Calculus 1 Quiz: Basic Trig Integrals

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Basic Trig IntegralsQuestion 1 of 20

Find the total area of the region enclosed by the curve $y=\sin(x)$ , the x-axis, and the lines $x=0$ and $x=3\pi/2$ .

Calculus 1 Quiz

Calculus 1 Quiz: Basic Trig Integrals

Practice Basic Trig Integrals in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Basic Trig Integrals, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find the total area of the region enclosed by the curve $y=\sin(x)$ , the x-axis, and the lines $x=0$ and $x=3\pi/2$ .

1
2
3 (correct answer)
0

Explanation: Total area requires integrating the absolute value of the function:

\int_0^{3\pi/2} |\sin(x)| dx

. The function

\sin(x)

is non-negative on

[0, \pi]

and non-positive on

[\pi, 3\pi/2]

. We must split the integral:

\int_0^\pi \sin(x) dx + \int_\pi^{3\pi/2} -\sin(x) dx

. The first integral is

[-\cos(x)]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 2

. The second integral is

[\cos(x)]_\pi^{3\pi/2} = \cos(3\pi/2) - \cos(\pi) = 0 - (-1) = 1

. The total area is the sum,

2+1=3

Question 2

Evaluate $\int \cos(x) \sec^2(\sin x) dx$ .

$\sin(x)\tan(\sin x) + C$
$\tan(\cos x) + C$
$\sec(\sin x) + C$
$\tan(\sin x) + C$ (correct answer)

Explanation: When you encounter an integral with a composite function structure like this, substitution is your go-to technique. The key insight is recognizing that

\cos(x)

is the derivative of

\sin(x)

, which appears inside the

\sec^2

function. Let's use substitution with

u = \sin(x)

, so

du = \cos(x)dx

. This transforms our integral into:

\int \sec^2(u) du

Since the antiderivative of

\sec^2(u)

\tan(u)

, we get:

\tan(u) + C = \tan(\sin x) + C

This confirms that D)

\tan(\sin x) + C

is correct. Now let's examine why the other options fail. Option A)

\sin(x)\tan(\sin x) + C

would result from incorrectly applying the product rule during integration, but that's not how antiderivatives work. Option B)

\tan(\cos x) + C

confuses the inner function—it would require

-\sin(x)dx

in the original integral, not

\cos(x)dx

. Option C)

\sec(\sin x) + C

mistakes the antiderivative of

\sec^2(u)

, which is

\tan(u)

, not

\sec(u)

. Study tip: When you see

\cos(x)

multiplied by a function of

\sin(x)

, immediately consider

u = \sin(x)

substitution. Similarly, watch for

\sin(x)

with functions of

\cos(x)

(use

u = \cos(x)

). Memorize that

\int \sec^2(u)du = \tan(u) + C

—this is a fundamental antiderivative that appears frequently in calculus problems.

Question 3

Evaluate $\int_{\pi/6}^{\pi/3} \frac{\sin(x)}{\cos^2(x)} dx$ .

$2 - \frac{2\sqrt{3}}{3}$ (correct answer)
$\frac{2\sqrt{3}}{3} - 2$
$\sqrt{3} - \frac{\sqrt{3}}{3}$
$1 - \sqrt{2}$

Explanation: The integrand can be rewritten as

\frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x)\tan(x)

. The integral is

\int_{\pi/6}^{\pi/3} \sec(x)\tan(x) dx

. The antiderivative of

\sec(x)\tan(x)

\sec(x)

. Evaluating the definite integral:

[\sec(x)]_{\pi/6}^{\pi/3} = \sec(\pi/3) - \sec(\pi/6) = 2 - \frac{2}{\sqrt{3}} = 2 - \frac{2\sqrt{3}}{3}

Question 4

Let $F(x) = \int_{x}^{\pi/4} \sec^2(t) \,dt$ . What is the value of $F'(\pi/4)$ ?

