \sin x} = 1 - \frac{1}{1 + \sin x} \), and \( \int_{0}^{\pi} \frac{dx}{1 + \sin x} = 2 \).
3. So \( I = \frac{\pi}{2}(\pi - 2) = \pi\left(\frac{\pi}{2} - 1\right) \).

Teacher's Note:
1. ∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx because sin(π − x) = sin x.
2. The standard value ∫₀^π dx/(1 + sin x) = 2 is used here.

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

Explanation:
1. Put x = sin θ, dx = cos θ dθ, √(1 − x²) = cos θ; θ: 0 → π/2.
2. \( \int_{0}^{\pi/2} \frac{\cos\theta}{\sin\theta + \cos\theta}d\theta = \frac{\pi}{4} \) (by θ → π/2 − θ and adding).

Teacher's Note:
1. x = sin θ removes √(1 − x²) and produces a sin/cos symmetric integral.
2. \( \int_{0}^{\pi/2} \frac{\cos\theta}{\sin\theta + \cos\theta}d\theta = \int_{0}^{\pi/2} \frac{\sin\theta}{\sin\theta + \cos\theta}d\theta \) and their sum is π/2.

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

Explanation:
1. \( \frac{1}{1 + \tan^3 x} = \frac{\cos^3 x}{\cos^3 x + \sin^3 x} \). Replace x by π/2 − x: \( I = \int_{0}^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x}dx \).
2. Adding: \( 2I = \frac{\pi}{2} \), so \( I = \frac{\pi}{4} \).

Teacher's Note:
1. The power 3 makes no difference: ∫₀^(π/2) dx/(1 + tanⁿx) = π/4 for every n.
2. Write tan x as sin x/cos x so that the sin ↔ cos symmetry becomes visible.