CBSE Class 12 Mathematics Integrals MCQs Set 07

Answer: (a) \( \log \frac{4}{3} \)

Explanation:
1. Put t = sin x (t: 0 → 1): \( \int_{0}^{1} \frac{dt}{(1 + t)(2 + t)} = \int_{0}^{1} \left(\frac{1}{1 + t} - \frac{1}{2 + t}\right)dt \).
2. \( = \left[\log\frac{1 + t}{2 + t}\right]_{0}^{1} = \log\frac{2}{3} - \log\frac{1}{2} = \log\frac{4}{3} \).

Teacher's Note:
1. cos x dx with a function of sin x calls for t = sin x.
2. For 1/((t + p)(t + q)), the partial fractions are (1/(q − p))[1/(t + p) − 1/(t + q)].

Answer: (c) \( \frac{5}{12} \)

Explanation:
1. \( \sec^7\theta\sin^3\theta = \tan^3\theta\sec^4\theta = \tan^3\theta(1 + \tan^2\theta)\sec^2\theta \).
2. Put t = tan θ (t: 0 → 1): \( \int_{0}^{1} (t^3 + t^5)\,dt = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \).

Teacher's Note:
1. Convert sinᵐ/cosⁿ with m + n even into powers of tan θ and sec²θ, then substitute t = tan θ.
2. Keep one sec²θ aside for dt and write the remaining sec²θ as 1 + tan²θ.

Answer: (d) \( \frac{2}{3} \)

Explanation:
1. sin 2x = 2 sin x cos x, so the integrand is 2 sin²x cos x.
2. Put t = sin x: \( \int_{0}^{1} 2t^2\,dt = \frac{2}{3} \).

Teacher's Note:
1. Expanding sin 2x first reveals a simple f²·f′ pattern.
2. Product-to-sum (cos x − cos 3x)/2 also works and gives (1 − (−1/3))/2 = 2/3.

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

Explanation:
1. Put t = sin θ, dt = cos θ dθ; θ = π/4 → t = 1/√2, θ = π/2 → t = 1.
2. \( \int_{1/\sqrt{2}}^{1} \frac{dt}{t^2} = \left[-\frac{1}{t}\right]_{1/\sqrt{2}}^{1} = -1 + \sqrt{2} = \sqrt{2} - 1 \).

Teacher's Note:
1. cos θ cosec²θ is the derivative of −cosec θ, so the integral is [−cosec θ] directly.
2. cosec(π/4) = √2 and cosec(π/2) = 1.

Answer: (b) sin (log 2)

Explanation:
1. Put t = log x, dt = dx/x; limits 0 to log 2.
2. \( \int_{0}^{\log 2} \cos t\,dt = \sin(\log 2) - \sin 0 = \sin(\log 2) \).

Teacher's Note:
1. A function of log x multiplied by 1/x always suggests t = log x.
2. The upper limit becomes log 2, not log 3; compute new limits carefully.

Answer: (b) \( \frac{\pi}{4} - \frac{1}{2} \log 2 \)

Explanation:
1. Put x = sin t: dx = cos t dt, \( (1 - x^2)^{3/2} = \cos^3 t \), limits 0 to π/4.
2. \( \int_{0}^{\pi/4} t\sec^2 t\,dt = [t\tan t + \log\cos t]_{0}^{\pi/4} \) (by parts).
3. \( = \frac{\pi}{4} + \log\frac{1}{\sqrt{2}} = \frac{\pi}{4} - \frac{1}{2}\log 2 \).

Teacher's Note:
1. \( \sin^{-1} x \) with \( \sqrt{1 - x^2} \) around suggests x = sin t, which turns the inverse function into t.
2. ∫ t sec²t dt by parts: t tan t − ∫ tan t dt = t tan t + log cos t.