$-1$
$2$
$0$
$-2$ (correct answer)

Explanation: When you see a derivative of an integral where the variable appears in the limits of integration, you need the Fundamental Theorem of Calculus Part I combined with the chain rule. For

F(x) = \int_{x}^{\pi/4} \sec^2(t) \,dt

, notice that the variable

x

appears in the lower limit. The general rule states that if

G(x) = \int_{a(x)}^{b(x)} f(t) \,dt

, then

G'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x)

. In our case,

a(x) = x

b(x) = \pi/4

, and

f(t) = \sec^2(t)

. Since

a'(x) = 1

and

b'(x) = 0

(constant), we get:

F'(x) = \sec^2(\pi/4) \cdot 0 - \sec^2(x) \cdot 1 = -\sec^2(x)

Therefore,

F'(\pi/4) = -\sec^2(\pi/4) = -(\sqrt{2})^2 = -2

, confirming answer D. Let's examine why the other options are wrong: A)

-1

likely comes from forgetting that

\sec(\pi/4) = \sqrt{2}

, not 1. B)

2

results from dropping the negative sign that comes from having

x

in the lower limit. C)

0

might stem from incorrectly thinking the derivative is zero because one limit is constant. Key strategy: When the variable is in the lower limit of integration, there's always a negative sign in front. Remember that

\sec(\pi/4) = \sqrt{2}

, so

\sec^2(\pi/4) = 2

. Practice recognizing when to apply the Fundamental Theorem versus basic antiderivative rules.

Question 5

The rate of change of the temperature of a chemical reaction is given by $R(t) = \pi \sin(\frac{\pi t}{6})$ degrees Celsius per minute, for $t \ge 0$ . If the initial temperature at $t=0$ is $10^{\circ}C$ , what is the temperature at $t=3$ minutes?

$6^{\circ}C$
$4^{\circ}C$
$10 + \pi^{\circ}C$
$16^{\circ}C$ (correct answer)

Explanation: When you encounter a rate of change problem, you're dealing with derivatives and integrals. Here,

R(t) = \pi \sin(\frac{\pi t}{6})

represents how fast temperature is changing, so to find the actual temperature, you need to integrate this rate function. To find the temperature function

T(t)

, integrate the rate:

T(t) = \int \pi \sin(\frac{\pi t}{6}) dt

. Using substitution with

u = \frac{\pi t}{6}

, so

du = \frac{\pi}{6} dt

, this becomes

T(t) = -6\cos(\frac{\pi t}{6}) + C

. Since the initial temperature is

10°C

t = 0

T(0) = -6\cos(0) + C = -6(1) + C = 10

, so

C = 16

. Therefore,

T(t) = -6\cos(\frac{\pi t}{6}) + 16

. At

t = 3

T(3) = -6\cos(\frac{\pi \cdot 3}{6}) + 16 = -6\cos(\frac{\pi}{2}) + 16 = -6(0) + 16 = 16°C

. Choice A (

6°C

) likely comes from incorrectly calculating

16 - 6\cos(\frac{\pi}{2}) = 16 - 6 = 10

, then subtracting 4 somehow. Choice B (

4°C

) might result from sign errors or confusing the integration constant. Choice C (

10 + \pi°C

) appears to add the coefficient

\pi

directly to the initial temperature without proper integration. Remember: when given a rate of change, always integrate to find the original function, then use initial conditions to determine the constant of integration. Don't forget that

\cos(\frac{\pi}{2}) = 0

Question 6

For $x$ in an interval where $\sin(x) \neq 0$ , find the indefinite integral $\int \frac{\sin(2x)}{\sin(x)} \,dx$

$\sin(x) + C$
$-2\sin(x) + C$
$2\sin(x) + C$ (correct answer)
$\ln|\sin(x)| + C$

Explanation: When you encounter a trigonometric integral like this, look for opportunities to use trigonometric identities to simplify before integrating. The key insight here is recognizing that

\sin(2x) = 2\sin(x)\cos(x)

. Let's substitute this identity into our integral:

\int \frac{\sin(2x)}{\sin(x)} \,dx = \int \frac{2\sin(x)\cos(x)}{\sin(x)} \,dx

Since we're told that

\sin(x) \neq 0

in our interval, we can safely cancel the

\sin(x)

terms:

