JEE Mathematics Indefinite Integration MCQs Set 18

Answer: (c) \( - \frac{\left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{- x}}{2} + c \)

Explanation:
1. Use ∫e^(ax) sin bx dx = e^(ax)(a sin bx - b cos bx)/(a² + b²) with a = -1, b = 1.
2. = e^(-x)(-sin x - cos x)/2.
3. Integral = -(e^(-x)/2)(sin x + cos x).
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The derivative of 3x² + 14x - 5 is 6x + 14 = 2(3x + 7), so the integrand is (1/2) f'(x)/f(x).
2. By the rule in R, the integral is (1/2) log|3x² + 14x - 5| + c. So A is true, R is true and R explains A.

Teacher's Note:
1. First decide whether the assertion and the reason are each true, then decide whether the reason actually explains the assertion.
2. Standard forms for quadratics under a root depend on the sign of the leading coefficient and of the discriminant.

Answer: (c) \( \sqrt{e^{2 x} - 1} - \sec^{-1}{\left(e^{x} \right)} + c \)

Explanation:
1. Put √(e^(2x) - 1) = u, so e^(2x) = u² + 1 and dx = u du/(u² + 1).
2. The integral is ∫u²/(u² + 1) du = u - tan⁻¹u.
3. Since tan⁻¹u = sec⁻¹(e^x) when u = √(e^(2x) - 1), integral = √(e^(2x) - 1) - sec⁻¹(e^x) + c.
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (d) \( \sin^{-1}{\left(2 \sin{\left(x \right)} - 1 \right)} + c \)

Explanation:
1. The integrand is √((1 + sin x)/sin x). Put sin x = t, so dx = dt/√(1 - t²) and cos x = √(1 - t²) = √(1 - t)√(1 + t).
2. The integral becomes ∫dt/√(t - t²) = ∫dt/√(1/4 - (t - 1/2)²).
3. This equals sin⁻¹(2t - 1).
4. Integral = sin⁻¹(2 sin x - 1) + c.
5. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the standard formula.
2. With a = 3, b = 4, c = 0: a² + b² = 25, which gives A. A is true and R explains it.

Teacher's Note:
1. First decide whether the assertion and the reason are each true, then decide whether the reason actually explains the assertion.
2. Standard forms for quadratics under a root depend on the sign of the leading coefficient and of the discriminant.

Answer: (b) \( g(1)=\frac{3^7}7 \)

Explanation:
1. x²⁷(1 + x + x²)⁶(6x² + 5x + 4) = (x⁴ + x⁵ + x⁶)⁶ (6x⁵ + 5x⁴ + 4x³), and 6x⁵ + 5x⁴ + 4x³ is the derivative of x⁶ + x⁵ + x⁴.
2. Put t = x⁶ + x⁵ + x⁴: g(x) = t⁷/7 + C, and g(0) = 0 gives C = 0.
3. g(1) = 3⁷/7, g(2) = 112⁷/7 and g(-1) = 1/7, so only g(1) = 3⁷/7 holds.

Teacher's Note:
1. Remember the standard integral formulas together with the conditions under which each holds.
2. Check every statement on its own, for example by differentiating the proposed result.

Answer: (a) \( e^{\tan^{-1}{\left(x \right)}} \sec^{-1}{\left(\sqrt{x^{2} + 1} \right)}^{2} + c \)

Explanation:
1. For x > 0, sec⁻¹√(1 + x²) = tan⁻¹x and cos⁻¹((1 - x²)/(1 + x²)) = 2 tan⁻¹x.
2. Put tan⁻¹x = t, so dx/(1 + x²) = dt.
3. The integral is ∫e^t(t² + 2t) dt = e^t t² (the form e^t[f + f'] with f = t²).
4. Integral = e^(tan⁻¹x)(tan⁻¹x)² = e^(tan⁻¹x)(sec⁻¹√(1 + x²))² + c.
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (c) \( 2 \sqrt{x \cos{\left(x \right)}} + c \)