Answer: (c) \( \pi \)

Explanation:
1. Divide by cos⁴θ: \( \frac{4\cdot 2\sin\theta\cos\theta}{\sin^4\theta + \cos^4\theta} = \frac{8\tan\theta\sec^2\theta}{\tan^4\theta + 1} \).
2. Put t = tan θ, then u = t²: \( \int_{0}^{1} \frac{8t\,dt}{1 + t^4} = 4\int_{0}^{1} \frac{du}{1 + u^2} \).
3. \( = 4\cdot\frac{\pi}{4} = \pi \).

Teacher's Note:
1. Dividing by cos⁴θ is the standard step when the denominator is a homogeneous expression of degree 4 in sin and cos.
2. t dt with 1 + t⁴ calls for u = t², giving the arctan form.

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

Explanation:
1. Multiply inside by (1 − x): \( \sqrt{\frac{1 - x}{1 + x}} = \frac{1 - x}{\sqrt{1 - x^2}} \).
2. \( \int_{0}^{1} \left(\frac{1}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}}\right)dx = \left[\sin^{-1} x + \sqrt{1 - x^2}\right]_{0}^{1} \).
3. \( = \left(\frac{\pi}{2} + 0\right) - (0 + 1) = \frac{\pi}{2} - 1 \).

Teacher's Note:
1. Rationalising √((1 − x)/(1 + x)) gives a numerator that splits into two standard integrals.
2. The integrand is between 0 and 1 on [0, 1], so the answer must be less than 1; π/2 − 1 ≈ 0.57 fits.

Answer: (c) \( \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \)

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

Teacher's Note:
1. For 1/(a + b cos x), the half-angle substitution t = tan(x/2) always works.
2. The value simplifies further to \( \frac{2}{\sqrt{3}}\cdot\frac{\pi}{6} = \frac{\pi}{3\sqrt{3}} \).

Answer: (a) \( \frac{\pi}{4} - \frac{1}{2} \log 2 \)

Explanation:
1. By parts: \( \int \tan^{-1} x\,dx = x\tan^{-1} x - \int \frac{x}{1 + x^2}dx = x\tan^{-1} x - \frac{1}{2}\log(1 + x^2) \).
2. \( \left[x\tan^{-1} x - \frac{1}{2}\log(1 + x^2)\right]_{0}^{1} = \frac{\pi}{4} - \frac{1}{2}\log 2 \).

Teacher's Note:
1. For an inverse trigonometric function alone, integrate by parts with 1 as the second function.
2. The integrand is below π/4 on [0, 1], so the answer must be less than π/4 ≈ 0.785; π/4 − ½ log 2 ≈ 0.44 fits.

Answer: (b) \( \frac{1}{2} - \frac{\sqrt{3}\,\pi}{12} \)

Explanation:
1. Put t = sin⁻¹x: x = sin t, \( \frac{dx}{\sqrt{1 - x^2}} = dt \), limits 0 to π/6.
2. \( \int_{0}^{\pi/6} t\sin t\,dt = [-t\cos t + \sin t]_{0}^{\pi/6} \).
3. \( = -\frac{\pi}{6}\cdot\frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1}{2} - \frac{\sqrt{3}\,\pi}{12} \).

Teacher's Note:
1. The combination \( \frac{dx}{\sqrt{1 - x^2}} \) is exactly d(sin⁻¹x), so substitute t = sin⁻¹x.
2. ∫ t sin t dt = −t cos t + sin t; evaluate the cos term with its sign carefully.

Answer: (a) π + 2

Explanation:
1. Multiply inside by (2 + x): \( \sqrt{\frac{2 + x}{2 - x}} = \frac{2 + x}{\sqrt{4 - x^2}} \).
2. \( \int_{0}^{2} \left(\frac{2}{\sqrt{4 - x^2}} + \frac{x}{\sqrt{4 - x^2}}\right)dx = \left[2\sin^{-1}\frac{x}{2} - \sqrt{4 - x^2}\right]_{0}^{2} \).
3. \( = (\pi - 0) - (0 - 2) = \pi + 2 \).

