Techniques of Integration MCQs in Calculus

By: Prof. Dr. Fazal Rehman Shamil | Last updated: February 12, 2025

Question 1:

\[
\text{Which of the following is the integral of } \int \sin(x) \, dx?
\]
\[
\text{(a) } \cos(x), \quad \text{(b) } -\cos(x), \quad \text{(c) } \sin(x), \quad \text{(d) } -\sin(x)
\]
Answer: B

Step by Step Solution

Solution:

Using the basic integration rule:
\[
\int \sin(x) \, dx = -\cos(x) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{-\cos(x)}.
\]

Question 2:

\[
\text{Use substitution to evaluate the integral } \int x \cos(x^2) \, dx.
\]
\[
\text{(a) } \frac{1}{2} \sin(x^2) + C, \quad \text{(b) } \frac{1}{2} \cos(x^2) + C, \quad \text{(c) } \frac{1}{2} \sin(x) + C, \quad \text{(d) } \cos(x) + C
\]
Answer: A

Step by Step Solution

Solution:

Using substitution, let \( u = x^2 \), then \( du = 2x dx \).
\[
\int x \cos(x^2) \, dx = \frac{1}{2} \int \cos(u) \, du.
\]
\[
= \frac{1}{2} \sin(u) + C = \frac{1}{2} \sin(x^2) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{2} \sin(x^2) + C}.
\]

Question 3:

\[
\text{Which method would you use to evaluate the integral } \int \frac{1}{x^2 + 1} \, dx?
\]
\[
\text{(a) } Substitution, \quad \text{(b) } Integration by parts, \quad \text{(c) } Partial fractions, \quad \text{(d) } Standard integral
\]
Answer: D

Step by Step Solution

Solution:

The given integral is a standard result:
\[
\int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) + C.
\]
\[
\text{Since this follows directly from a standard formula, the correct answer is } \boxed{\text{Standard integral}}.
\]

Question 4:

\[
\text{Evaluate the integral using integration by parts: } \int x e^x \, dx.
\]
\[
\text{(a) } x e^x – e^x + C, \quad \text{(b) } x e^x + e^x + C, \quad \text{(c) } e^x + C, \quad \text{(d) } e^x – x e^x + C
\]
Answer: A

Step by Step Solution

Solution:

Using integration by parts, let:
\[
u = x, \quad dv = e^x dx.
\]
Then, \( du = dx \) and \( v = e^x \), so applying integration by parts:
\[
\int x e^x \, dx = x e^x – \int e^x \, dx.
\]
\[
= x e^x – e^x + C.
\]
\[
\text{Thus, the correct answer is } \boxed{x e^x – e^x + C}.
\]

Question 5:

\[
\text{Find the integral of } \int \frac{1}{\sqrt{1 – x^2}} \, dx.
\]
\[
\text{(a) } \arcsin(x) + C, \quad \text{(b) } \arccos(x) + C, \quad \text{(c) } \ln(1 – x^2) + C, \quad \text{(d) } \ln(x) + C
\]
Answer: A

Step by Step Solution

Solution:

The given integral is a standard result:
\[
\int \frac{1}{\sqrt{1 – x^2}} \, dx = \arcsin(x) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\arcsin(x) + C}.
\]

Question 6:

\[
\text{Which technique is most useful to solve the integral } \int \frac{1}{x(x+1)} \, dx?
\]
\[
\text{(a) } \text{Substitution}, \quad \text{(b) } \text{Integration by parts}, \quad \text{(c) } \text{Partial fractions}, \quad \text{(d) } \text{Trigonometric substitution}
\]
Answer: C

Step by Step Solution

Solution:

The given integral contains a rational function that can be decomposed into simpler fractions:
\[
\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}.
\]
Solving for \( A \) and \( B \) using partial fractions, we integrate separately.
\[
\text{Thus, the correct answer is } \boxed{\text{Partial fractions}}.
\]

Question 7:

\[
\text{Find the integral } \int \ln(x) \, dx.
\]
\[
\text{(a) } x \ln(x) – x + C, \quad \text{(b) } x \ln(x) + x + C, \quad \text{(c) } x \ln(x) – C, \quad \text{(d) } \ln(x) + C
\]
Answer: A

Step by Step Solution

Solution:

Using integration by parts, let:
\[
u = \ln(x), \quad dv = dx.
\]
Then, \( du = \frac{1}{x} dx \) and \( v = x \), applying integration by parts:
\[
\int \ln(x) \, dx = x \ln(x) – x + C.
\]
\[
\text{Thus, the correct answer is } \boxed{x \ln(x) – x + C}.
\]

Question 8:

\[
\text{Evaluate the integral using trigonometric substitution: } \int \frac{dx}{\sqrt{1 – x^2}}.
\]
\[
\text{(a) } \arcsin(x) + C, \quad \text{(b) } \arccos(x) + C, \quad \text{(c) } \ln(x) + C, \quad \text{(d) } x + C
\]
Answer: A

Step by Step Solution

Solution:

Using the standard integral formula:
\[
\int \frac{dx}{\sqrt{1 – x^2}} = \arcsin(x) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\arcsin(x) + C}.
\]

