JEE Mathematics Indefinite Integration MCQs Set 22

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 = 4: b² - a² = 7, b + a = 7 and b - a = 1, so the result is (1/√7) log|(√7 + tan(x/2))/(√7 - tan(x/2))|.
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) Both I and II are true

Explanation:
1. I: put tan(x/2) = t, so cos x = (1 - t²)/(1 + t²) and dx = 2dt/(1 + t²). The integral is ∫2 dt/(7 - t²) = (1/√7) log|(√7 + t)/(√7 - t)|. I is true.
2. II: cos x - sin x = √2 cos(x + π/4), so the integral is (1/√2)∫sec(x + π/4) dx = (1/√2) log|tan(π/4 + (x + π/4)/2)| = (1/√2) log|tan(x/2 + 3π/8)|. 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) \( k=-\frac49 \)

Explanation:
1. Put ⁴√x + 1 = t, so x^(-3/4) dx = 4 dt.
2. ∫4 dt/t^10 = -(4/9) t^(-9).
3. So k = -4/9.
4. 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: (b) \( \left(\frac13,\frac13\right) \)

Explanation:
1. (x² + 2)/((x² + 1)(x² + 4)) = (1/3)/(x² + 1) + (2/3)/(x² + 4).
2. ∫ = (1/3) tan⁻¹x + (2/3)(1/2) tan⁻¹(x/2) = (1/3) tan⁻¹x + (1/3) tan⁻¹(x/2).
3. So (A, B) = (1/3, 1/3).
4. Differentiating this result returns the integrand, so option (b) 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) \( \frac{\sqrt{2} \sin^{-1}{\left(\frac{\sqrt{2} x}{x^{2} + 1} \right)}}{2} + c \)

Explanation:
1. Divide numerator and denominator by x²: (1/x² - 1) dx/((x + 1/x)√(x² + 1/x²)).
2. Put x + 1/x = t, so -(1 - 1/x²) dx = ... , and x² + 1/x² = t² - 2.
3. The integral becomes -∫dt/(t√(t² - 2)) = (1/√2) sin⁻¹(√2/t) with t = (x² + 1)/x.
4. Integral = (1/√2) sin⁻¹(√2 x/(x² + 1)).
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: (c) \( l^{r+1}(x)+c \)

Explanation:
1. d/dx[l^(k+1)(x)] = (l^k)'(x)/l^k(x).
2. So d/dx[l^(r+1)(x)] = 1/(x l(x) l²(x) ... l^r(x)).
3. Hence the integral is l^(r+1)(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: (b) \( \frac12 \)

Explanation:
1. Since f is a primitive, f'(x) = (tan x - sin x)/x³.
2. tan x - sin x = tan x (1 - cos x), so f'(x) = (tan x/x) · ((1 - cos x)/x²).
3. As x → 0, tan x/x → 1 and (1 - cos x)/x² → 1/2.
4. The limit is 1/2.

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) \( \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}} + c \)

Explanation:
1. Put x = a tan θ: dx = a sec²θ dθ and (a² + x²)^(3/2) = a³sec³θ.
2. The integral is (1/a²)∫cos θ dθ = (1/a²) sin θ.
3. With sin θ = x/√(a² + x²): Integral = x/(a²√(a² + x²)).
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) \( \sqrt{2} \sqrt{\left(\tan{\left(\alpha \right)} + \tan{\left(x \right)}\right) \sec{\left(\alpha \right)}} + c \)

Explanation:
1. sin(2x + α) + sin α = 2 sin(x + α) cos x = 2cos²x (tan x cos α + sin α).
2. So the integrand is (1/√2) sec²x/√(tan x cos α + sin α).
3. Put tan x cos α + sin α = t², so sec²x dx = 2t dt/cos α.
4. Integral = (1/√2)(2/cos α)√(tan x cos α + sin α) = √(2 sec α (tan x + tan α)) + c.
5. 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) \( x \)

Explanation:
1. Put 1 + 2cos α/x + 1/x² = t, so the integrand becomes (cos α/x² + 1/x³) dx/(1 + 2cos α/x + 1/x²)^(3/2) after dividing by x³.
2. dt = -2(cos α/x² + 1/x³) dx, so the integral is -(1/2)∫t^(-3/2) dt = t^(-1/2).
3. t^(-1/2) = x/√(x² + 2x cos α + 1) for x > 0.
4. Hence f(x) = x.
5. 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) \( \frac{3 \left(x + 1\right)^{\frac{2}{3}}}{2} - 3 \sqrt[3]{x + 1} + 3 \log{\left(\left|{\sqrt[3]{x + 1} + 1}\right| \right)} + c \)

Explanation:
1. Put x + 1 = t³, so dx = 3t² dt.
2. ∫3t²/(1 + t) dt = 3∫(t - 1 + 1/(1 + t)) dt = 3t²/2 - 3t + 3 log|1 + t|.
3. Integral = (3/2)(x + 1)^(2/3) - 3(x + 1)^(1/3) + 3 log|1 + ³√(x + 1)|.
4. 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) \( (1,2) \)

Explanation:
1. f'(x) = e^x (x - 1)(x - 2).
2. f decreases where f'(x) < 0, i.e. (x - 1)(x - 2) < 0 (since e^x > 0).
3. This holds for 1 < x < 2.
4. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Use the definition that the integral is the function whose derivative is the integrand.
2. Differentiating the options is the fastest check.

