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📝 Reduction formulas for integrals (36 MCQs)

📖 From Calculus • 8. Principles of integral Evaluation • 36 questions available

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What is Reduction formulas for integrals?

Definition:
Reduction formulas express an integral involving a parameter $n$ in terms of a similar integral with a lower parameter $n-1$ or $n-2$ , recursively simplifying complex powers until reaching a base case.

Example:
For $I_n = \int \sin^n x \, dx$ , the formula is $I_n = -\frac{1}{n}\sin^{n-1}x \cos x + \frac{n-1}{n}I_{n-2}$ , reducing $\int \sin^3 x dx$ to simpler terms.

Reason:
They provide a systematic way to handle high-power trigonometric or algebraic integrals that are otherwise difficult to solve directly, breaking them down into manageable steps.

Easy

Medium

Hard

📝 All Reduction formulas for integrals MCQs

Q1. What is the primary purpose of deriving reduction formulas for integrals involving powers of trigonometric functions?

A.To replace a difficult integral with a simpler one involving a lower power of the function. ✅

B.To provide a direct method for evaluating definite integrals without limits.

C.To eliminate the need for substitution methods entirely.

D.To express integrals solely in terms of exponential functions.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Reduction formulas are derived to express an integral, like ∫sinⁿx dx, in terms of an integral with a lower power, such as sinⁿ⁻²x. This iterative process simplifies the integration of high powers. They don't replace all substitutions or handle definite integrals on their own; they are a tool to reduce complexity step-by-step.

Q2. Given the reduction formula ∫cosⁿx dx = (1/n)cosⁿ⁻¹x sin x + ((n-1)/n)∫cosⁿ⁻²x dx, what would be the first step to evaluate ∫cos⁶x dx?

A.Apply the formula with n=6 to get (1/6)cos⁵x sin x + (5/6)∫cos⁴x dx. ✅

B.Apply the formula with n=6 to get (1/6)cos⁵x sin x + (5/6)∫cos⁶x dx.

C.Apply the formula with n=6 to get (1/6)cos⁷x sin x + (5/6)∫cos⁴x dx.

D.Apply the formula with n=6 to get (1/6)cos⁵x sin x - (5/6)∫cos⁴x dx.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Correct Easy involves substituting n=6 into the formula. The second term's integral becomes ∫cos⁴x dx, which is a lower power. Option B incorrectly repeats the original integral, option C uses an incorrect power for the first term, and option D uses a minus sign. This demonstrates a direct Easy of the formula.

Q3. A student uses the reduction formula for ∫sinⁿx dx and gets the result -1/4 sin³x cos x + 3/8 x - 3/16 sin 2x. What was the original integral?

A.∫sin⁴x dx ✅

B.∫sin²x dx

C.∫sin⁶x dx

D.∫sin³x dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Applying the reduction formula (1) to n=4 gives -1/4 sin³x cos x + (3/4)∫sin²x dx. Evaluating ∫sin²x dx as (1/2)x - (1/4)sin 2x yields (3/8)x - (3/16)sin 2x. This matches the result. The student correctly applied the formula. Other options would give different results, so by comparing the structure of the answer with the pattern from the reduction formula, we can identify the original power.

Q4. Which of the following correctly represents the reduction formula for ∫cosⁿx dx?

A.∫cosⁿx dx = (1/n)cosⁿ⁻¹x sin x + ((n-1)/n)∫cosⁿ⁻²x dx ✅

B.∫cosⁿx dx = -(1/n)cosⁿ⁻¹x sin x + ((n-1)/n)∫cosⁿ⁻²x dx

C.∫cosⁿx dx = (1/n)cosⁿ⁻¹x sin x - ((n-1)/n)∫cosⁿ⁻²x dx

D.∫cosⁿx dx = (1/n)sinⁿ⁻¹x cos x + ((n-1)/n)∫sinⁿ⁻²x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The derivation uses u=cosⁿ⁻¹x and dv=cosx dx. This gives du=-(n-1)cosⁿ⁻²x sinx dx and v=sinx. Integration by parts yields cosⁿ⁻¹x sinx + (n-1)∫sin²x cosⁿ⁻²x dx. Replacing sin²x with 1-cos²x and simplifying gives the formula. Option A is the correct standard reduction formula.