Question 9:

\[
\text{What is the integral of } \int \sec^2(x) \, dx?
\]
\[
\text{(a) } \cos(x) + C, \quad \text{(b) } \sin(x) + C, \quad \text{(c) } \sec(x) + C, \quad \text{(d) } \tan(x) + C
\]
Answer: D

Step by Step Solution

Solution:

Using the standard result:
\[
\int \sec^2(x) \, dx = \tan(x) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\tan(x) + C}.
\]

Question 10:

\[
\text{Evaluate the integral using the method of partial fractions: } \int \frac{1}{x^2 – 1} \, dx.
\]
\[
\text{(a) } \frac{1}{2} \ln\left| \frac{x-1}{x+1} \right| + C, \quad \text{(b) } \ln(x) + C, \quad \text{(c) } \frac{1}{2} \ln(x) + C, \quad \text{(d) } \ln(x^2 – 1) + C
\]
Answer: A

Step by Step Solution

Solution:

Using partial fraction decomposition:
\[
\frac{1}{x^2 – 1} = \frac{A}{x-1} + \frac{B}{x+1}.
\]
Solving for \( A \) and \( B \), we get:
\[
\frac{1}{x^2 – 1} = \frac{1}{2} \left( \frac{1}{x-1} – \frac{1}{x+1} \right).
\]
Integrating both terms:
\[
\int \frac{1}{x^2 – 1} \, dx = \frac{1}{2} \ln \left| \frac{x-1}{x+1} \right| + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{1}{2} \ln\left| \frac{x-1}{x+1} \right| + C}.
\]

Question 11:

\[
\text{Find the integral of } \int e^{-x^2} \, dx.
\]
\[
\text{(a) } \frac{\sqrt{\pi}}{2} \text{ (Error function)}, \quad \text{(b) } e^{-x^2} + C, \quad \text{(c) } \ln(x) + C, \quad \text{(d) } x^2 e^{-x^2} + C
\]
Answer: A

Step by Step Solution

Solution:

The given integral does not have an elementary function solution. Instead, it is expressed in terms of the error function:
\[
\int e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \operatorname{erf}(x) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\frac{\sqrt{\pi}}{2} \text{ (Error function)}}.
\]

Question 12:

\[
\text{Find the integral of } \int \tan(x) \, dx.
\]
\[
\text{(a) } \ln|\cos(x)| + C, \quad \text{(b) } -\ln|\sin(x)| + C, \quad \text{(c) } -\ln|\cos(x)| + C, \quad \text{(d) } \ln|\sin(x)| + C
\]
Answer: C

Step by Step Solution

Solution:

Using the identity:
\[
\tan(x) = \frac{\sin(x)}{\cos(x)}
\]
We use substitution, let \( u = \cos(x) \), then \( du = -\sin(x) dx \), so:
\[
\int \tan(x) \, dx = \int \frac{\sin(x)}{\cos(x)} dx = -\int \frac{du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C.
\]
\[
\text{Thus, the correct answer is } \boxed{-\ln|\cos(x)| + C}.
\]

Question 13:

\[
\text{Find the integral of } \int x \sin(x) \, dx \text{ using integration by parts.}
\]
\[
\text{(a) } x \cos(x) – \sin(x) + C, \quad \text{(b) } \cos(x) – x \sin(x) + C, \quad \text{(c) } x \sin(x) – \cos(x) + C, \quad \text{(d) } -x \cos(x) + \sin(x) + C
\]
Answer: D

Step by Step Solution

Solution:

Using integration by parts, let:
\[
u = x, \quad dv = \sin(x) dx
\]
Then, we compute:
\[
du = dx, \quad v = -\cos(x)
\]
Applying the integration by parts formula:
\[
\int u dv = uv – \int v du
\]
\[
\int x \sin(x) dx = -x \cos(x) + \int \cos(x) dx
\]
\[
= -x \cos(x) + \sin(x) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{-x \cos(x) + \sin(x) + C}.
\]

Question 14:

\[
\text{Use substitution to evaluate the integral } \int \frac{2x}{x^2 + 1} \, dx.
\]
\[
\text{(a) } \ln|x| + C, \quad \text{(b) } \ln(x^2 + 1) + C, \quad \text{(c) } \arctan(x) + C, \quad \text{(d) } \arcsin(x) + C
\]
Answer: C

Step by Step Solution

Solution:

Let:
\[
u = x^2 + 1, \quad du = 2x dx
\]
Then:
\[
\int \frac{2x}{x^2 + 1} dx = \int \frac{du}{u} = \ln|u| + C = \ln(x^2 + 1) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\arctan(x) + C}.
\]

Question 15:

\[
\text{What is the integral of } \int \frac{dx}{x \ln(x)}?
\]
\[
\text{(a) } \ln(x) + C, \quad \text{(b) } \frac{1}{\ln(x)} + C, \quad \text{(c) } \ln(\ln(x)) + C, \quad \text{(d) } \ln(x \ln(x)) + C
\]
Answer: C

Step by Step Solution

Solution:

Using substitution, let:
\[
u = \ln(x), \quad du = \frac{dx}{x}
\]
Then the integral simplifies to:
\[
\int \frac{du}{u} = \ln|u| + C = \ln(\ln(x)) + C.
\]
\[
\text{Thus, the correct answer is } \boxed{\ln(\ln(x)) + C}.
\]

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