Explanation:
1. The integrand is (cos x - x sin x)/√(x cos x).
2. Since d/dx(x cos x) = cos x - x sin x, this is (x cos x)'/√(x cos x).
3. Integral = 2√(x cos x) + c.
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) \( \frac{\log{\left(x^{2} + 2 \right)}}{2} + \frac{\sqrt{2} \tan^{-1}{\left(\frac{\sqrt{2} x}{2} \right)}}{2} + c \)

Explanation:
1. The numerator factorises: x⁵ + x⁴ + 4x³ + 4x² + 4x + 4 = (x² + 2)²(x + 1).
2. So the integrand is (x + 1)/(x² + 2).
3. ∫x/(x² + 2) dx = (1/2) log(x² + 2) and ∫dx/(x² + 2) = (1/√2) tan⁻¹(x/√2).
4. Integral = (1/2) log|x² + 2| + (1/√2) tan⁻¹(x/√2) + c.
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Split rational functions into partial fractions, or complete the square for quadratics.
2. Compare coefficients carefully to avoid algebra slips.

Answer: (d) 2

Explanation:
1. 5 tan x/(tan x - 2) = 5 sin x/(sin x - 2cos x).
2. Write 5 sin x = A(sin x - 2cos x) + B(cos x + 2 sin x): A + 2B = 5 and -2A + B = 0, so A = 1 and B = 2.
3. The integral is x + 2 log|sin x - 2cos x| + k, so a = 2.
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Convert products and powers of trigonometric functions to sums using identities, or substitute for the simplest function.
2. Check the final form against the options by differentiation.

Answer: (d) does not exist

Explanation:
1. √(x - 3) is real only for x ≥ 3.
2. For x ≥ 3, ln x ≥ ln 3 > 1, but sin⁻¹ and cos⁻¹ are defined only for arguments in [-1, 1].
3. So the integrand is not defined for any real x and the integral does not exist.
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Learn the standard integrals thoroughly; most one-line questions reduce to them after a small rewrite.
2. Always verify by differentiating the answer when options look similar.

Answer: (a) \( - 2 \log{\left(\frac{\sqrt{1 - x} + 1}{\sqrt{x}} \right)} + 2 \cos^{-1}{\left(\sqrt{x} \right)} + c \)

Explanation:
1. Put x = cos²θ, so dx = -2 sin θ cos θ dθ and √((1 - √x)/(1 + √x)) = √((1 - cos θ)/(1 + cos θ)) = tan(θ/2).
2. The integrand becomes -2 tan(θ/2) · sin θ cos θ/cos²θ dθ = -2 (1 - cos θ)/cos θ dθ = -2(sec θ - 1) dθ.
3. Integral = -2[log|sec θ + tan θ| - θ].
4. With sec θ = 1/√x, tan θ = √(1 - x)/√x and θ = cos⁻¹√x: integral = -2[log((1 + √(1 - x))/√x) - cos⁻¹√x] + c.
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is the standard formula.
2. With a = 5, b = 6, c = 0: a² + b² = 61, which gives A. A is true and R explains it.

Teacher's Note:
1. First decide whether the assertion and the reason are each true, then decide whether the reason actually explains the assertion.
2. Standard forms for quadratics under a root depend on the sign of the leading coefficient and of the discriminant.

Answer: (b) \( -\cos x\log(\sec x+\tan x) \)

Explanation:
1. By parts with u = log(sec x + tan x) and dv = sin x dx, v = -cos x.
2. ∫ = -cos x log(sec x + tan x) + ∫cos x · sec x dx.
3. = -cos x log(sec x + tan x) + x + c, so f(x) = -cos x log(sec x + tan x).
4. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (b) \( \frac{\left(\left(-2 + \sqrt{3}\right) \sin{\left(x \right)} + \left(1 + 2 \sqrt{3}\right) \cos{\left(x \right)}\right) e^{2 x}}{5} + c \)

Explanation:
1. Use ∫e^(2x) cos x dx = e^(2x)(2cos x + sin x)/5 and ∫e^(2x) sin x dx = e^(2x)(2 sin x - cos x)/5.
2. √3(2cos x + sin x) - (2 sin x - cos x) = (2√3 + 1) cos x + (√3 - 2) sin x.
3. Integral = (e^(2x)/5)[(2√3 + 1) cos x + (√3 - 2) sin x].
4. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (c) \( -\frac3{16} \)

Explanation:
1. sin 3x cos 5x = (1/2)(sin 8x - sin 2x).
2. ∫ = -cos 8x/16 + cos 2x/4 + C.
3. At x = 0: -1/16 + 1/4 + C = 0, so C = -3/16.
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Convert products and powers of trigonometric functions to sums using identities, or substitute for the simplest function.
2. Check the final form against the options by differentiation.

