Integral tables with reduction formulas: 40 Practice MCQs with Answers and Explanation

Q1. Using reduction formula for $\int \sin^4 x \, dx$ , what is the coefficient of $x$ in the final antiderivative?

A.

\frac{3}{8}

✅

B.

\frac{1}{4}

C.

\frac{3}{4}

D.

\frac{1}{8}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The reduction formula $\int \sin^n x \, dx = -\frac{1}{n}\sin^{n-1}x \cos x + \frac{n-1}{n} \int \sin^{n-2}x \, dx$ applied recursively for $n=4$ yields $\frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$ . The coefficient of the linear term $x$ comes from reducing the integral down to $\int dx$ , which is $\frac{3}{8}$ .

Q2. A student evaluates $\int \cos^5 x \, dx$ using reduction formula and gets $\frac{1}{5}\cos^4 x\sin x + \frac{4}{5}\left(\frac{1}{3}\cos^2 x\sin x + \frac{2}{3}\sin x\right) + C$ . What simplification error likely occurred?

A.The coefficient of

\sin x

should be

\frac{8}{15}

, not

\frac{8}{15}

but the student forgot to distribute.

B.The

\frac{4}{5}

multiplier was applied incorrectly to all terms. ✅

C.The student used the formula for

\sin^n x

instead of

\cos^n x

.

D.The reduction formula should reduce exponent by 1, not by 2.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The correct antiderivative is $\frac{1}{5}\cos^4 x\sin x + \frac{4}{5}\left(\frac{1}{3}\cos^2 x\sin x + \frac{2}{3}\int \cos x dx\right)$ , which simplifies to $\frac{1}{5}\cos^4 x\sin x + \frac{4}{15}\cos^2 x\sin x + \frac{8}{15}\sin x + C$ . The student likely did not distribute $\frac{4}{5}$ through the entire bracket, leading to incorrect coefficients.

Q3. An engineer needs to compute $\int_0^{\pi/2} \sin^6 x \, dx$ . Using reduction formula, what is the exact value?

A.

\frac{5\pi}{32}

✅

B.

\frac{3\pi}{16}

C.

\frac{5\pi}{16}

D.

\frac{3\pi}{32}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying $\int_0^{\pi/2} \sin^n x dx = \frac{n-1}{n} \int_0^{\pi/2} \sin^{n-2} x dx$ , for $n=6$ : $\frac{5}{6} \times \frac{3}{4} \times \frac{1}{2} \times \frac{\pi}{2} = \frac{5\pi}{32}$ . This is a direct Easy of reduction formulas in evaluating definite integrals over symmetric limits.

Q4. Which of the following integrals requires reduction formula Easy to evaluate efficiently?

A.

\int x e^{x^2} dx

B.

\int \sin^3 x \cos^2 x dx

C.

\int \sec^3 x dx

✅

D.

\int \frac{1}{x^2+1} dx

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: $\int \sec^3 x dx$ is classically evaluated using the reduction formula $\int \sec^n x dx = \frac{\sec^{n-2}x \tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2}x dx$ . While $\sin^3 x \cos^2 x$ can be handled by u-substitution, $\sec^3 x$ typically requires reduction or integration by parts with a clever choice of $u$ and $dv$ .

Q5. If $I_n = \int \sin^n x \, dx$ , then the reduction formula $I_n = -\frac{1}{n}\sin^{n-1}x\cos x + \frac{n-1}{n} I_{n-2}$ is derived using:

A.Integration by parts with

u = \sin^{n-1}x

and

dv = \sin x \, dx

✅

B.Substitution

u = \cos x

C.Trigonometric identity

\sin^2 x = 1 - \cos^2 x

D.Direct integration of

\sin^n x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The reduction formula is derived by writing $\sin^n x = \sin^{n-1}x \sin x$ and applying integration by parts with $u = \sin^{n-1}x$ , $dv = \sin x \, dx$ . This gives $I_n = -\sin^{n-1}x \cos x + (n-1)\int \sin^{n-2}x \cos^2 x dx$ . Then using $\cos^2 x = 1 - \sin^2 x$ and simplifying yields the formula.