\int \frac{2\sin(x)\cos(x)}{\sin(x)} \,dx = \int 2\cos(x) \,dx = 2\sin(x) + C

This confirms that C) $2\sin(x) + C$ is correct. Let's examine why the other options are wrong. A) $\sin(x) + C$ is missing the factor of 2 that comes from the double angle formula. B) $-2\sin(x) + C$ has the wrong sign—this would be the result if you mistakenly thought the derivative of

\sin(x)

was

-\cos(x)

. D) $\ln|\sin(x)| + C$ would be correct if the integrand were

\frac{\cos(x)}{\sin(x)} = \cot(x)

, but that's not what we have here. Study tip: When you see trigonometric integrals involving double angles like

\sin(2x)

\cos(2x)

, immediately think about using double angle formulas to simplify. This often transforms a complicated-looking integral into something much more manageable.

Question 7

Evaluate $\int_{0}^{\sqrt{\pi}} x \sin(x^2) \,dx$

$2$
$1$ (correct answer)
$-1$
$1/2$

Explanation: When you encounter an integral like this where the integrand contains a function and its derivative (or a multiple of it), think substitution. The key insight here is recognizing that

x

is closely related to the derivative of

x^2

. Let's use the substitution

u = x^2

. Then

du = 2x \, dx

, which means

x \, dx = \frac{1}{2} du

. We also need to change our limits of integration. When

x = 0

, we have

u = 0^2 = 0

. When

x = \sqrt{\pi}

, we have

u = (\sqrt{\pi})^2 = \pi

. Substituting everything:

\int_{0}^{\sqrt{\pi}} x \sin(x^2) \,dx = \int_{0}^{\pi} \sin(u) \cdot \frac{1}{2} \,du = \frac{1}{2} \int_{0}^{\pi} \sin(u) \,du

Now we can integrate:

\frac{1}{2} \int_{0}^{\pi} \sin(u) \,du = \frac{1}{2}[-\cos(u)]_{0}^{\pi} = \frac{1}{2}[-\cos(\pi) - (-\cos(0))] = \frac{1}{2}[-(-1) + 1] = \frac{1}{2}[2] = 1

So the answer is B) 1. Choice A) 2 would result from forgetting the

\frac{1}{2}

factor from the substitution. Choice C)

-1

comes from sign errors when evaluating

-\cos(\pi) + \cos(0)

. Choice D)

\frac{1}{2}

would occur if you incorrectly computed

-\cos(\pi) + \cos(0) = 1

instead of 2. Strategy tip: When you see

x \sin(x^2),x \cos(x^2)

, or similar patterns, immediately consider

u

-substitution with

u

equal to the "inside" function.

Question 8

Let $g(x) = \int_0^x \cos(t) dt$ .

For what value of $x$ in the interval $(0, \pi)$ does $g(x)$ have a local maximum?

$\frac{\pi}{4}$
$\frac{\pi}{2}$ (correct answer)
$\frac{3\pi}{4}$
$\pi$

Explanation: According to the Fundamental Theorem of Calculus Part 1, the derivative of

g(x)

g'(x) = \cos(x)

. A local maximum occurs at a critical point where the derivative changes sign from positive to negative. We need to find where

\cos(x)=0

and check the sign change. In the interval

(0, \pi)

\cos(x)=0

only at

x=\pi/2

. For

x

(0, \pi/2)

\cos(x) > 0

. For

x

(\pi/2, \pi)

\cos(x) < 0

. Since

g'(x)

changes from positive to negative at

x=\pi/2

g(x)

has a local maximum there.

Question 9

Let $I_1 = \int_0^{\pi/4} \sec^2(x) dx$ and $I_2 = \int_0^{\pi/4} \cos(x) dx$ .

Which of the following statements is true?

$I_1 < I_2$
$I_1 = I_2$
$I_1 = \sqrt{2} I_2$ (correct answer)
$I_1 = 2 I_2$

Explanation: First, evaluate

I_1

I_1 = [\tan(x)]_0^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1

. Next, evaluate

I_2

I_2 = [\sin(x)]_0^{\pi/4} = \sin(\pi/4) - \sin(0) = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}

. Now we compare

I_1

and

I_2

. We have

I_1=1

and

I_2=\frac{\sqrt{2}}{2}

. Does

I_1 = \sqrt{2} I_2

? Substituting the values, we check if

1 = \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{2} = 1

. This is true.