Answer: (c) \( \frac{\tan^{3}{\left(x \right)}}{3} + 2 \tan{\left(x \right)} - \cot{\left(x \right)} + c \)

Explanation:
1. Divide numerator and denominator by cos⁶x: the integrand is sec⁶x/tan²x dx.
2. Put tan x = t: sec⁶x dx = (1 + t²)² dt, so the integral is ∫(1 + t²)²/t² dt = ∫(t² + 2 + 1/t²) dt.
3. = t³/3 + 2t - 1/t = (1/3)tan³x + 2 tan x - cot x.
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Convert products and powers of trigonometric functions to sums using identities, or substitute for the simplest function.
2. Check the final form against the options by differentiation.

Answer: (d) I-c, II-d, III-a, IV-b

Explanation:
1. I: multiply by (1 - cos x): ∫(1 - cos x)/sin²x dx = -cot x + cosec x, which is (c).
2. II: multiply by (1 + cos x): ∫(1 + cos x)/sin²x dx = -cot x - cosec x, which is (d).
3. III: multiply by (1 + sin x): ∫(1 + sin x)/cos²x dx = tan x + sec x, which is (a).
4. IV: multiply by (1 - sin x): ∫(1 - sin x)/cos²x dx = tan x - sec x, which is (b).

Teacher's Note:
1. Multiplying by the conjugate converts 1/(1 ± cos x) and 1/(1 ± sin x) into forms that integrate at once.
2. Check each pairing by differentiating the result where in doubt.

Answer: (c) \( - \frac{\sinh^{-1}{\left(\frac{1}{x^{3}} \right)}}{3} + c \)

Explanation:
1. Write x√(x⁶ + 1) = x⁴√(1 + (1/x³)²) and put 1/x³ = t, so dx/x⁴ = -dt/3.
2. ∫ = -(1/3)∫dt/√(1 + t²) = -(1/3) sinh⁻¹t.
3. Integral = -(1/3) sinh⁻¹(1/x³).
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) \( 4 \)

Explanation:
1. Write x + 2 = -(1/2)(4 - 2x) + 4.
2. ∫-(1/2)(4 - 2x)/√(4x - x²) dx = -√(4x - x²).
3. ∫4 dx/√(4 - (x - 2)²) = 4 sin⁻¹((x - 2)/2).
4. So k = 4.
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Rationalise or simplify the integrand algebraically before integrating.
2. Substitution of a power of x often removes fractional exponents.

Answer: (c) \( \frac\pi4x+x\cot^{-1}x+\frac12\log(1+x^2)+c \)

Explanation:
1. For x > 0, cot⁻¹((x - 1)/(x + 1)) = π/2 - tan⁻¹((x - 1)/(x + 1)) = π/2 - (tan⁻¹x - π/4) = 3π/4 - tan⁻¹x.
2. ∫tan⁻¹x dx = x tan⁻¹x - (1/2) log(1 + x²).
3. Integral = 3πx/4 - x tan⁻¹x + (1/2) log(1 + x²).
4. Since tan⁻¹x = π/2 - cot⁻¹x, this equals πx/4 + x cot⁻¹x + (1/2) log(1 + x²).
5. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (d) \( \frac{\left(x + 2\right)^{10}}{10} - \frac{\left(x + 2\right)^{8}}{8} + c \)

Explanation:
1. Put x + 2 = t: (x + 1)(x + 3) = (t - 1)(t + 1) = t² - 1.
2. ∫(t² - 1)t⁷ dt = ∫(t⁹ - t⁷) dt = t¹⁰/10 - t⁸/8.
3. Integral = (x + 2)¹⁰/10 - (x + 2)⁸/8 + c.
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.