JEE Mathematics Indefinite Integration MCQs Set 20

Answer: (b) \( \sqrt{x^{2} + 1} \sqrt{\frac{1}{x^{2} - 1}} + c \)

Explanation:
1. Put x² = u: the integral is ∫2x dx/... = ∫du/((1 - u)√(u² - 1)).
2. Writing (1 - u)√(u² - 1) = -(u - 1)^(3/2)(u + 1)^(1/2) and putting √((u + 1)/(u - 1)) = s, the integral reduces to a standard form.
3. Integral = √((x² + 1)/(x² - 1)) + c.
4. Differentiating this result returns the integrand, so option (b) 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: (c) Both I and II are true

Explanation:
1. I: divide numerator and denominator by cos²x and put tan x = t: ∫sec²x dx/(a² tan²x + b²) = ∫dt/(a²t² + b²) = (1/(ab)) tan⁻¹(at/b). So I is true.
2. II: write 3cos x + 2 sin x = A(4 sin x + 5 cos x) + B(4 cos x - 5 sin x). Then 5A + 4B = 3 and 4A - 5B = 2, giving A = 23/41 and B = 2/41.
3. So the integral is (23/41)x + (2/41) log|4 sin x + 5 cos x| + c, and II is true.

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

Explanation:
1. Let f(x) = ax³ + bx² + cx + d. f(0) = 1 gives d = 1, and x = 0 being a critical point gives c = 0.
2. No local extremum at x = 0 means f'' also vanishes there, so b = 0. Then f(1) = 2 gives a = 1, so f(x) = x³ + 1.
3. (x³ + 1)/(x² + 1) = x - x/(x² + 1) + 1/(x² + 1).
4. Integral = x²/2 - (1/2) log(x² + 1) + tan⁻¹x + 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) \( -\frac16 \)

Explanation:
1. x⁴ + 5x² + 4 = (x² + 1)(x² + 4), and 1/((x² + 1)(x² + 4)) = (1/3)[1/(x² + 1) - 1/(x² + 4)].
2. ∫ = (1/3) tan⁻¹x - (1/6) tan⁻¹(x/2).
3. So A = 1/3 and B = -1/6.
4. Differentiating this result returns the integrand, so option (d) 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: (c) \( \tan{\left(x \right)} - \tan^{-1}{\left(x \right)} + c \)

Explanation:
1. x² + sin²x = (1 + x²) - cos²x, so the integrand is sec²x - 1/(1 + x²).
2. Integral = tan x - tan⁻¹x.
3. 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) \( \frac{12 \left(\sqrt[4]{x} + 1\right)^{\frac{7}{3}}}{7} - 3 \left(\sqrt[4]{x} + 1\right)^{\frac{4}{3}} + c \)

Explanation:
1. Put 1 + x^(1/4) = z³, so x^(1/4) = z³ - 1, x = (z³ - 1)⁴ and dx/√x = 4x^(1/4)... = 12z²(z³ - 1) dz.
2. The integral is ∫z · 12 z²(z³ - 1) dz = 12∫(z⁶ - z³) dz = 12(z⁷/7 - z⁴/4).
3. Integral = 12[(1 + ⁴√x)^(7/3)/7 - (1 + ⁴√x)^(4/3)/4].
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.

Answer: (d) \( e^{2 x} \sin{\left(3 x \right)} + c \)

Explanation:
1. Use ∫e^(kx)[k f(x) + f'(x)] dx = e^(kx) f(x) with k = 2 and f = sin 3x.
2. Integral = e^(2x) sin 3x.
3. Differentiating this result returns the integrand, so option (d) 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) \( \tan^{-1}{\left(\tan{\left(x \right)} - \cot{\left(x \right)} \right)} + c \)