Teacher's Note:
1. Rationalise so that the denominator becomes √(a² − x²), then split.
2. The lower-limit term −√4 = −2 contributes +2; missing it gives π.

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

Explanation:
1. Put x = sin θ, \( \frac{dx}{\sqrt{1 - x^2}} = d\theta \); limits 0 to π/2.
2. The integral becomes \( \int_{0}^{\pi/2} \log\sin\theta\,d\theta \).
3. This standard integral equals \( -\frac{\pi}{2}\log 2 \).

Teacher's Note:
1. Remember \( \int_{0}^{\pi/2} \log\sin x\,dx = \int_{0}^{\pi/2} \log\cos x\,dx = -\frac{\pi}{2}\log 2 \).
2. log x < 0 on (0, 1), so the answer must be negative; this leaves only (c) and (d).

Answer: (d) \( \frac{\pi}{8} \)

Explanation:
1. Put t = sin²x, dt = 2 sin x cos x dx; limits 0 to 1.
2. \( \frac{1}{2}\int_{0}^{1} \frac{dt}{1 + t^2} = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8} \).

Teacher's Note:
1. sin⁴x = (sin²x)², so t = sin²x makes the denominator 1 + t².
2. The factor ½ from dt = 2 sin x cos x dx halves π/4 to π/8.

Answer: (d) \( \frac{1}{ab} \)

Explanation:
1. ax + b(1 − x) = (a − b)x + b. \( \int \frac{dx}{[(a - b)x + b]^2} = -\frac{1}{(a - b)[(a - b)x + b]} \).
2. \( \left[-\frac{1}{(a - b)((a - b)x + b)}\right]_{0}^{1} = \frac{1}{a - b}\left(\frac{1}{b} - \frac{1}{a}\right) = \frac{1}{a - b}\cdot\frac{a - b}{ab} \).
3. \( = \frac{1}{ab} \).

Teacher's Note:
1. Rewrite the bracket as a linear expression px + q before integrating its power.
2. Quick check: with a = b the integrand is 1/a², whose integral over [0, 1] is 1/a² = 1/(ab).

Answer: (a) − log 2

Explanation:
1. Put t = πx/2, dx = (2/π)dt; limits 0 to π/2.
2. \( \frac{2}{\pi}\int_{0}^{\pi/2} \log\sin t\,dt = \frac{2}{\pi}\left(-\frac{\pi}{2}\log 2\right) = -\log 2 \).

Teacher's Note:
1. Scale the variable so that the standard result ∫₀^(π/2) log sin t dt = −(π/2) log 2 can be used.
2. The integrand is negative throughout (0, 1), so only − log 2 or −(π/2) log 2 can be right; the factor 2/π decides.

Answer: (d) None of these

Explanation:
1. Note \( \frac{d}{dx}(x^3 + 3e^x + 4) = 3x^2 + 3e^x = 3(x^2 + e^x) \).
2. So the integral \( = \left[\frac{(x^3 + 3e^x + 4)^2}{6}\right]_{0}^{1} = \frac{(3e + 5)^2 - 7^2}{6} = \frac{(3e - 2)(3e + 12)}{6} \).
3. This is about 20.7, which matches none of (a), (b), (c).

Teacher's Note:
1. Look for one factor being a constant multiple of the derivative of the other: ∫ f·f′ dx = f²/2.
2. The integrand exceeds 7 on [0, 1], so any answer near 1 is impossible.

Answer: (a) 8/5

Explanation:
1. Put t = sin x, dt = cos x dx, cos²x = 1 − t²; limits −1 to 1.
2. \( \int_{-1}^{1} (1 - t^2)(1 + t)^2\,dt = \int_{-1}^{1} (1 + 2t - 2t^3 - t^4)\,dt \).
3. Odd terms vanish: \( 2\int_{0}^{1} (1 - t^4)\,dt = 2\left(1 - \frac{1}{5}\right) = \frac{8}{5} \).

