CBSE Class 12 Mathematics Integrals MCQs Set 09

Answer: (a) \( \pi \)

Explanation:
1. By parts: \( \int x\sin x\,dx = -x\cos x + \sin x \).
2. \( [-x\cos x + \sin x]_{0}^{\pi} = -\pi(-1) + 0 - 0 = \pi \).

Teacher's Note:
1. Take x as the first function so that its derivative 1 simplifies the remaining integral.
2. Also ∫₀^π x sin x dx = (π/2)∫₀^π sin x dx = (π/2)(2) = π by the a − x property.

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

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

Teacher's Note:
1. ∫₀^(π/2) dx/(1 + tanⁿx) = π/4 for any real n, including n = ½.
2. This is the same symmetry as in ∫₀^(π/2) √cos x/(√sin x + √cos x) dx.

Answer: (a) (a + b)π/4

Explanation:
1. \( \int_{0}^{\pi/2} \cos^2 x\,dx = \int_{0}^{\pi/2} \sin^2 x\,dx = \frac{\pi}{4} \) (they are equal and add up to π/2).
2. So the value is \( a\cdot\frac{\pi}{4} + b\cdot\frac{\pi}{4} = \frac{(a + b)\pi}{4} \).

Teacher's Note:
1. On [0, π/2], sin²x and cos²x have equal integrals by the a − x property, each π/4.
2. Check with a = b = 1: the integrand is 1 and the integral is π/2 = 2π/4.

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

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

Teacher's Note:
1. This is the classic example of the property ∫₀^a f(x) dx = ∫₀^a f(a − x) dx.
2. The two companion integrals are equal, and their sum is just the length of the interval.

Answer: (a) a

Explanation:
1. Replace x by 2a − x: \( I = \int_{0}^{2a} \frac{f(2a - x)}{f(2a - x) + f(x)}dx \).
2. Adding the two forms: \( 2I = \int_{0}^{2a} 1\,dx = 2a \).
3. So I = a.

Teacher's Note:
1. The answer is half the length of the interval; this pattern appears in many disguises.
2. The method works for any f, as long as the denominator f(x) + f(2a − x) is never zero.

Answer: (c) 0

Explanation:
1. Replace x by π/2 − x: cos x and sin x swap, the denominator is unchanged and the numerator changes sign.
2. So I = −I, which gives I = 0.

Teacher's Note:
1. If the a − x substitution turns the integrand into its negative, the integral is zero.
2. Graphically, the curve is antisymmetric about x = π/4, so the areas above and below cancel.

Answer: (c) 2

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

Teacher's Note:
1. By symmetry about π/2, ∫₀^π |cos x| dx = 2∫₀^(π/2) cos x dx = 2.
2. Without the modulus the value would be 0, a common distractor.

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

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

Teacher's Note:
1. The same result holds for any power: ∫₀^(π/2) cosⁿx/(sinⁿx + cosⁿx) dx = π/4.
2. Recognising the pattern lets you write the answer without any integration.

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

Explanation:
1. \( \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} = \frac{\cos x}{\cos x + \sin x} \) (multiply top and bottom by √(sin x cos x)).
2. By x → π/2 − x and adding, \( 2I = \frac{\pi}{2} \), so \( I = \frac{\pi}{4} \).

Teacher's Note:
1. Simplifying √cot x and √tan x reduces this to the familiar cos x/(sin x + cos x).
2. Directly, x → π/2 − x swaps cot and tan, which already shows the two companion integrals are equal.

Answer: (a) \( -\left(\frac{\pi}{2}\right) \log 2 \)

Explanation:
1. Let I = ∫₀^(π/2) log sin x dx = ∫₀^(π/2) log cos x dx. Then \( 2I = \int_{0}^{\pi/2} \log\frac{\sin 2x}{2}dx = \int_{0}^{\pi/2} \log\sin 2x\,dx - \frac{\pi}{2}\log 2 \).
2. With t = 2x, \( \int_{0}^{\pi/2} \log\sin 2x\,dx = \frac{1}{2}\int_{0}^{\pi} \log\sin t\,dt = \int_{0}^{\pi/2} \log\sin t\,dt = I \).
3. So 2I = I − (π/2) log 2, giving \( I = -\frac{\pi}{2}\log 2 \).