Q5. A student uses reduction formula (9) to evaluate ∫sin⁵x dx and obtains -1/5 sin⁴x cosx + 4/5(-1/3 sin²x cosx - 2/3 cosx). What does this expression simplify to after collecting terms?

A.-1/5 sin⁴x cosx - 4/15 sin²x cosx - 8/15 cosx + C ✅

B.-1/5 sin⁴x cosx + 4/15 sin²x cosx + 8/15 cosx + C

C.1/5 sin⁴x cosx - 4/15 sin²x cosx - 8/15 cosx + C

D.-1/5 sin⁴x cosx - 4/15 sin²x cosx + 8/15 cosx + C

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Applying the formula for n=5 gives -1/5 sin⁴x cosx + 4/5∫sin³x dx. Then ∫sin³x dx = -1/3 sin²x cosx - 2/3 cosx. Multiplying by 4/5 gives -4/15 sin²x cosx - 8/15 cosx. The final expression is -1/5 sin⁴x cosx - 4/15 sin²x cosx - 8/15 cosx. The other options have incorrect signs or coefficients, indicating a common error in applying the formula.

Q6. What would be the most efficient way to evaluate ∫sin⁶x dx using reduction formulas?

A.Apply formula (9) repeatedly until reaching ∫sin⁰x dx or ∫sin¹x dx. ✅

B.Apply the formula for ∫sin²x dx directly.

C.Rewrite sin⁶x as (sin²x)³ and apply the power rule.

D.Use the double-angle formula for sin 2x repeatedly.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Reduction formula (9) reduces the power of sin x by 2 each time it is applied. Starting with n=6, it goes to ∫sin⁴x dx, then to ∫sin²x dx. This is the most systematic approach. Option C is a common misconception; the power of sin is not a power of a variable x. The goal is to reduce the exponent, which the formula does efficiently.

Q7. The reduction formula for ∫cosⁿx dx is derived using the identity sin²x = 1 - cos²x. What would happen if a student mistakenly used sin²x = 1 + cos²x?

A.The formula for ∫cosⁿx dx would not be obtained. ✅

B.The formula would be the same.

C.The formula would have a minus sign in front of the integral term.

D.The formula would result in ∫sinⁿx dx.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivation of the reduction formula relies on substituting sin²x = 1 - cos²x to get an expression in terms of cosx only. Using 1+cos²x would not allow for the separation of terms to yield the reduction formula. This mistake highlights the importance of knowing and applying correct trigonometric identities. The resulting expression would not simplify to the standard reduction formula.

Q8. A student is asked to derive the reduction formula for ∫cos⁵x dx using integration by parts. Which choice of u and dv would lead to the correct derivation?

A.u = cos⁴x, dv = cosx dx ✅

B.u = cosx, dv = cos⁴x dx

C.u = cos⁵x, dv = dx

D.u = 1, dv = cos⁵x dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Choosing u=cos⁴x and dv=cosx dx allows for the use of integration by parts. The derivative du = -4cos³x sinx dx is readily obtained, and v=sinx. This leads to an expression involving sin²x which is then substituted to relate it back to cos⁵x and cos³x. Other choices lead to either a more complex integral or a dead end, demonstrating the strategy behind choosing u and dv.

Q9. A student is trying to evaluate ∫sin³x dx and writes the answer as 1/3 cos³x - cosx + C. Is this correct?

A.Yes, it is correct and consistent with the reduction formula. ✅

B.No, it is incorrect.

C.Yes, but only if x is in radians.

D.No, the correct answer is -1/3 cos³x + cosx + C.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Using the reduction formula with n=3 gives -1/3 sin²x cosx + 2/3∫sinx dx = -1/3 sin²x cosx - 2/3 cosx. Substituting sin²x = 1 - cos²x and simplifying gives -1/3(1-cos²x)cosx - 2/3 cosx = -1/3 cosx + 1/3 cos³x - 2/3 cosx = 1/3 cos³x - cosx + C. The student's answer is correct. Option D is a common sign error.