Teacher's Note:
1. An odd power of cos x suggests t = sin x, keeping one cos x for dt.
2. After substitution the limits are symmetric, so drop the odd terms at once.

Answer: (b) 1

Explanation:
1. An antiderivative of \( \frac{1}{1 + \sin x} \) is tan x − sec x \( = -\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) \).
2. \( \left[-\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right]_{0}^{\pi/2} = -\tan 0 + \tan\frac{\pi}{4} = 1 \).

Teacher's Note:
1. Multiplying by (1 − sin x)/(1 − sin x) turns the integrand into sec²x − sec x tan x.
2. Using the tan(π/4 − x/2) form avoids the undefined values of tan and sec at π/2.

Answer: (b) \( \frac{1}{a - b} \log \frac{a}{b} \)

Explanation:
1. Put t = a sin²x + b cos²x: dt = (2a sin x cos x − 2b cos x sin x)dx = (a − b) sin 2x dx; t: b → a.
2. \( \frac{1}{a - b}\int_{b}^{a} \frac{dt}{t} = \frac{1}{a - b}\log\frac{a}{b} \).

Teacher's Note:
1. The numerator sin 2x is, up to a constant, the derivative of the denominator; the integral is a logarithm.
2. At x = 0 the denominator is b and at π/2 it is a; getting the limits the right way round fixes the sign.

Answer: (b) 2

Explanation:
1. Put t = π ln x, dt = (π/x)dx; x = 1 → t = 0, x = e³⁷ → t = 37π.
2. \( \int_{0}^{37\pi} \sin t\,dt = [-\cos t]_{0}^{37\pi} = -\cos 37\pi + 1 = 1 + 1 = 2 \).

Teacher's Note:
1. The factor π/x is exactly the derivative of π ln x, so the whole integrand becomes sin t dt.
2. cos(odd multiple of π) = −1; this is what makes the answer 2 rather than 0.

Answer: (d) log 4

Explanation:
1. Put x = t², dx = 2t dt; limits 0 to 1. Then x + √x = t² + t = t(t + 1).
2. \( \int_{0}^{1} \frac{2t\,dt}{t(t + 1)} = 2[\log(t + 1)]_{0}^{1} = 2\log 2 = \log 4 \).

Teacher's Note:
1. Substituting x = t² clears √x and makes the integrand rational.
2. Write 2 log 2 as log 4 to match the options.

Answer: (a) 2

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

Teacher's Note:
1. A function of √x times \( \frac{1}{\sqrt{x}} \) always suggests t = √x.
2. Note √(π²/4) = π/2 for the new upper limit.

Answer: (a) \( 2(3\sqrt{3} - \pi)/3 \)

Explanation:
1. Put x = 2 sec θ: dx = 2 sec θ tan θ dθ, √(x² − 4) = 2 tan θ; θ: 0 → π/3.
2. Integrand becomes \( \frac{2\tan\theta}{2\sec\theta}\cdot 2\sec\theta\tan\theta = 2\tan^2\theta = 2(\sec^2\theta - 1) \).
3. \( 2[\tan\theta - \theta]_{0}^{\pi/3} = 2\left(\sqrt{3} - \frac{\pi}{3}\right) = \frac{2(3\sqrt{3} - \pi)}{3} \).

Teacher's Note:
1. For √(x² − a²), the substitution x = a sec θ removes the root.
2. x = 4 gives sec θ = 2, i.e. θ = π/3.

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

Explanation:
1. Put t = sin x, dt = cos x dx; limits 0 to 1.
2. \( \int_{0}^{1} \frac{dt}{1 + t^2} = \tan^{-1} 1 = \frac{\pi}{4} \).

Teacher's Note:
1. cos x dx with a function of sin x signals t = sin x.
2. The integrand is at most 1 on an interval of length π/2, so values π and 2π are impossible.