Teacher's Note:
1. This standard result is used in many other problems; remember ∫₀^(π/2) log sin x dx = −(π/2) log 2.
2. The value is negative because log sin x ≤ 0 on the interval.

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

Explanation:
1. Let f(x) be the integrand. Replacing x by π/2 − x gives \( \frac{e^{(\pi/2 - x)^2}}{e^{(\pi/2 - x)^2} + e^{x^2}} \), and f(x) + f(π/2 − x) = 1.
2. So \( 2I = \int_{0}^{\pi/2} 1\,dx = \frac{\pi}{2} \), and \( I = \frac{\pi}{4} \).

Teacher's Note:
1. The structure g(x)/(g(x) + g(a − x)) on [0, a] always integrates to a/2.
2. There is no elementary antiderivative of e^(x²); symmetry is the only route.

Answer: (d) 0

Explanation:
1. Replace x by π/2 − x: tan x becomes cot x, so \( I = \int_{0}^{\pi/2} \log\cot x\,dx = -\int_{0}^{\pi/2} \log\tan x\,dx = -I \).
2. So 2I = 0, i.e. I = 0.

Teacher's Note:
1. log tan x = log sin x − log cos x, and those two integrals over [0, π/2] are equal.
2. The graph of log tan x is antisymmetric about x = π/4.

Answer: (c) 0

Explanation:
1. Replace x by π/2 − x: sin 2x is unchanged and log tan x becomes log cot x = −log tan x.
2. So I = −I, giving I = 0.

Teacher's Note:
1. Check how each factor behaves under x → a − x; one even and one odd factor give zero.
2. sin(π − 2x) = sin 2x keeps the first factor the same.

Answer: (c) \( \frac{\pi}{8} \log 2 \)

Explanation:
1. Replace x by π/4 − x: \( 1 + \tan\left(\frac{\pi}{4} - x\right) = 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{2}{1 + \tan x} \).
2. So \( I = \int_{0}^{\pi/4} [\log 2 - \log(1 + \tan x)]\,dx = \frac{\pi}{4}\log 2 - I \).
3. Hence \( I = \frac{\pi}{8}\log 2 \).

Teacher's Note:
1. The identity (1 + tan x)(1 + tan(π/4 − x)) = 2 is the key to this famous integral.
2. The integrand is positive, so options with log(½) (negative) are ruled out.

Answer: (d) 0

Explanation:
1. Replace θ by 2π − θ: sin 2θ becomes −sin 2θ while cos θ is unchanged.
2. So I = −I, which gives I = 0.

Teacher's Note:
1. Use ∫₀^(2a) f(x) dx = 0 when f(2a − x) = −f(x).
2. The values of a and b do not matter, as long as the denominator is never zero.

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

Explanation:
1. Using x → π − x: \( I = \frac{\pi}{2}\int_{0}^{\pi} \sin^3 x\,dx \).
2. \( \int_{0}^{\pi} \sin^3 x\,dx = \int_{0}^{\pi} (1 - \cos^2 x)\sin x\,dx = \left[-\cos x + \frac{\cos^3 x}{3}\right]_{0}^{\pi} = \frac{4}{3} \).
3. So \( I = \frac{\pi}{2}\cdot\frac{4}{3} = \frac{2\pi}{3} \).

Teacher's Note:
1. ∫₀^π x f(sin x) dx = (π/2)∫₀^π f(sin x) dx removes the factor x.
2. For sin³x, write sin²x = 1 − cos²x and substitute t = cos x.

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

Explanation:
1. Replace x by a − x: \( I = \int_{0}^{a} f(a - x)g(a - x)\,dx = \int_{0}^{a} f(x)[2 - g(x)]\,dx \).
2. So \( I = 2\int_{0}^{a} f(x)\,dx - I \), giving \( I = \int_{0}^{a} f(x)\,dx \).

Teacher's Note:
1. Use the given symmetry conditions after the a − x substitution, then solve for I.
2. A quick check: f = 1, g = 1 satisfy the conditions and give I = a = ∫₀^a 1 dx.