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

Explanation:
1. R is the standard result for this type of integral.
2. With a = 2, b = 3, c = 4, d = 5: (ac + bd)/(c² + d²) = 23/41 and (ad - bc)/(c² + d²) = (10 - 12)/41 = -2/41.
3. This gives 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) \( \log{\left(\left|{\frac{\sqrt{\sec{\left(2 x \right)}}}{\sec{\left(\alpha - x \right)} \sec{\left(\alpha + x \right)}}}\right| \right)} + c \)

Explanation:
1. Since 2x = (x + α) + (x - α), tan 2x = (tan(x + α) + tan(x - α))/(1 - tan(x + α)tan(x - α)).
2. So tan(x - α)tan(x + α)tan 2x = tan 2x - tan(x + α) - tan(x - α).
3. Integral = (1/2)log|sec 2x| - log|sec(x + α)| - log|sec(x - α)|.
4. = log|√(sec 2x)/(sec(x + α) sec(x - α))| + c.
5. 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) \( \frac{-n}{m+1}I_{m,n-1} \)

Explanation:
1. By parts with u = (log x)^n and dv = x^m dx: I_(m,n) = x^(m+1)(log x)^n/(m + 1) - (n/(m + 1))∫x^m (log x)^(n-1) dx.
2. So f(x) = -(n/(m + 1)) I_(m,n-1).
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: (b) \( \sqrt{2} \tan^{-1}{\left(\frac{\sqrt{2} \left(\tan{\left(x \right)} - 1\right)}{2 \sqrt{\tan{\left(x \right)}}} \right)} + c \)

Explanation:
1. Put tan x = t², so dx = 2t dt/(1 + t⁴).
2. The integral is ∫(1/t + t)·2t/(1 + t⁴) dt = 2∫(1 + t²)/(1 + t⁴) dt = 2∫(1 + 1/t²)/(t² + 1/t²) dt.
3. With u = t - 1/t: 2∫du/(u² + 2) = √2 tan⁻¹(u/√2).
4. Integral = √2 tan⁻¹((tan x - 1)/√(2 tan x)) + c.
5. 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: (d) \( \left(x - \sec{\left(x \right)}\right) e^{\sin{\left(x \right)}} + c \)

Explanation:
1. Split: ∫x e^(sin x) cos x dx - ∫e^(sin x) sec x tan x dx.
2. By parts on the first part with u = x and v = e^(sin x): x e^(sin x) - ∫e^(sin x) dx.
3. By parts on the second with u = e^(sin x) and v = sec x: e^(sin x) sec x - ∫e^(sin x) cos x sec x dx = e^(sin x) sec x - ∫e^(sin x) dx.
4. Subtracting, integral = e^(sin x)(x - sec x) + c.
5. 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) \( \frac{\left(n x^{n} + 1\right)^{1 - \frac{1}{n}}}{n \left(n - 1\right)} + c \)

Explanation:
1. f(x) = x/(1 + xⁿ)^(1/n) gives f(f(x)) = x/(1 + 2xⁿ)^(1/n), and in general the n-fold composition is g(x) = x/(1 + nxⁿ)^(1/n).
2. So xⁿ⁻² g(x) = xⁿ⁻¹/(1 + nxⁿ)^(1/n).
3. Put 1 + nxⁿ = t, so dt = n² xⁿ⁻¹ dx.
4. ∫t^(-1/n) dt/n² = (1/n²)·t^(1 - 1/n)/(1 - 1/n) = t^(1 - 1/n)/(n(n - 1)).
5. Integral = (1/(n(n - 1)))(1 + nxⁿ)^(1 - 1/n) + K.
6. 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: (d) \( \log{\left(\left|{\frac{\sqrt{2 x e^{\sin{\left(x \right)}} + 1} - 1}{\sqrt{2 x e^{\sin{\left(x \right)}} + 1} + 1}}\right| \right)} + c \)

Explanation:
1. For x > 0, √(2x³e^(sin x) + x²) = x√(2x e^(sin x) + 1).
2. Put 2x e^(sin x) + 1 = t², so 2e^(sin x)(1 + x cos x) dx = 2t dt.
3. The integral becomes ∫dt/(x e^(sin x)) with x e^(sin x) = (t² - 1)/2, i.e. ∫2 dt/(t² - 1) = log|(t - 1)/(t + 1)|.
4. Integral = log|(√(2x e^(sin x) + 1) - 1)/(√(2x e^(sin x) + 1) + 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: (d) \( - \sin{\left(x \right)} - \frac{\sin{\left(2 x \right)}}{2} + c \)

Explanation:
1. By 2 cos A cos B = cos(A + B) + cos(A - B): (cos x + cos 2x)(1 - 2cos 3x) = -(cos 4x + cos 5x).
2. So the integrand equals -(cos x + cos 2x).
3. Integral = -sin x - (1/2) sin 2x + c.
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) \( -\frac\pi4 \)

Explanation:
1. 1 + sin x = 1 + cos(π/2 - x) = 2cos²(π/4 - x/2) = 2cos²(x/2 - π/4).
2. So the integrand is (1/2) sec²(x/2 - π/4).
3. ∫ = tan(x/2 - π/4) + b, so k = -π/4.
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.