Explanation:
1. Divide numerator and denominator by cos⁶x: sec⁶x/(1 + tan⁶x).
2. Put tan x = t: ∫(1 + t²)²/(1 + t⁶) dt = ∫(1 + t²)/(1 - t² + t⁴) dt, since 1 + t⁶ = (1 + t²)(1 - t² + t⁴).
3. Divide by t²: ∫(1 + 1/t²)/(t² - 1 + 1/t²) dt, and with u = t - 1/t this is ∫du/(u² + 1).
4. Integral = tan⁻¹(tan x - cot x) + c.
5. Differentiating this result returns the integrand, so option (a) 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: (b) \( \sqrt{2} \log{\left(\left|{\tan{\left(\frac{x}{4} + \frac{3 \pi}{8} \right)}}\right| \right)} + c \)

Explanation:
1. 1 - sin x = (cos(x/2) - sin(x/2))² = 2cos²(π/4 + x/2), and cos(π/4 + x/2) > 0 for x in (π/4, π/2).
2. ∫dx/(√2 cos(π/4 + x/2)) = (1/√2)∫sec(π/4 + x/2) dx = (1/√2)·2 log|tan(π/4 + (π/4 + x/2)/2)|.
3. Integral = √2 log|tan(3π/8 + x/4)|.
4. Differentiating this result returns the integrand, so option (b) 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: (a) \( \frac{x^{8}}{x^{8} + x + 1} + c \)

Explanation:
1. Divide numerator and denominator by x¹⁶: the numerator becomes (7/x⁸ + 8/x⁹) and the denominator (1/x⁸ + 1/x⁷ + 1)².
2. Put 1/x⁸ + 1/x⁷ + 1 = t, so dt = -(8/x⁹ + 7/x⁸) dx.
3. ∫ = -∫dt/t² = 1/t = x⁸/(1 + x + x⁸).
4. Integral = x⁸/(1 + x + x⁸) + 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: (c) \( \frac{\sqrt{x^{4} + x^{2} + 1}}{x} + c \)

Explanation:
1. Divide numerator and denominator by x³: (x - 1/x³)/√(x² + 1 + 1/x²)·... = (x - 1/x³) dx/√(x² + 1 + 1/x²).
2. Put x² + 1 + 1/x² = t, so (1/2)dt = (x - 1/x³) dx.
3. ∫dt/(2√t) = √t = √(x⁴ + x² + 1)/x.
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: (c) \( a=1,\ b=-1 \)

Explanation:
1. Let √(1 + x²) = t, so x dx/√(1 + x²) = dt and x + √(1 + x²) = t + √(t² - 1).
2. I = ∫ln(t + √(t² - 1)) dt = t ln(t + √(t² - 1)) - ∫t · (1/√(t² - 1)) dt.
3. ∫t dt/√(t² - 1) = √(t² - 1) = x.
4. So I = √(1 + x²) ln(x + √(1 + x²)) - x + c, giving a = 1 and b = -1.
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) \( f(1)=\frac1{\alpha-2\beta} \)

Explanation:
1. Since 1012 + 3012 = 2·2012, the integrand is x (1 + x²)^(-1012/2012) (2 + x²)^(-3012/2012) = [(1 + x²)/(2 + x²)]^(-1012/2012) · x/(2 + x²)².
2. Put t = (1 + x²)/(2 + x²), so dt = 2x dx/(2 + x²)².
3. The integral is (1/2)∫t^(-1012/2012) dt = (503/500) t^(250/503) = (503/500)(1 - 1/(2 + x²))^(250/503).
4. So α = 503, β = 250 and f(x) = 1/(2 + x²). Then f(1) = 1/3 = 1/(α - 2β), while f(√2) = 1/4 and f(√3) = 1/5 do not match the other options.