Q6. Evaluate $\int \tan^5 x \, dx$ using reduction formula $\int \tan^n x dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2}x dx$ . What is the result?

A.

\frac{1}{4}\tan^4 x - \frac{1}{2}\tan^2 x + \ln|\sec x| + C

✅

B.

\frac{1}{4}\tan^4 x - \frac{1}{2}\tan^2 x - \ln|\sec x| + C

C.

\frac{1}{4}\tan^4 x + \frac{1}{2}\tan^2 x - \ln|\sec x| + C

D.

\frac{1}{4}\tan^4 x + \frac{1}{2}\tan^2 x + \ln|\sec x| + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Apply reduction formula twice: $I_5 = \frac{1}{4}\tan^4 x - I_3$ , and $I_3 = \frac{1}{2}\tan^2 x - I_1$ , with $I_1 = \int \tan x dx = \ln|\sec x|$ . Combining gives $\frac{1}{4}\tan^4 x - \frac{1}{2}\tan^2 x + \ln|\sec x| + C$ . The alternating signs are crucial.

Q7. A student attempting to evaluate $\int \sin^5 x \, dx$ using reduction formula stops after one Easy and writes $-\frac{1}{5}\sin^4 x\cos x + \frac{4}{5} \int \sin^3 x dx$ . What is the next step?

A.Apply reduction formula again to

\int \sin^3 x dx

B.Substitute

u = \cos x

C.Use identity

\sin^3 x = (1-\cos^2 x)\sin x

D.Both A and C are valid next steps ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: After first reduction, $\int \sin^3 x dx$ can be evaluated either by applying the reduction formula again (which gives $-\frac{1}{3}\sin^2 x\cos x + \frac{2}{3}\int \sin x dx$ ) or by substitution $u = \cos x$ . Both methods are correct and will yield equivalent results. The student's work is correct so far.

Q8. Which reduction formula is correctly stated for $\int \sec^n x \, dx$ ?

A.

\int \sec^n x dx = \frac{\sec^{n-1}x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}x dx

B.

\int \sec^n x dx = \frac{\sec^{n-2}x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}x dx

✅

C.

\int \sec^n x dx = \frac{\sec^{n-2}x \tan x}{n-1} - \frac{n-2}{n-1} \int \sec^{n-2}x dx

D.

\int \sec^n x dx = \frac{\sec^{n-1}x \tan x}{n-1} - \frac{n-2}{n-1} \int \sec^{n-2}x dx

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The standard reduction formula for $\sec^n x$ is $\int \sec^n x dx = \frac{\sec^{n-2}x \tan x}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}x dx$ . This is derived by writing $\sec^n x = \sec^{n-2}x \sec^2 x$ and applying integration by parts with $u = \sec^{n-2}x$ , $dv = \sec^2 x dx$ . The signs and powers must match exactly.

Q9. What is $\int \cos^6 x \, dx$ using reduction formula?

A.

\frac{1}{6}\cos^5 x\sin x + \frac{5}{24}\cos^3 x\sin x + \frac{5}{16}x + \frac{5}{32}\sin 2x + C

B.

\frac{1}{6}\cos^5 x\sin x + \frac{5}{6}\left(\frac{1}{4}\cos^3 x\sin x + \frac{3}{8}x + \frac{3}{16}\sin 2x\right) + C

C.Both are equivalent ✅

D.Neither is correct

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Applying reduction formula recursively: $I_6 = \frac{1}{6}\cos^5 x\sin x + \frac{5}{6}I_4$ , $I_4 = \frac{1}{4}\cos^3 x\sin x + \frac{3}{4}I_2$ , $I_2 = \frac{1}{2}x + \frac{1}{4}\sin 2x$ . Option B shows the expanded form, while option A shows the simplified form. Both are mathematically equivalent.

Q10. Using reduction formula, express $\int_0^{\pi} \sin^8 x \, dx$ in terms of $\int_0^{\pi} \sin^6 x \, dx$ .

A.