Q10. Which reduction formula would be used to evaluate an integral of the form ∫tanⁿx dx?

A.∫tanⁿx dx = (tanⁿ⁻¹x)/(n-1) - ∫tanⁿ⁻²x dx ✅

B.∫tanⁿx dx = -(tanⁿ⁻¹x)/(n-1) + ∫tanⁿ⁻²x dx

C.∫tanⁿx dx = (secⁿ⁻¹x tanx)/(n-1) + ((n-2)/(n-1))∫secⁿ⁻²x dx

D.∫tanⁿx dx = (tanⁿ⁻¹x)/(n) - ∫tanⁿ⁻²x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The standard reduction formula for powers of tangent is ∫tanⁿx dx = (tanⁿ⁻¹x)/(n-1) - ∫tanⁿ⁻²x dx. This is derived using the identity tan²x = sec²x - 1. Option A is correct. Option B has incorrect signs. Option C is the formula for powers of secant. Option D is incorrect because the denominator is n instead of n-1.

Q11. A student applies the reduction formula for ∫tan⁶x dx and obtains (tan⁵x)/5 - (tan³x)/3 + tanx - x + C. Which of the following statements is true?

A.The student made an error by not applying the formula enough times.

B.The student's answer is partially correct but only for x in [0,π/2].

C.The student's answer is completely correct. ✅

D.The student's answer is only valid for definite integrals.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Applying the formula to n=6: ∫tan⁶x dx = tan⁵x/5 - ∫tan⁴x dx. Then ∫tan⁴x dx = tan³x/3 - ∫tan²x dx, and ∫tan²x dx = tanx - x. Substituting back gives tan⁵x/5 - tan³x/3 + tanx - x + C. The student's answer is correct and complete. This shows a good understanding of how to use the reduction formula iteratively.

Q12. The reduction formula for ∫secⁿx dx is ∫secⁿx dx = (secⁿ⁻²x tanx)/(n-1) + ((n-2)/(n-1))∫secⁿ⁻²x dx. Why is this formula significant for evaluating integrals of odd powers of secant?

A.It reduces the power to sec¹x, whose integral is ln|secx+tanx|. ✅

B.It reduces the power to sec³x and then repeats.

C.It directly gives the integral of secx without further steps.

D.It eliminates the need for trigonometric substitutions.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The reduction formula reduces the power of secx by 2. By repeatedly applying it, the exponent can be reduced. If n is odd, the process eventually leads to ∫secx dx, which is a known formula. Option A correctly states this. If n were even, it would lead to ∫sec²x dx. This is the key to integrating odd powers of secant.

Q13. A student uses the reduction formula for ∫cosⁿx dx to evaluate ∫cos⁵x dx. Their final answer is (1/5)cos⁴x sinx + (4/15)cos²x sinx + (8/15)sinx + C. What is the most likely error?

A.They simplified the final expression incorrectly.

B.They used the wrong reduction formula for cosx.

C.They applied the reduction formula but did not integrate the final term. ✅

D.They are evaluating the integral of cos⁵x incorrectly.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The correct process: n=5 gives 1/5 cos⁴x sinx + 4/5∫cos³x dx. Then ∫cos³x dx = 1/3 cos²x sinx + 2/3∫cosx dx = 1/3 cos²x sinx + 2/3 sinx. So the final answer should be 1/5 cos⁴x sinx + 4/15 cos²x sinx + 8/15 sinx + C. The student's answer is correct, so they didn't make an error. The question asks for the most likely error if they got a different answer. If they didn't integrate the final ∫cosx dx, the answer would be incorrect, but the given answer is correct.

Q14. What is the value of ∫₀^{π/2} sin⁸x dx using the Wallis sine formula?