Answer: (a) (1/2) log 3

Explanation:
1. Put t = tan(x/2): \( 1 + 2\sin x + \cos x = \frac{(1 + t^2) + 4t + (1 - t^2)}{1 + t^2} = \frac{2 + 4t}{1 + t^2} \), \( dx = \frac{2\,dt}{1 + t^2} \).
2. \( \int_{0}^{1} \frac{2\,dt}{2 + 4t} = \int_{0}^{1} \frac{dt}{1 + 2t} = \frac{1}{2}[\log(1 + 2t)]_{0}^{1} = \frac{1}{2}\log 3 \).

Teacher's Note:
1. With t = tan(x/2), the t² terms cancel here because 1 + cos x already contains 1 + t² and 1 − t².
2. Limits: x = π/2 gives t = tan(π/4) = 1.

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

Explanation:
1. Put x = tan θ: dx = sec²θ dθ, \( (x^2 + 1)^{3/2} = \sec^3\theta \); θ: 0 → π/4.
2. \( \int_{0}^{\pi/4} \cos\theta\,d\theta = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \).

Teacher's Note:
1. For powers of (x² + a²), use x = a tan θ.
2. An antiderivative to remember: \( \int \frac{dx}{(x^2 + 1)^{3/2}} = \frac{x}{\sqrt{x^2 + 1}} \).

Answer: (d) \( \pi/8 \)

Explanation:
1. Put t = x⁴, dt = 4x³ dx; limits 0 to 1. Then x⁸ = t².
2. \( \frac{1}{4}\int_{0}^{1} \frac{dt}{\sqrt{1 - t^2}} = \frac{1}{4}\cdot\frac{\pi}{2} = \frac{\pi}{8} \).

Teacher's Note:
1. x³ is (up to a constant) the derivative of x⁴, and x⁸ = (x⁴)²; this pairing decides the substitution.
2. Do not forget the factor ¼ from dt = 4x³ dx.

Answer: (c) 1

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

Teacher's Note:
1. The substitution t = sin⁻¹x is natural because dx/√(1 − x²) is its differential.
2. At t = π/2, cos t = 0, so the −t cos t term vanishes.

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

Explanation:
1. Put t = tan(x/2) (t: 0 → ∞): \( \int_{0}^{\infty} \frac{2\,dt}{(a + b) + (a - b)t^2} \).
2. \( = \frac{2}{a - b}\cdot\sqrt{\frac{a - b}{a + b}}\left[\tan^{-1}\left(t\sqrt{\frac{a - b}{a + b}}\right)\right]_{0}^{\infty} = \frac{2}{\sqrt{a^2 - b^2}}\cdot\frac{\pi}{2} \).
3. \( = \frac{\pi}{\sqrt{a^2 - b^2}} \).

Teacher's Note:
1. The condition a > b > 0 keeps a + b cos x positive, so the integral is finite.
2. Check with b = 0: the integral is π/a, matching π/√(a² − 0).

Answer: (c) \( \pi^2/8 \)

Explanation:
1. Put t = sin⁻¹x, \( dt = \frac{dx}{\sqrt{1 - x^2}} \); limits 0 to π/2.
2. \( \int_{0}^{\pi/2} t\,dt = \frac{1}{2}\cdot\frac{\pi^2}{4} = \frac{\pi^2}{8} \).

Teacher's Note:
1. The integrand is f·f′ with f = sin⁻¹x, so the antiderivative is (sin⁻¹x)²/2.
2. Squaring π/2 gives π²/4; halving gives π²/8.

Answer: (b) \( -ae^{-b} \)

Explanation:
1. Put t = b − s: dt = −ds; t = b − 1 → s = 1, t = b → s = 0; \( e^{-t} = e^{-b}e^{s} \), t − b − 1 = −(s + 1).
2. \( \int_{1}^{0} \frac{e^{-b}e^{s}}{-(s + 1)}(-ds) = e^{-b}\int_{1}^{0} \frac{e^s\,ds}{s + 1} = -e^{-b}\int_{0}^{1} \frac{e^s\,ds}{s + 1} \).
3. \( = -ae^{-b} \).