\frac{7}{8} \int_0^{\pi} \sin^6 x dx

✅

B.

\frac{7}{8} \int_0^{\pi} \sin^6 x dx

C.

\frac{7}{8} \int_0^{\pi} \sin^6 x dx

D.

\frac{7}{8} \int_0^{\pi} \sin^6 x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For definite integrals from 0 to $\pi$ , $\int_0^{\pi} \sin^n x dx = \frac{n-1}{n} \int_0^{\pi} \sin^{n-2} x dx$ . This follows from the reduction formula because the boundary term $-\frac{1}{n}\sin^{n-1}x\cos x$ evaluates to 0 at both 0 and $\pi$ . Thus $I_8 = \frac{7}{8}I_6$ .

Q11. A common error when applying $\int \tan^n x dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2}x dx$ is:

A.Forgetting the minus sign between terms

B.Using

\sec^2 x

instead of

\tan^2 x

in the reduction

C.Misapplying the coefficient

\frac{1}{n-1}

D.All of the above ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Each listed error is common: the minus sign is often forgotten, students sometimes replace $\tan^{n-2}x$ with $\sec^{n-2}x$ , and the coefficient $\frac{1}{n-1}$ is sometimes miswritten. The correct formula includes a minus sign and the coefficient $\frac{1}{n-1}$ on the $\tan^{n-1}x$ term.

Q12. What is the reduction formula for $\int \cot^n x \, dx$ that is analogous to tangent?

A.

\int \cot^n x dx = -\frac{\cot^{n-1}x}{n-1} - \int \cot^{n-2}x dx

✅

B.

\int \cot^n x dx = \frac{\cot^{n-1}x}{n-1} - \int \cot^{n-2}x dx

C.

\int \cot^n x dx = -\frac{\cot^{n-1}x}{n-1} + \int \cot^{n-2}x dx

D.

\int \cot^n x dx = \frac{\cot^{n-1}x}{n-1} + \int \cot^{n-2}x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For $\cot^n x$ , the reduction formula is $\int \cot^n x dx = -\frac{\cot^{n-1}x}{n-1} - \int \cot^{n-2}x dx$ . The negative sign on the first term comes from the derivative of $\cot x$ being $-\csc^2 x$ , and the minus sign between terms is analogous to the tangent formula.

Q13. If $I_n = \int \sin^n x \, dx$ and $n$ is even, repeated Easy of reduction formula will eventually express $I_n$ in terms of:

A.

\int \sin^0 x \, dx = x

✅

B.

\int \sin^1 x \, dx = -\cos x

C.

\int \sin^2 x \, dx = \frac{1}{2}x - \frac{1}{4}\sin 2x

D.

\int \sin^2 x \, dx = \frac{1}{2}x + \frac{1}{4}\sin 2x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: When $n$ is even, repeated reduction by 2 eventually reaches $n=0$ , giving $\int \sin^0 x dx = \int 1 dx = x$ . When $n$ is odd, it reaches $n=1$ , giving $-\cos x$ . This is a key insight for evaluating integrals of sine and cosine powers.

Q14. Evaluate $\int_0^{\pi/2} \cos^7 x \, dx$ using reduction formula.

A.

\frac{16}{35}

✅

B.

\frac{8}{35}

C.

\frac{32}{35}

D.

\frac{24}{35}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For odd $n$ , $\int_0^{\pi/2} \cos^n x dx = \frac{2\cdot4\cdot6\cdots(n-1)}{3\cdot5\cdot7\cdots n}$ . For $n=7$ , $\frac{2\cdot4\cdot6}{3\cdot5\cdot7} = \frac{48}{105} = \frac{16}{35}$ . This is the Wallis cosine formula derived from reduction formula.

Q15. A reduction formula $I_n = \frac{n-1}{n} I_{n-2}$ is used to evaluate $I_{10} = \int_0^{\pi/2} \cos^{10} x dx$ . Starting from $I_0 = \pi/2$ , what is $I_{10}$ ?

A.

\frac{63\pi}{256}

✅

B.

\frac{63\pi}{512}

C.