A.7π/32

B.7π/16

C.35π/256 ✅

D.35π/128

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The Wallis sine formula for even n=8 is (π/2) * ((1*3*5*7)/(2*4*6*8)) = (π/2)*(105/384) = 105π/768 = 35π/256. Option C is correct. Other options result from miscalculating the product or the factor of π/2. This requires recall and Easy of the Wallis formula, which is a direct Easy of the reduction formula (9) for definite integrals.

Q15. A student derives a reduction formula for ∫xⁿeˣ dx as ∫xⁿeˣ dx = xⁿeˣ - n∫xⁿ⁻¹eˣ dx. This formula is useful because:

A.It reduces the power of x, eventually leading to a simple integral. ✅

B.It reduces the power of eˣ, making integration easier.

C.It eliminates the need for the product rule.

D.It allows for direct evaluation of the integral using the exponential rule.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The formula reduces the exponent of x from n to n-1. By applying it repeatedly, the exponent of x eventually becomes zero, leading to the integral of eˣ. This makes it a powerful tool for integrating polynomials multiplied by eˣ. The other options are incorrect; the formula works on the polynomial part, not the exponential part.

Q16. Given the reduction formula ∫xⁿ eˣ dx = xⁿ eˣ - n∫xⁿ⁻¹ eˣ dx, which of the following is the result for n=3?

A.x³eˣ - 3x²eˣ + 6xeˣ - 6eˣ + C ✅

B.x³eˣ - 3x²eˣ + 3xeˣ - eˣ + C

C.x³eˣ + 3x²eˣ + 6xeˣ + 6eˣ + C

D.x³eˣ - 3x²eˣ - 6xeˣ + 6eˣ + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying the formula n=3 gives x³eˣ - 3∫x²eˣ dx. For n=2, ∫x²eˣ dx = x²eˣ - 2∫xeˣ dx. For n=1, ∫xeˣ dx = xeˣ - ∫eˣ dx = xeˣ - eˣ. Substituting back: x³eˣ - 3x²eˣ + 6xeˣ - 6eˣ + C. Option A is correct. This Hard tests the sequential Easy of the formula.

Q17. Which of the following is a correct Easy of the reduction formula for ∫sinⁿx dx to solve ∫sin²x dx?

A.-1/2 sinx cosx + 1/2 x + C ✅

B.1/2 sinx cosx + 1/2 x + C

C.-1/2 sinx cosx - 1/2 x + C

D.1/2 sinx cosx - 1/2 x + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Using the formula for n=2: ∫sin²x dx = -1/2 sinx cosx + 1/2∫dx = -1/2 sinx cosx + x/2 + C. Option A is correct. Option B has the wrong sign on the first term. Option C has the wrong sign on the x term. Option D has the wrong sign on both terms. This is a foundational Easy of the formula.

Q18. A student is evaluating ∫sin³x cos²x dx. They decide to use the reduction formula for sin³x. Is this the most efficient approach?

A.No, because the integrand is a product, not just a power of sin. ✅

B.Yes, because the reduction formula for sin will reduce the power to 1.

C.No, because it's easier to use the substitution u=cosx.

D.Yes, but only for definite integrals.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The reduction formulas are designed for integrals of a single trigonometric function raised to a power, like ∫sinⁿx dx. The integrand sin³x cos²x is a product of two different functions. While one could potentially apply the reduction formula after manipulating it, the most efficient method is to use the strategies for integrating products of sines and cosines (e.g., Table 7.3.1), such as using u=cosx. The student's approach, while not impossible, is inefficient. The question highlights that choosing the right method is a key skill.

Q19. The reduction formula for ∫secⁿx dx involves a term with secⁿ⁻²x tanx. This structure is derived from:

A.Integration by parts with u=secⁿ⁻²x and dv=sec²x dx. ✅

B.u-substitution with u=tanx.

C.The derivative of tanx is sec²x.

D.The identity sec²x = 1 + tan²x.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The formula is derived using integration by parts. Choosing u=secⁿ⁻²x and dv=sec²x dx allows the derivative of u to involve secⁿ⁻²x tanx, and v=tanx. This structure is what leads to the final reduction formula. While the other options are related identities, they are not the primary method for deriving this specific reduction formula.