Teacher's Note:
1. When a given integral must be reused, choose a substitution that makes the new limits match the known ones.
2. Track both minus signs (from dt and from the denominator) and then the reversal of limits.

Answer: (a) 6

Explanation:
1. \( \frac{(x - x^3)^{1/3}}{x^4} = \frac{x\left(\frac{1}{x^2} - 1\right)^{1/3}}{x^4} = \frac{1}{x^3}\left(\frac{1}{x^2} - 1\right)^{1/3} \).
2. Put \( t = \frac{1}{x^2} - 1 \), \( dt = -\frac{2}{x^3}dx \); x = 1/3 → t = 8, x = 1 → t = 0.
3. \( \frac{1}{2}\int_{0}^{8} t^{1/3}\,dt = \frac{1}{2}\cdot\frac{3}{4}\left[t^{4/3}\right]_{0}^{8} = \frac{3}{8}\cdot 16 = 6 \).

Teacher's Note:
1. Take the highest power of x out of the bracket so that the remaining expression has a simple derivative.
2. 8^(4/3) = (8^(1/3))⁴ = 2⁴ = 16.

Answer: (c) \( \frac{1}{b^2} \log\left(\frac{a^2 + b^2}{a^2}\right) \)

Explanation:
1. Put t = a² + b² sin²x, dt = b² sin 2x dx; t: a² → a² + b².
2. \( \frac{1}{b^2}\int_{a^2}^{a^2 + b^2} \frac{dt}{t} = \frac{1}{b^2}\log\frac{a^2 + b^2}{a^2} \).

Teacher's Note:
1. d(sin²x) = sin 2x dx, so sin 2x in the numerator pairs with sin²x in the denominator.
2. The ratio of the denominator's end values gives the log; at x = 0 it is a² and at π/2 it is a² + b².

Answer: (b) 1 − log 2

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

Teacher's Note:
1. Write t/(1 + t) as 1 − 1/(1 + t) before integrating.
2. The integrand never exceeds ½ on [0, π/2], so the value is below π/4; this rules out 3 ± log 2 and 3/2 − log 2.

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

Explanation:
1. Put x = 2 sec θ: dx = 2 sec θ tan θ dθ, √(x² − 4) = 2 tan θ, x⁴ = 16 sec⁴θ; θ: 0 → π/3.
2. Integrand becomes \( \frac{2\tan\theta\cdot 2\sec\theta\tan\theta}{16\sec^4\theta} = \frac{1}{4}\sin^2\theta\cos\theta \).
3. \( \frac{1}{4}\left[\frac{\sin^3\theta}{3}\right]_{0}^{\pi/3} = \frac{1}{12}\cdot\frac{3\sqrt{3}}{8} = \frac{\sqrt{3}}{32} \).

Teacher's Note:
1. x = a sec θ is the standard substitution for √(x² − a²).
2. The integrand is positive, so the negative option is impossible.

Answer: (c) log 4

Explanation:
1. Put t = √(eᵘ − 1) (u the variable): eᵘ = t² + 1, du = 2t dt/(t² + 1).
2. \( \int \frac{du}{\sqrt{e^u - 1}} = \int \frac{2\,dt}{t^2 + 1} = 2\tan^{-1}\sqrt{e^u - 1} \); at u = log 2 this is 2·(π/4) = π/2.
3. So \( 2\tan^{-1}\sqrt{e^x - 1} - \frac{\pi}{2} = \frac{\pi}{6} \), giving \( \tan^{-1}\sqrt{e^x - 1} = \frac{\pi}{3} \).
4. eˣ − 1 = 3, so eˣ = 4 and x = log 4.

Teacher's Note:
1. Substituting the entire square root as t is the quickest way to rationalise √(eˣ − 1).
2. Check: at x = log 4 the upper value is 2 tan⁻¹√3 = 2π/3, and 2π/3 − π/2 = π/6.