\frac{63\pi}{128}

D.

\frac{31\pi}{64}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Apply recursively: $I_{10} = \frac{9}{10}\cdot\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{9\cdot7\cdot5\cdot3\cdot1}{10\cdot8\cdot6\cdot4\cdot2} \cdot \frac{\pi}{2} = \frac{945}{3840} \cdot \frac{\pi}{2} = \frac{63}{256}\pi$ .

Q16. Which of the following is NOT a correct Easy of a reduction formula?

A.

\int \sec^3 x dx = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln|\sec x + \tan x| + C

B.

\int \tan^4 x dx = \frac{1}{3}\tan^3 x - \tan x + x + C

C.

\int \sin^2 x dx = \frac{1}{2}x - \frac{1}{4}\sin 2x + C

D.

\int \cos^5 x dx = \frac{1}{5}\cos^4 x\sin x + \frac{4}{5}\left(\frac{1}{3}\cos^2 x\sin x + \frac{2}{3}\cos x\right) + C

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The last term in option D should be $\frac{2}{3}\sin x$ , not $\frac{2}{3}\cos x$ . The reduction formula for $\cos^n x$ eventually reduces to $\int \cos x dx = \sin x$ . This is a common sign/function error when reducing.

Q17. How does the reduction formula $I_n = -\frac{1}{n}\sin^{n-1}x\cos x + \frac{n-1}{n}I_{n-2}$ relate to the derivative of $\sin^{n-1}x\cos x$ ?

A.The derivative of

\sin^{n-1}x\cos x

equals

n\sin^{n-2}x - n\sin^n x

✅

B.The derivative of

\sin^{n-1}x\cos x

equals

\sin^{n-2}x - \sin^n x

C.The derivative of

\sin^{n-1}x\cos x

equals

-\sin^{n-2}x + \sin^n x

D.The derivative of

\sin^{n-1}x\cos x

equals

-n\sin^{n-2}x + n\sin^n x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Derivative: $\frac{d}{dx}(\sin^{n-1}x\cos x) = (n-1)\sin^{n-2}x\cos^2 x - \sin^n x = (n-1)\sin^{n-2}x(1-\sin^2 x) - \sin^n x = (n-1)\sin^{n-2}x - n\sin^n x$ . Rearranging gives $\sin^n x = -\frac{1}{n}\frac{d}{dx}(\sin^{n-1}x\cos x) + \frac{n-1}{n}\sin^{n-2}x$ , which integrates to the formula.

Q18. A problem requires evaluating $\int \sec^5 x \, dx$ . Which sequence of reduction steps is correct?

A.

I_5 = \frac{1}{4}\sec^3 x\tan x + \frac{3}{4}I_3

,

I_3 = \frac{1}{2}\sec x\tan x + \frac{1}{2}I_1

B.

I_5 = \frac{1}{4}\sec^3 x\tan x + \frac{3}{4}I_3

,

I_3 = \frac{1}{2}\sec x\tan x + \frac{1}{2}I_1

✅

C.

I_5 = \frac{1}{4}\sec^3 x\tan x + \frac{3}{4}I_3

,

I_3 = \frac{1}{2}\sec x\tan x + \frac{1}{2}I_1

D.

I_5 = \frac{1}{4}\sec^3 x\tan x + \frac{3}{4}I_3

,

I_3 = \frac{1}{2}\sec x\tan x + \frac{1}{2}I_1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The correct reduction: $I_5 = \frac{\sec^3 x\tan x}{4} + \frac{3}{4}I_3$ , and $I_3 = \frac{\sec x\tan x}{2} + \frac{1}{2}I_1$ , where $I_1 = \int \sec x dx = \ln|\sec x + \tan x|$ . The coefficients and powers must be carefully tracked.

Q19. What is the primary advantage of using a reduction formula over direct integration?

A.It transforms a complex integral into a simpler form that can be evaluated recursively. ✅

B.It always produces a shorter answer.

C.It eliminates the need for trigonometric identities.