Question 10

Let $A = \int_0^{\pi/2} \sin(x) dx$ .

What is the value of $\int_0^{2\pi} |\sin(x)| dx$ in terms of A?

A
2A
4A (correct answer)
0

Explanation: First, let's find the value of A:

A = \int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = -\cos(\pi/2) - (-\cos(0)) = 0 - (-1) = 1

. The integral

\int_0^{2\pi} |\sin(x)| dx

represents the area under two full arches of the sine curve. The area of one arch is

\int_0^\pi \sin(x) dx = [-\cos(x)]_0^\pi = -(-1) - (-1) = 2

. Since there are two identical arches between 0 and

2\pi

, the total area is

2 \times 2 = 4

. Since

A=1

, the value of the integral is

4A

Question 11

Evaluate $\int \frac{\sin(2x)}{\sin(x)} dx$ .

$2\sin(x) + C$ (correct answer)
$-\frac{1}{2}\cos(2x) + C$
$2\ln|\sin(x)| + C$
$-2\sin(x) + C$

Explanation: This problem requires using the double-angle identity for sine,

\sin(2x) = 2\sin(x)\cos(x)

. Substituting this into the integrand gives

\int \frac{2\sin(x)\cos(x)}{\sin(x)} dx

. For

\sin(x) \neq 0

, this simplifies to

\int 2\cos(x) dx

. The antiderivative is

2\sin(x) + C

Question 12

Evaluate $\int (\tan^2(x) + \sec^2(x)) dx$ .

$x + C$
$2\tan(x) - x + C$ (correct answer)
$2\sec(x)\tan(x) - x + C$
$\frac{1}{3}\tan^3(x) + \tan(x) + C$

Explanation: To integrate

\tan^2(x)

, we use the identity

\tan^2(x) = \sec^2(x) - 1

. Substituting this into the integrand gives

\int ((\sec^2(x) - 1) + \sec^2(x)) dx = \int (2\sec^2(x) - 1) dx

. We can integrate this term by term. The integral of

2\sec^2(x)

2\tan(x)

, and the integral of

-1

-x

. Therefore, the result is

2\tan(x) - x + C

Question 13

Evaluate the definite integral $\int_{0}^{\pi/6} \sin(3x) \,dx$

$1/3$ (correct answer)
$1$
$-1/3$
$-1$

Explanation: To evaluate the integral, we use a u-substitution. Let

u = 3x

, so

du = 3\,dx

, or

dx = du/3

. We must also change the limits of integration. When

x=0

u=3(0)=0

. When

x=\pi/6

u=3(\pi/6)=\pi/2

. The integral becomes:

\int_{0}^{\pi/2} \sin(u) \frac{du}{3} = \frac{1}{3} \int_{0}^{\pi/2} \sin(u) \,du

Now, we integrate and apply the Fundamental Theorem of Calculus:

\frac{1}{3} [-\cos(u)]_{0}^{\pi/2} = \frac{1}{3} (-\cos(\pi/2) - (-\cos(0))) = \frac{1}{3} (-(0) - (-1)) = \frac{1}{3}(1) = \frac{1}{3}

Question 14

What is the value of the definite integral $\int_{0}^{3\pi/2} |\cos(x)| \,dx$ ?

$3$ (correct answer)
$-1$
$1$
$2$

Explanation: The function

|\cos(x)|

is the absolute value of

\cos(x)

. We must split the integral based on the intervals where

\cos(x)

is positive or negative. In the interval

[0, 3\pi/2]

\cos(x) \ge 0

for

x \in [0, \pi/2]

and

\cos(x) \le 0

for

x \in [\pi/2, 3\pi/2]

. So, we split the integral:

\int_{0}^{3\pi/2} |\cos(x)| \,dx = \int_{0}^{\pi/2} \cos(x) \,dx + \int_{\pi/2}^{3\pi/2} -\cos(x) \,dx