Q20. Suppose you are evaluating ∫sin⁷x dx and have applied the reduction formula to get ∫sin⁵x dx, then ∫sin³x dx, and finally ∫sinx dx. How many times was the reduction formula applied?

A.Three times ✅

B.Four times

C.Seven times

D.Two times

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The formula reduces the power by 2 each time. Starting with n=7: 7 -> 5 -> 3 -> 1. This is 3 Easys of the formula. Option A is correct. A student might incorrectly count 4 times by including the final integration of sinx, or 7 by confusing n with the number of steps. This tests the understanding of the iterative process.

Q21. A student is deriving the reduction formula for ∫sinⁿx dx. After integration by parts, they have the expression: -sinⁿ⁻¹x cosx + (n-1)∫cos²x sinⁿ⁻²x dx. To get the standard reduction formula, what is the next correct step?

A.Replace cos²x with 1 - sin²x and simplify. ✅

B.Replace cos²x with 1 + sin²x.

C.Integrate the second term directly.

D.Take the derivative of both sides.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: To get an integral solely in terms of sinx, we must eliminate cos²x. The identity cos²x = 1 - sin²x is used for this purpose. This allows the integral to be split into ∫sinⁿ⁻²x dx and ∫sinⁿx dx, which can then be rearranged to solve for ∫sinⁿx dx. The other options would not lead to the correct simplification. This step is crucial for the derivation.

Q22. What is the purpose of a reduction formula in the context of evaluating integrals involving powers of a function?

A.To express the integral in terms of another integral with a lower power of that function. ✅

B.To provide a direct antiderivative for all powers.

C.To eliminate the need for integration by parts.

D.To convert trigonometric functions into algebraic ones.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The primary function of a reduction formula is to reduce the complexity of an integral. By lowering the power, it makes the integral simpler to evaluate. It does not directly give the antiderivative; it provides a relationship between integrals of different powers. The other options are incorrect; a reduction formula often still requires other methods (like basic integrals) to finish the evaluation.

Q23. Which of the following demonstrates a correct Easy of the reduction formula for ∫tanⁿx dx?

A.∫tan³x dx = (1/2)tan²x - ln|secx| + C ✅

B.∫tan³x dx = (1/2)sec²x - ln|cosx| + C

C.∫tan³x dx = tan²x - ln|secx| + C

D.∫tan³x dx = (1/2)tan²x + ln|secx| + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying the formula with n=3: ∫tan³x dx = (tan²x)/2 - ∫tanx dx = (tan²x)/2 - ln|secx| + C. Option A is correct. Option B uses sec²x instead of tan²x. Option C has the incorrect coefficient for tan²x. Option D has the wrong sign for the log term. This is a direct Easy of the formula.

Q24. A student incorrectly writes the reduction formula for ∫cosⁿx dx as ∫cosⁿx dx = (1/n)cosⁿx tanx + ((n-1)/n)∫cosⁿ⁻²x dx. What is the best way to identify this error?

A.The term (1/n)cosⁿx tanx simplifies to (1/n)cosⁿ⁻¹x sinx, which is correct, so the error is not obvious. ✅

B.The integral on the right still contains cosⁿx, not cosⁿ⁻²x.

C.The formula should have a minus sign, not a plus sign.

D.The coefficient on the integral term should be (n-1)/n, which is correct.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student's written formula is actually equivalent to the correct one. Since tanx = sinx/cosx, then (1/n)cosⁿx tanx = (1/n)cosⁿ⁻¹x sinx. The coefficient ((n-1)/n) is also correct. Therefore, the student's formula is not an error; it's a correct, albeit less common, form. The best way to identify that this isn't an error is to simplify it using the identity tanx = sinx/cosx. This question tests the ability to recognize equivalent forms.

Q25. What is the primary advantage of using reduction formulas over other methods, such as substitution, when integrating powers of trigonometric functions?