Answer: (b) \( \pi \)

Explanation:
1. Let F(x) = f(cos²x + tan²x); F(π − x) = F(x). By x → π − x: \( I = \int_{0}^{\pi} (\pi - x)F(x)\,dx \), so \( I = \frac{\pi}{2}\int_{0}^{\pi} F(x)\,dx \).
2. F(π − x) = F(x) also gives \( \int_{0}^{\pi} F = 2\int_{0}^{\pi/2} F \).
3. So \( I = \pi\int_{0}^{\pi/2} F(x)\,dx \) and k = π.

Teacher's Note:
1. cos²(π − x) = cos²x and tan²(π − x) = tan²x, which is why the property applies.
2. Two properties are combined: ∫₀^π x F = (π/2)∫₀^π F and ∫₀^(2a) F = 2∫₀^a F.

Answer: (a) \( \pi \tan \frac{\pi}{8} \)

Explanation:
1. The limits add to π, so replace φ by π − φ: \( I = \int_{\pi/4}^{3\pi/4} \frac{\pi - \phi}{1 + \sin\phi}d\phi \); adding, \( 2I = \pi\int_{\pi/4}^{3\pi/4} \frac{d\phi}{1 + \sin\phi} \).
2. An antiderivative of \( \frac{1}{1 + \sin\phi} \) is \( -\tan\left(\frac{\pi}{4} - \frac{\phi}{2}\right) \); from π/4 to 3π/4 this gives \( \tan\frac{\pi}{8} + \tan\frac{\pi}{8} = 2\tan\frac{\pi}{8} \).
3. So 2I = 2π tan(π/8), i.e. \( I = \pi\tan\frac{\pi}{8} \).

Teacher's Note:
1. ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx works for any limits, not just from 0.
2. tan(π/4 − 3π/8) = tan(−π/8) = −tan(π/8); the minus signs combine to give 2 tan(π/8).

Answer: (c) 0

Explanation:
1. Replace x by π − x: cos²x is unchanged and cos 3(π − x) = −cos 3x, so cos⁵3x changes sign.
2. So I = −I, giving I = 0.

Teacher's Note:
1. Use ∫₀^(2a) f = 0 when f(2a − x) = −f(x); here 2a = π.
2. An odd power keeps the minus sign; an even power would have removed it.

Answer: (b) \( \int_{-a}^{a} f(-x)\,dx \)

Explanation:
1. Put x = −t: \( \int_{-a}^{a} f(x)\,dx = \int_{a}^{-a} f(-t)(-dt) = \int_{-a}^{a} f(-t)\,dt \).
2. Hence \( \int_{-a}^{a} f(x)\,dx = \int_{-a}^{a} f(-x)\,dx \) for every f.

Teacher's Note:
1. Options (a) and (c) hold only for even and odd f respectively; the question asks for what is always true.
2. Reflecting x to −x over a symmetric interval never changes the integral.

Answer: (c) The value of r

Explanation:
1. \( \ln\frac{1 + x}{1 - x} \) is odd, so its integral over [−½, ½] is 0.
2. \( \ln\left(\frac{1 - x}{1 + x}\right)^{-2} = 2\ln\frac{1 + x}{1 - x} \), also odd, so that term also gives 0.
3. The constant r gives r × 1 = r. The value is r, so it depends only on r.

Teacher's Note:
1. \( \ln\frac{1 + x}{1 - x} \) is a standard odd function: replacing x by −x inverts the fraction.
2. Over a symmetric interval, only the even parts of the integrand survive.

Answer: (a) 0

Explanation:
1. Let \( f(x) = \log\frac{2 - \sin x}{2 + \sin x} \). Then \( f(-x) = \log\frac{2 + \sin x}{2 - \sin x} = -f(x) \), so f is odd.
2. The integral of an odd function over [−π/2, π/2] is 0.

Teacher's Note:
1. log(p/q) = −log(q/p): swapping numerator and denominator changes the sign.
2. Always test for odd/even symmetry first when the limits are −a and a.

Answer: (c) 0

Explanation:
1. \( f(x) = \log\frac{1 + x}{1 - x} \) satisfies f(−x) = −f(x), so it is odd.
2. Hence the integral over [−1, 1] is 0.

Teacher's Note:
1. The integrand is unbounded near x = ±1, but the integral still converges; by oddness the value is 0.
2. Odd functions over symmetric limits give 0 without any evaluation.