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: (b) \( 1 \)

Explanation:
1. The integrand is cos²x/(x sin x + cos x)² = (x cos x/(x sin x + cos x)²)(cos x/x).
2. By parts with v' = x cos x/(x sin x + cos x)², v = -1/(x sin x + cos x): I = -cos x/(x(x sin x + cos x)) - ∫dx/x² = -cos x/(x(x sin x + cos x)) + 1/x.
3. Simplifying, I = sin x/(x sin x + cos x) = tan x/(x tan x + 1).
4. Hence k = 1.
5. 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) c, a, b

Explanation:
1. I: ∫dx/√(a² - x²) = sin⁻¹(x/a), so I matches (c).
2. II: ∫dx/√(x² + a²) = sinh⁻¹(x/a), so II matches (a).
3. III: ∫dx/√(x² - a²) = cosh⁻¹(x/a), so III matches (b).
4. With a = 2 the matching is I-c, II-a, III-b.

Teacher's Note:
1. Learn the standard integrals thoroughly; matching questions reduce to recalling them accurately.
2. Check each pairing by differentiating the result where in doubt.

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

Explanation:
1. R is the standard result (substitute tan(x/2) = t and complete the square).
2. With a = 5, b = 4: √(a² - b²) = 3, so the result is (2/3) tan⁻¹((4 + 5 tan(x/2))/3).
3. This is A, so 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: (c) I-c, II-b, III-a

Explanation:
1. Put √x = t, so dx/√x = 2 dt.
2. I: 2∫sin t dt = -2cos √x, which is (c).
3. II: 2∫cos t dt = 2 sin √x, which is (b).
4. III: 2∫e^t dt = 2e^(√x), which is (a).

Teacher's Note:
1. Learn the standard integrals thoroughly; matching questions reduce to recalling them accurately.
2. Check each pairing by differentiating the result where in doubt.

Answer: (d) \( x \log{\left(x \right)}^{2} + c \)

Explanation:
1. Put log x = t, so x = e^t and dx = e^t dt: ∫(2t + t²)e^t dt = e^t t² (the form e^t[f + f'] with f = t²).
2. Integral = x(log x)².
3. Differentiating this result returns the integrand, so option (d) 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. R is the standard result for a > b (substitute tan(x/2) = t).
2. With a = 3, b = 2: a² - b² = 5 and (a - b)/(a + b) = 1/5.
3. This gives (2/√5) tan⁻¹((1/√5) tan(x/2)), which is 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: (d) \( -\frac83\log|1-x|+\frac23x+c \)

Explanation:
1. Put t = (3x - 4)/(3x + 4), so x = (4t + 4)/(3 - 3t).
2. f(t) = x + 2 = (4t + 4)/(3(1 - t)) + 2 = (10 - 2t)/(3(1 - t)) = (2/3)[1 + 4/(1 - t)].
3. ∫f(t) dt = (2/3)t - (8/3) log|1 - t|.
4. Hence ∫f(x) dx = (2/3)x - (8/3) log|1 - x| + c.

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) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Put tan x = t: ∫t⁴ dt = t⁵/5, so A is true.
2. R is the power rule for integration (valid for n ≠ -1) and it is exactly the rule used in A with f(x) = tan x and n = 4.

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: (d) Neither I nor II

Explanation:
1. Differentiate sin^(m+1)x cos^(n-1)x: it equals (m + 1) sin^m x cos^n x - (n - 1) sin^(m+2)x cos^(n-2)x = (m + n) sin^m x cos^n x - (n - 1) sin^m x cos^(n-2)x.
2. Integrating and rearranging gives (m + n) I(m,n) = sin^(m+1)x cos^(n-1)x + (n - 1) I(m,n-2), so Statement I is correct.
3. Similarly, differentiating sin^(m-1)x cos^(n+1)x gives -(m + n) sin^m x cos^n x + (m - 1) sin^(m-2)x cos^n x, which proves Statement II.
4. Both statements are correct, so neither is incorrect.

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) \( - 3 x + \tan{\left(x \right)} - \cot{\left(x \right)} + c \)

Explanation:
1. sin⁶x + cos⁶x = 1 - 3sin²x cos²x.
2. The integrand is sec²x cosec²x - 3 = sec²x + cosec²x - 3.
3. Integral = tan x - cot x - 3x.
4. Differentiating this result returns the integrand, so option (a) 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.