D.It avoids integration by parts entirely.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Reduction formulas systematically reduce the power of the function, making it possible to evaluate higher powers by repeated Easy. They often require integration by parts and identities, but provide a structured approach that is particularly useful for symbolic computation and pattern recognition.

Q20. Evaluate $\int_0^{\pi/4} \tan^6 x \, dx$ using reduction formula and the identity $\tan^2 x = \sec^2 x - 1$ .

A.

\frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4}

✅

B.

\frac{1}{5} - \frac{1}{3} + 1 + \frac{\pi}{4}

C.

\frac{1}{5} + \frac{1}{3} + 1 - \frac{\pi}{4}

D.

\frac{1}{5} + \frac{1}{3} - 1 + \frac{\pi}{4}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: $I_6 = \frac{1}{5}\tan^5 x - I_4$ , $I_4 = \frac{1}{3}\tan^3 x - I_2$ , $I_2 = \tan x - x$ . At $x = \pi/4$ , $\tan x = 1$ , so $I_6 = \frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4}$ . The alternating signs and evaluating at 0 (where tan=0) is key.

Q21. When using reduction formula $\int \cos^n x dx = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}\int \cos^{n-2}x dx$ , the term $\frac{1}{n}\cos^{n-1}x\sin x$ is often zero for definite integrals from 0 to $\pi/2$ . Why?

A.Because

\sin 0 = 0

and

\cos(\pi/2) = 0

✅

B.Because

\sin(\pi/2) = 0

and

\cos 0 = 0

C.Because the term simplifies to a constant

D.Because the term integrates to zero

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: At $x=0$ , $\sin 0 = 0$ , so the term is 0. At $x=\pi/2$ , $\cos(\pi/2) = 0$ , so the term is 0. Thus the boundary term vanishes for this special interval, simplifying the evaluation to $\int_0^{\pi/2} \cos^n x dx = \frac{n-1}{n} \int_0^{\pi/2} \cos^{n-2} x dx$ .

Q22. A student evaluates $\int \tan^4 x \, dx$ and gets $\frac{1}{3}\tan^3 x + \tan x - x + C$ . Is this correct?

A.Yes, it is correct.

B.No, the sign before

\tan x

should be negative. ✅

C.No, the coefficient of

\tan^3 x

should be

\frac{1}{3}

.

D.No, the term

-x

should be

+x

.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The correct antiderivative is $\frac{1}{3}\tan^3 x - \tan x + x + C$ . The student likely used the reduction formula incorrectly: $I_4 = \frac{1}{3}\tan^3 x - I_2$ , and $I_2 = \tan x - x$ , so $I_4 = \frac{1}{3}\tan^3 x - \tan x + x + C$ .

Q23. If $I_n = \int_0^{\pi/2} \sin^n x \, dx$ , what is $I_{2n}$ in terms of $I_0$ ?

A.

\frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}

✅

B.

\frac{(2n-1)!!}{(2n)!!} \cdot \pi

C.

\frac{(2n)!!}{(2n-1)!!} \cdot \frac{\pi}{2}

D.

\frac{(2n)!!}{(2n-1)!!} \cdot \pi

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This is the Wallis sine formula: $\int_0^{\pi/2} \sin^{2n} x dx = \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} \cdot \frac{\pi}{2}$ . This is derived by repeated Easy of the reduction formula $I_n = \frac{n-1}{n}I_{n-2}$ .

Q24. Which of the following integrals would be best approached using a reduction formula?

A.

\int \sin^4 x \cos^2 x dx

B.

\int \tan^3 x \sec^3 x dx

C.

\int \sec^4 x \tan^2 x dx

D.

\int \sin^{10} x dx

✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: While options A, B, and C can be handled by u-substitution or other methods, $\int \sin^{10} x dx$ is most efficiently evaluated using the sine reduction formula. The reduction formula systematically reduces the power of sine, making it ideal for high even powers.

Q25. How would you evaluate $\int \sec^7 x \, dx$ using reduction formula?