Answer: (d) p and s

Explanation:
1. qx is odd, so \( \int_{-2}^{2} qx\,dx = 0 \).
2. \( \int_{-2}^{2} (px^2 + s)\,dx = 2\int_{0}^{2} (px^2 + s)\,dx = \frac{16p}{3} + 4s \).
3. So the value depends on p and s, and both must be known.

Teacher's Note:
1. Over symmetric limits, odd-degree terms vanish, so their coefficients are irrelevant.
2. Both even terms px² and s contribute, so knowing only one of them is not enough.

Answer: (b) 0

Explanation:
1. Let g(x) = sin x f(cos x). Then g(−x) = −sin x f(cos x) = −g(x), so g is odd.
2. Hence \( \int_{-a}^{a} \sin x\,f(\cos x)\,dx = 0 \).

Teacher's Note:
1. sin x is odd and f(cos x) is even (cos is even), so their product is odd.
2. Odd × even = odd, and odd × odd = even; these rules settle most symmetric-limit questions.

Answer: (c) 0

Explanation:
1. sin x is odd; \( \frac{e^{-\cos^2 x}}{1 + \cos^2 x} \) is even. So the integrand is odd.
2. The integral over [−π/2, π/2] is therefore 0.

Teacher's Note:
1. Identify the odd factor; everything built only from cos x is even.
2. Symmetric limits plus an odd integrand mean the answer is 0 immediately.

Answer: (a) 0

Explanation:
1. x²/(x² + 1) is even and sin 2x is odd, so the integrand is odd.
2. Hence the integral over [−3, 3] is 0.

Teacher's Note:
1. Even × odd = odd.
2. There is no need to attempt integration by parts; check symmetry first.

Answer: (a) 0

Explanation:
1. cos x is even and \( \ln\frac{1 + x}{1 - x} \) is odd, so the integrand is odd.
2. Hence the integral over [−½, ½] is 0.

Teacher's Note:
1. ln((1 + x)/(1 − x)) changes sign when x is replaced by −x.
2. Even × odd = odd, which vanishes over symmetric limits.

Answer: (a) 0

Explanation:
1. cos x is even and \( \log\frac{1 - x}{1 + x} \) is odd, so their product is odd.
2. Hence the integral over [−½, ½] is 0.

Teacher's Note:
1. log((1 − x)/(1 + x)) = −log((1 + x)/(1 − x)); both forms are odd.
2. Recognising the symmetry saves an otherwise very hard integration.

Answer: (c) 0

Explanation:
1. 3 sin x + sin³x is odd (both terms are odd).
2. So its integral over [−π/2, π/2] is 0.

Teacher's Note:
1. Odd powers of sin x are odd functions.
2. Over symmetric limits, check oddness before any computation.

Answer: (d) 0

Explanation:
1. \( f(x) = \log\frac{2 - x}{2 + x} \) satisfies f(−x) = −f(x), so it is odd.
2. Hence the integral over [−1, 1] is 0.

Teacher's Note:
1. The pattern log((a − x)/(a + x)) is always odd.
2. Symmetric limits make the answer 0 at once.

Answer: (b) 0

Explanation:
1. x|x| is odd: (−x)|−x| = −x|x|.
2. Hence \( \int_{-1}^{1} x|x|\,dx = 0 \).

Teacher's Note:
1. x is odd and |x| is even, so the product is odd.
2. Directly: ∫₋₁⁰ (−x²) dx + ∫₀¹ x² dx = −1/3 + 1/3 = 0.

Answer: (d) 4

Explanation:
1. |x| is even, so \( \int_{-2}^{2} |x|\,dx = 2\int_{0}^{2} |x|\,dx \).
2. On [0, 2], |x| = x, so the value is \( 2\left[\frac{x^2}{2}\right]_{0}^{2} = 2\cdot 2 = 4 \).

Teacher's Note:
1. For an even function, ∫₋ₐᵃ f = 2∫₀ᵃ f.
2. Geometrically, the graph forms two triangles of area 2 each.

Answer: (a) 0

Explanation:
1. Let \( f(x) = \sqrt{1 + x + x^2} - \sqrt{1 - x + x^2} \). Then \( f(-x) = \sqrt{1 - x + x^2} - \sqrt{1 + x + x^2} = -f(x) \).
2. So f is odd and its integral over [−1, 1] is 0.