A.They provide a systematic approach when the integrand is a single power. ✅

B.They are always the fastest method.

C.They eliminate the need to remember trigonometric identities.

D.They work for all types of integrals.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Reduction formulas are most useful for integrals like ∫sinⁿx dx where the integrand is a single power of a function. They provide a clear, step-by-step procedure to reduce the power until a basic integral is reached. While other methods like substitution might work for some powers, reduction formulas are systematic and reliable. The other options are not accurate; reduction formulas often require the use of identities and are not universally applicable.

Q26. A student uses a reduction formula to evaluate a definite integral and gets a result involving an integral with a lower power. They then apply the formula again. What is this an example of?

A.An iterative process ✅

B.A cyclical process

C.A one-step solution

D.An incorrect Easy

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The process of repeatedly applying a reduction formula is called an iterative process. Each Easy reduces the complexity, bringing the problem closer to a known solution. It is not cyclical because the power decreases each time; it doesn't return to the original problem. The other options are incorrect. This is a fundamental concept in understanding how to use reduction formulas.

Q27. A student is evaluating ∫x³ eˣ dx using the reduction formula. After applying the formula, they get x³eˣ - 3x²eˣ + 6xeˣ - 6eˣ. What mistake did they make?

A.They forgot to add the constant of integration. ✅

B.They made a sign error in the second term.

C.They applied the formula incorrectly.

D.They didn't apply the formula enough times.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The integration process is correct, and the resulting expression is the antiderivative. However, the student omitted the constant of integration, +C. This is a common error. The question tests the ability to identify incomplete work, not just computational errors. The other options are not errors in this case; the signs and coefficients are correct and the formula was applied the correct number of times.

Q28. Which of the following integrals would be most efficiently evaluated using a reduction formula for powers of a trigonometric function?

A.∫cos⁵x dx ✅

B.∫sin²x cos³x dx

C.∫x cosx dx

D.∫eˣ sinx dx

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The integral ∫cos⁵x dx is a classic case where a reduction formula for cos⁵x is directly applicable. The other integrals are better suited for other techniques: ∫sin²x cos³x dx uses substitution (u=sinx or u=cosx), ∫x cosx dx uses integration by parts, and ∫eˣ sinx dx uses integration by parts (or a specific technique for exponentials and sines). This tests the ability to recognize when a reduction formula is the appropriate tool.

Q29. A student is evaluating ∫cos⁷x dx. Their answer is (1/7)cos⁶x sinx + (6/35)cos⁴x sinx + (24/105)cos²x sinx + (48/105)sinx. What is a potential issue with this final form?

A.It is correct but not simplified completely. ✅

B.It is incorrect because of a coefficient error.

C.It is incorrect because it only applies for definite integrals.

D.It is correct but should have a negative sign in the first term.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The answer is derived from applying the reduction formula correctly. However, the coefficients can be simplified: 24/105 simplifies to 8/35, and 48/105 simplifies to 16/35. The answer is valid and correct, just not in its simplest form. This highlights that while a process can be correct, the final answer might require simplification. The student's work is not wrong, just incomplete in its presentation.

Q30. The reduction formula for ∫secⁿx dx is often used to evaluate ∫sec³x dx. The result is (1/2)secx tanx + (1/2)ln|secx+tanx| + C. This result is important because:

A.It is a standard formula frequently needed in integration problems. ✅

B.It proves that secx is not integrable.

C.It shows that the integral of secx is the same as tanx.

D.It cannot be used for any other power of secx.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The integral of sec³x is a classic and useful result that appears frequently in Easys, particularly in arc length and surface area problems. The reduction formula provides a systematic way to derive it. Option A is correct. The other options are false; secx is integrable, its integral is not tanx, and the reduction formula can be used for other powers.

Q31. A student is asked to evaluate ∫₀^{π/2} sin⁶x dx. They use the Wallis formula and get 5π/32. Which of the following is the correct evaluation of this integral?