A.Apply reduction formula twice, ending with

\int \sec x dx

B.Apply reduction formula twice, ending with

\int \sec^3 x dx

C.Apply reduction formula three times, ending with

\int \sec x dx

✅

D.Apply reduction formula once, then substitute

u = \tan x

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: $I_7 = \frac{\sec^5 x\tan x}{6} + \frac{5}{6}I_5$ , $I_5 = \frac{\sec^3 x\tan x}{4} + \frac{3}{4}I_3$ , $I_3 = \frac{\sec x\tan x}{2} + \frac{1}{2}I_1$ , and $I_1 = \int \sec x dx$ . The reduction formula for odd powers eventually reduces to $\sec x$ .

Q26. What is the result of applying reduction formula for $\int \sin^3 x \, dx$ and then simplifying using $\sin^2 x = 1 - \cos^2 x$ ?

A.

\frac{1}{3}\cos^3 x - \cos x + C

✅

B.

-\frac{1}{3}\cos^3 x + \cos x + C

C.

\frac{1}{3}\sin^3 x - \sin x + C

D.

-\frac{1}{3}\sin^3 x + \sin x + C

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: $I_3 = -\frac{1}{3}\sin^2 x\cos x + \frac{2}{3}\int \sin x dx = -\frac{1}{3}\sin^2 x\cos x - \frac{2}{3}\cos x + C$ . Using $\sin^2 x = 1 - \cos^2 x$ , this becomes $-\frac{1}{3}(1-\cos^2 x)\cos x - \frac{2}{3}\cos x = -\frac{1}{3}\cos x + \frac{1}{3}\cos^3 x - \frac{2}{3}\cos x = \frac{1}{3}\cos^3 x - \cos x + C$ .

Q27. A student evaluating $\int_0^{\pi/2} \sin^{10} x dx$ using reduction formula obtains $\frac{63\pi}{512}$ . Is this correct?

A.Yes, because

I_{10} = \frac{9}{10}\cdot\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2}

✅

B.No, the final factor should be

\pi

, not

\pi/2

C.No, the product should be

\frac{10}{9}\cdot\frac{8}{7}\cdot\frac{6}{5}\cdot\frac{4}{3}\cdot\frac{2}{1}

D.No, the product should be

\frac{9}{10}\cdot\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The reduction formula for even powers from 0 to $\pi/2$ is $I_{2n} = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$ . For $n=5$ , $\frac{9\cdot7\cdot5\cdot3\cdot1}{10\cdot8\cdot6\cdot4\cdot2} \cdot \frac{\pi}{2} = \frac{945}{3840} \cdot \frac{\pi}{2} = \frac{63}{256} \cdot \frac{\pi}{2} = \frac{63\pi}{512}$ .

Q28. What is the reduction formula for $\int \csc^n x \, dx$ analogous to $\sec^n x$ ?

A.

\int \csc^n x dx = -\frac{\csc^{n-2}x\cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2}x dx

✅

B.

\int \csc^n x dx = \frac{\csc^{n-2}x\cot x}{n-1} - \frac{n-2}{n-1}\int \csc^{n-2}x dx

C.

\int \csc^n x dx = \frac{\csc^{n-1}x\cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2}x dx

D.

\int \csc^n x dx = -\frac{\csc^{n-1}x\cot x}{n-1} - \frac{n-2}{n-1}\int \csc^{n-2}x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The standard reduction formula for $\csc^n x$ is $\int \csc^n x dx = -\frac{\csc^{n-2}x\cot x}{n-1} + \frac{n-2}{n-1}\int \csc^{n-2}x dx$ . The negative sign comes from the derivative of $\csc x$ being $-\csc x\cot x$ .

Q29. Evaluate $\int_0^{\pi/2} \sin^5 x \, dx$ using reduction formula.

A.

\frac{8}{15}

✅

B.

\frac{16}{15}

C.

\frac{4}{15}

D.

\frac{2}{15}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: For odd $n$ , $\int_0^{\pi/2} \sin^{2n+1} x dx = \frac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdots(2n+1)}$ . For $n=2$ (since $2n+1 = 5$ ), $\frac{2\cdot4}{3\cdot5} = \frac{8}{15}$ . This is the Wallis sine formula for odd powers.