Teacher's Note:
1. Replacing x by −x swaps the two square roots, so the difference changes sign.
2. No integration of the square roots is needed.

Answer: (a) 0

Explanation:
1. sin³x is odd and cos²x is even, so sin³x cos²x is odd.
2. Hence the integral over [−1, 1] is 0.

Teacher's Note:
1. Odd × even = odd.
2. The limits are −1 and 1 (not angles like ±π/2), but the symmetry argument works the same way.

Answer: (d) 0

Explanation:
1. sin¹¹x is an odd function, since sin(−x) = −sin x and 11 is odd.
2. Hence the integral over [−1, 1] is 0.

Teacher's Note:
1. The product formulas in (a) and (b) are reduction-formula values for ∫₀^(π/2) sinⁿx, which do not apply here.
2. Over symmetric limits, odd powers of sin x always integrate to zero.

Answer: (a) 0

Explanation:
1. f(x) + f(−x) is even, and g(x) − g(−x) is odd.
2. Their product is odd, so its integral over [−π, π] is 0.

Teacher's Note:
1. f(x) + f(−x) is always even and f(x) − f(−x) is always odd, for any function f.
2. The one-to-one condition is not needed; only the symmetry matters.

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

Explanation:
1. x tan⁻¹x is even (odd × odd), so \( I = 2\int_{0}^{1} x\tan^{-1} x\,dx \).
2. By parts: \( \int x\tan^{-1} x\,dx = \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}\int \frac{x^2}{1 + x^2}dx = \frac{x^2}{2}\tan^{-1} x - \frac{1}{2}(x - \tan^{-1} x) \).
3. From 0 to 1: \( \frac{\pi}{8} - \frac{1}{2}\left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4} - \frac{1}{2} \), so \( I = \frac{\pi}{2} - 1 \).

Teacher's Note:
1. The product of two odd functions is even, so the integral does not vanish here.
2. Write x²/(1 + x²) = 1 − 1/(1 + x²) to finish the integration by parts.

Answer: (c) \( 2\int_{0}^{\pi/2} f(\sin x)\,dx \)

Explanation:
1. cos(−x) = cos x, so f(cos x) is an even function of x (whatever f is).
2. Hence \( \int_{-\pi/2}^{\pi/2} f(\cos x)\,dx = 2\int_{0}^{\pi/2} f(\cos x)\,dx \).
3. By x → π/2 − x, \( \int_{0}^{\pi/2} f(\cos x)\,dx = \int_{0}^{\pi/2} f(\sin x)\,dx \), so the value is \( 2\int_{0}^{\pi/2} f(\sin x)\,dx \).

Teacher's Note:
1. The oddness of f is a distraction: what matters is whether the integrand is odd or even in x, and f(cos x) is even.
2. On [0, π/2], sin x and cos x can always be interchanged inside a definite integral.

Answer: (d) \( 2\pi \)

Explanation:
1. Expand: \( (\cos px - \sin qx)^2 = \cos^2 px + \sin^2 qx - 2\cos px\sin qx \).
2. cos px sin qx is odd, so its integral over [−π, π] is 0.
3. For a non-zero integer p, \( \int_{-\pi}^{\pi} \cos^2 px\,dx = \pi \); similarly \( \int_{-\pi}^{\pi} \sin^2 qx\,dx = \pi \). Total \( = 2\pi \).

Teacher's Note:
1. Over a full period, the average of cos² or sin² is ½, so the integral over length 2π is π.
2. The cross term vanishes by oddness, whether or not p = q.

Answer: (c) \( 2\int_{0}^{1} \frac{-x^2}{3 - |x|}\,dx \)

Explanation:
1. Split: \( \frac{\sin x}{3 - |x|} \) is odd, \( \frac{-x^2}{3 - |x|} \) is even.
2. The odd part integrates to 0 over [−1, 1]; the even part doubles: \( 2\int_{0}^{1} \frac{-x^2}{3 - |x|}dx \).

Teacher's Note:
1. Split any integrand into odd and even parts when the limits are symmetric.
2. |x| is even, so 3 − |x| does not affect the parity of each numerator term.