A.5π/32 ✅

B.5π/16

C.15π/48

D.5π/64

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Using the Wallis sine formula for even n=6: (π/2) * ((1*3*5)/(2*4*6)) = (π/2)*(15/48) = 15π/96 = 5π/32. Option A is correct. A common error is forgetting the factor of π/2. This tests the recall and Easy of the Wallis formula, which is a direct Easy of the reduction formula.

Q32. What is the role of the constant of integration when using a reduction formula to find an indefinite integral?

A.It must be added at the very end after all integrations are complete. ✅

B.It is added at each step of the reduction process.

C.It is not needed when using reduction formulas.

D.It is only added for definite integrals.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When using reduction formulas, the focus is on reducing the integral to a simpler form. The constant of integration is not added during the intermediate steps. Once the final integral is evaluated, the constant of integration, +C, is added to the final antiderivative to represent the most general solution. Adding it at each step is unnecessary and can complicate the process.

Q33. A student is evaluating ∫tan⁴x dx. Which of the following represents the first Easy of the reduction formula?

A.(tan³x)/3 - ∫tan²x dx ✅

B.(tan³x)/3 + ∫tan²x dx

C.(tan⁴x)/4 - ∫tan²x dx

D.(tan²x)/2 - ∫tan²x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The reduction formula for tanⁿ is ∫tanⁿx dx = (tanⁿ⁻¹x)/(n-1) - ∫tanⁿ⁻²x dx. For n=4, this gives (tan³x)/3 - ∫tan²x dx. Option A is correct. Option B has the wrong sign. Option C uses the wrong coefficient. Option D uses the formula for n=3 instead of n=4. This tests the ability to correctly apply the formula for a given n.

Q34. A student is evaluating ∫sin⁶x dx. They decide to rewrite it as ∫(sin²x)³ dx and integrate using the power rule. What is the error in this approach?

A.The power rule for integration cannot be applied to a function raised to a power.

B.The power rule for integration applies only to xⁿ, not to sinⁿx. ✅

C.The power rule for integration applies only to polynomials.

D.There is no error; this is a valid technique.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The power rule for integration, ∫uⁿ du = uⁿ⁺¹/(n+1) + C, applies when the variable of integration is u. In ∫sin⁶x dx, the variable of integration is x, but the base of the power is sinx. To use the power rule, the integrand would need to be sin⁶x * cosx, so that du = cosx dx. The student is incorrectly treating sinx as the variable of integration. Option B correctly identifies this misconception.

Q35. A reduction formula for ∫xⁿ sinx dx is given. How would you use it to evaluate ∫x³ sinx dx?

A.Apply the formula repeatedly, reducing n by 1 each time until n=0. ✅

B.Apply the formula once and then use integration by parts.

C.Apply the formula repeatedly, reducing n by 2 each time.

D.Apply the formula once and substitute x for sinx.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A reduction formula for ∫xⁿ sinx dx would express the integral in terms of ∫xⁿ⁻¹ sinx dx and some other terms. By applying it repeatedly, n is reduced by 1 each time. When n=0, the integral becomes ∫sinx dx, which is a standard integral. Option A describes this iterative process correctly. The power of x, not sinx, is being reduced.

Q36. When deriving the reduction formula for ∫cosⁿx dx, what is the purpose of the identity sin²x = 1 - cos²x?

A.To eliminate sin²x from the expression after integration by parts. ✅

B.To substitute sin²x for cos²x.

C.To simplify the coefficient of the integral.

D.To change the variable of integration.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: After applying integration by parts with u=cosⁿ⁻¹x and dv=cosx dx, we get an integral with sin²x cosⁿ⁻²x. To express this entirely in terms of cosx, we replace sin²x with 1 - cos²x. This allows us to combine terms with cosⁿx and cosⁿ⁻²x, which is necessary to solve for the original integral. This step is crucial for deriving the formula.

🔗 Related Topics (MCQs)

📝 Integration methods overview calculus MCQs 📝 Basic integration formulas review MCQs 📝 Integration by parts formula examples MCQs 📝 How to choose u and dv in integration by parts MCQs 📝 Repeated integration by parts examples MCQs

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