Q30. A reduction formula expresses $I_n$ in terms of $I_{n-2}$ . What is the primary risk when applying this formula repeatedly?

A.Accumulation of arithmetic errors in coefficients

B.Forgetting the boundary term in definite integrals

C.Misapplying the formula for even vs odd n

D.All of the above ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: When applying reduction formulas repeatedly, each of these is a potential error: coefficients can become complex, boundary terms must be evaluated at each step, and the base case depends on parity. Careful bookkeeping is essential to avoid mistakes.

Q31. What is the value of $\int_0^{\pi} \cos^6 x \, dx$ using reduction formula?

A.

\frac{5\pi}{16}

✅

B.

\frac{5\pi}{8}

C.

\frac{3\pi}{16}

D.

\frac{3\pi}{8}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For $\int_0^{\pi} \cos^n x dx$ , the reduction formula gives the same result as $\int_0^{\pi/2} \cos^n x dx$ multiplied by 2? Actually, $\int_0^{\pi} \cos^{2n} x dx = 2\int_0^{\pi/2} \cos^{2n} x dx = \frac{(2n-1)!!}{(2n)!!} \cdot \pi$ . For $n=3$ , $\frac{5\cdot3\cdot1}{6\cdot4\cdot2} \cdot \pi = \frac{15}{48}\pi = \frac{5\pi}{16}$ .

Q32. If reduction formula for $\sin^n x$ is applied to $n=0$ , what is $I_0$ ?

A.

x + C

✅

B.

-\cos x + C

C.

\sin x + C

D.

0

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: When $n=0$ , $\int \sin^0 x dx = \int 1 dx = x + C$ . This serves as the base case for even powers. The reduction formula reduces even powers down to $I_0$ , which is $x$ .

Q33. Which of the following expressions is a correct simplification of $\int \sin^7 x \cos^2 x dx$ using reduction formulas?

A.

\frac{1}{7}\cos^7 x - \frac{2}{5}\cos^5 x + \frac{1}{3}\cos^3 x - \cos x + C

✅

B.

\frac{1}{7}\cos^7 x - \frac{2}{5}\cos^5 x + \frac{1}{3}\cos^3 x - \cos x + C

C.

\frac{1}{7}\cos^7 x - \frac{2}{5}\cos^5 x + \frac{1}{3}\cos^3 x - \cos x + C

D.

\frac{1}{7}\cos^7 x - \frac{2}{5}\cos^5 x + \frac{1}{3}\cos^3 x - \cos x + C

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using $u = \cos x$ or $\sin^7 x = \sin x(1-\cos^2 x)^3$ , the integral becomes $\int (1-u^2)^3 u^2 du$ . Expanding: $\int (u^2 - 3u^4 + 3u^6 - u^8) du = \frac{1}{3}u^3 - \frac{3}{5}u^5 + \frac{3}{7}u^7 - \frac{1}{9}u^9 + C$ . Substituting back gives the correct expression.

Q34. A table of integrals gives $\int \tan^n x dx = \frac{\tan^{n-1}x}{n-1} - \int \tan^{n-2}x dx$ . Which is the corresponding reduction for $\int \cot^n x dx$ ?

A.

\int \cot^n x dx = -\frac{\cot^{n-1}x}{n-1} - \int \cot^{n-2}x dx

✅

B.

\int \cot^n x dx = -\frac{\cot^{n-1}x}{n-1} + \int \cot^{n-2}x dx

C.

\int \cot^n x dx = \frac{\cot^{n-1}x}{n-1} - \int \cot^{n-2}x dx

D.

\int \cot^n x dx = \frac{\cot^{n-1}x}{n-1} + \int \cot^{n-2}x dx

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The correct reduction formula for $\cot^n x$ is $\int \cot^n x dx = -\frac{\cot^{n-1}x}{n-1} - \int \cot^{n-2}x dx$ . The negative sign on the first term comes from the derivative of $\cot x$ being $-\csc^2 x$ .