Answer: (b) \( \frac{1}{3} \)

Explanation:
1. Put t = x⁴, dt = 4x³ dx: \( \frac{1}{4}\int_{0}^{1} \frac{t\,dt}{\sqrt{1 - t}} \).
2. Put u = 1 − t: \( \frac{1}{4}\int_{0}^{1} \frac{1 - u}{\sqrt{u}}du = \frac{1}{4}\left[2\sqrt{u} - \frac{2}{3}u^{3/2}\right]_{0}^{1} \).
3. \( = \frac{1}{4}\cdot\frac{4}{3} = \frac{1}{3} \).

Teacher's Note:
1. Split x⁷ as x⁴·x³ so that x³ dx goes into dt.
2. A second substitution u = 1 − t turns the integrand into simple powers of u.

Answer: (c) (1/2) log 3

Explanation:
1. Divide numerator and denominator by x²: \( \frac{\frac{1}{x^2} - 1}{x^2 + 1 + \frac{1}{x^2}} = -\frac{\left(1 - \frac{1}{x^2}\right)}{\left(x + \frac{1}{x}\right)^2 - 1} \).
2. Put u = x + 1/x, du = (1 − 1/x²)dx; x: 0 → 1 gives u: ∞ → 2.
3. \( \int_{\infty}^{2} \frac{-du}{u^2 - 1} = \int_{2}^{\infty} \frac{du}{u^2 - 1} = \frac{1}{2}\left[\log\frac{u - 1}{u + 1}\right]_{2}^{\infty} = 0 - \frac{1}{2}\log\frac{1}{3} = \frac{1}{2}\log 3 \).

Teacher's Note:
1. For (1 ∓ x²)/(polynomial in x² with symmetric coefficients), divide by x² and substitute x ± 1/x.
2. The sign of the numerator decides which combination to use: 1 − 1/x² pairs with x + 1/x.

Answer: (c) \( k = \frac{1}{12} \)

Explanation:
1. Write 13 = 13(sin²θ + cos²θ): the denominator is \( 9\sin^2\theta + 4\cos^2\theta \).
2. Using \( \int_{0}^{\pi/2} \frac{d\theta}{a^2\cos^2\theta + b^2\sin^2\theta} = \frac{\pi}{2ab} \) with a = 2, b = 3: value \( = \frac{\pi}{12} \).
3. So πk = π/12, i.e. \( k = \frac{1}{12} \).

Teacher's Note:
1. Replace the constant by itself times (sin²θ + cos²θ) to make the denominator homogeneous.
2. The result π/(2ab) comes from dividing by cos²θ and putting t = tan θ.

Answer: (c) \( \pi/12 \)

Explanation:
1. \( x^2 - 2x + 10 = (x - 1)^2 + 9 \).
2. \( \int_{1}^{4} \frac{dx}{(x - 1)^2 + 3^2} = \frac{1}{3}\left[\tan^{-1}\frac{x - 1}{3}\right]_{1}^{4} = \frac{1}{3}\cdot\frac{\pi}{4} = \frac{\pi}{12} \).

Teacher's Note:
1. Complete the square to reach 1/(u² + a²) and use (1/a) tan⁻¹(u/a).
2. The factor 1/3 is easy to forget; without it you would get π/4.

Answer: (d) e − 1

Explanation:
1. Put t = sin x, dt = cos x dx; limits 0 to 1.
2. \( \int_{0}^{1} e^t\,dt = e - 1 \).

Teacher's Note:
1. cos x is the derivative of the exponent sin x, so the integral is e^(sin x) directly.
2. e^(sin x) at π/2 is e, and at 0 it is 1.

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

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

Teacher's Note:
1. Shifting the variable and the limits by the same amount leaves the integral unchanged.
2. The letter used for the variable of integration does not matter, so ∫ f(t) dt = ∫ f(x) dx.