Q35. If $\int \sin^8 x \, dx = \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + \frac{1}{128}\sin 8x + C$ , what is $\int_0^{\pi} \sin^8 x \, dx$ ?

A.

\frac{3\pi}{8}

✅

B.

\frac{3\pi}{4}

C.

\frac{3\pi}{16}

D.

\frac{3\pi}{2}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The antiderivative contains terms $\sin 2x, \sin 4x, \sin 8x$ , all of which evaluate to 0 at both 0 and $\pi$ . Only the linear term $\frac{3}{8}x$ contributes: $\frac{3}{8}(\pi - 0) = \frac{3\pi}{8}$ .

Q36. What is the error in applying reduction formula $\int \sec^n x dx = \frac{\sec^{n-2}x\tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2}x dx$ for $n=2$ ?

A.It gives

\int \sec^2 x dx = \tan x + 0 \cdot \int 1 dx = \tan x + C

, which is correct.

B.It gives

\int \sec^2 x dx = \frac{\tan x}{1} + 0 \cdot \int 1 dx = \tan x + C

, which is correct.

C.It gives

\int \sec^2 x dx = \tan x + 0

, but

\int \sec^2 x dx

is actually

\tan x

, so no error.

D.All of the above confirm the formula works for

n=2

. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The reduction formula is valid for $n \ge 2$ . For $n=2$ , $\frac{\sec^0 x \tan x}{1} + \frac{0}{1}\int 1 dx = \tan x + C$ , which is correct. The formula breaks down for $n=1$ because the denominator becomes 0, but $n=2$ is fine.

Q37. In the derivation of $\int \cos^n x dx = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}\int \cos^{n-2}x dx$ , what is the key trigonometric identity used?

A.

\sin^2 x = 1 - \cos^2 x

✅

B.

\cos^2 x = 1 - \sin^2 x

C.

\sin 2x = 2\sin x\cos x

D.

\cos^2 x - \sin^2 x = \cos 2x

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: After applying integration by parts, one obtains $\int \cos^n x dx = \cos^{n-1}x\sin x + (n-1)\int \cos^{n-2}x\sin^2 x dx$ . Replacing $\sin^2 x$ with $1 - \cos^2 x$ and simplifying yields the reduction formula.

Q38. Which method is most efficient for evaluating $\int_0^{\pi/2} \sin^{100} x \, dx$ ?

A.Using reduction formula repeatedly ✅

B.Using integration by parts

C.Using substitution

u = \cos x

D.Using a CAS

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Reduction formula $I_n = \frac{n-1}{n}I_{n-2}$ gives a simple product formula. For $n=100$ , it is efficient to compute $\frac{99!!}{100!!} \cdot \frac{\pi}{2}$ . Integration by parts would be extremely tedious, and while a CAS is fast, the reduction formula provides insight and a closed form.

Q39. The reduction formula $I_n = -\frac{1}{n}\sin^{n-1}x\cos x + \frac{n-1}{n}I_{n-2}$ for $\sin^n x$ is valid for:

A.

n \ge 2

✅

B.

n \ge 0

C.

n \ge 1

D.

n \ge 3

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The formula is valid for $n \ge 2$ . For $n=0$ or $n=1$ , the base cases are handled separately. The formula reduces $n$ by 2, so it requires $n-2 \ge 0$ for the base case to be reached, hence $n \ge 2$ .

Q40. What is $\int \sec^4 x \, dx$ using reduction formula?

A.

\frac{1}{3}\sec^2 x\tan x + \frac{2}{3}\tan x + C

B.

\frac{1}{3}\sec^2 x\tan x + \frac{2}{3}\tan x + C

✅

C.

\frac{1}{3}\sec^2 x\tan x - \frac{2}{3}\tan x + C

D.

\frac{1}{3}\sec^2 x\tan x - \frac{2}{3}\tan x + C

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: $I_4 = \frac{\sec^2 x\tan x}{3} + \frac{2}{3}I_2$ , and $I_2 = \tan x$ . Thus $I_4 = \frac{1}{3}\sec^2 x\tan x + \frac{2}{3}\tan x + C$ . This can also be verified by differentiating the result.

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