Evaluate each part:

\int_{0}^{\pi/2} \cos(x) \,dx = [\sin(x)]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1

\int_{\pi/2}^{3\pi/2} -\cos(x) \,dx = -[\sin(x)]_{\pi/2}^{3\pi/2} = -(\sin(3\pi/2) - \sin(\pi/2)) = -(-1 - 1) = -(-2) = 2

The total value is the sum of the parts:

1 + 2 = 3

. Distractor B is the result of integrating

\cos(x)

without the absolute value. Distractor C is the area of only the first positive section. Distractor D is the area from

0

\pi

Question 15

What is the average value of the function $f(x) = \sin(x)$ on the closed interval $[0, \pi]$ ?

$-2/\pi$
$2$
$0$
$2/\pi$ (correct answer)

Explanation: When you encounter a question about the average value of a function, you're working with the Mean Value Theorem for Integrals. The average value of a continuous function

f(x)

over an interval

[a,b]

is given by the formula:

\text{Average value} = \frac{1}{b-a}\int_a^b f(x)\,dx

For

f(x) = \sin(x)

[0,\pi]

, we need to calculate:

\frac{1}{\pi-0}\int_0^\pi \sin(x)\,dx = \frac{1}{\pi}\int_0^\pi \sin(x)\,dx

The antiderivative of

\sin(x)

-\cos(x)

, so:

\frac{1}{\pi}[-\cos(x)]_0^\pi = \frac{1}{\pi}[-\cos(\pi) - (-\cos(0))] = \frac{1}{\pi}[-(-1) + 1] = \frac{2}{\pi}

This confirms answer choice D is correct. Looking at the wrong answers: A)

-2/\pi

results from a sign error when evaluating the antiderivative—forgetting that

-\cos(\pi) = -(-1) = 1

, not

-1

. B)

2

comes from correctly finding the numerator (2) but forgetting to divide by

\pi

. C)

0

might seem tempting because

\sin(x)

has both positive and negative values over many intervals, but on

[0,\pi]

, sine is entirely non-negative. Remember: the average value formula always requires both the definite integral AND division by the interval length. Also, visualize the function—

\sin(x)

forms a complete "hump" above the x-axis from 0 to

\pi

, so the average should be positive and reasonable in magnitude.

Question 16

Evaluate the definite integral $\int_{0}^{\pi/3} \tan(x) \,dx$

$-\ln(2)$
$\ln(2)$ (correct answer)
$3$
$\ln(2+\sqrt{3})$

Explanation: When evaluating definite integrals involving trigonometric functions like tangent, you need to find the antiderivative first, then apply the Fundamental Theorem of Calculus. To find

\int \tan(x) \,dx

, recall that

\tan(x) = \frac{\sin(x)}{\cos(x)}

. Using substitution with

u = \cos(x)

, so

du = -\sin(x) \,dx

, we get:

\int \tan(x) \,dx = \int \frac{\sin(x)}{\cos(x)} \,dx = -\int \frac{1}{u} \,du = -\ln|u| + C = -\ln|\cos(x)| + C

Now applying the limits from 0 to

\pi/3

\int_{0}^{\pi/3} \tan(x) \,dx = [-\ln|\cos(x)|]_{0}^{\pi/3} = -\ln|\cos(\pi/3)| - (-\ln|\cos(0)|)

Since

\cos(\pi/3) = 1/2

and

\cos(0) = 1

= -\ln(1/2) + \ln(1) = -\ln(1/2) + 0 = -\ln(2^{-1}) = \ln(2)

This confirms answer (B) $\ln(2)$ is correct. (A) $-\ln(2)$ results from forgetting the negative sign when setting up the substitution or incorrectly applying the limits. (C) 3 likely comes from confusing this with a different trigonometric integral or making arithmetic errors. (D) $\ln(2+\sqrt{3})$ might arise from incorrectly trying to use

\tan(\pi/3) = \sqrt{3}

directly in the calculation instead of properly integrating. Study tip: Always remember that

\int \tan(x) \,dx = -\ln|\cos(x)| + C

. Practice the u-substitution method for trigonometric ratios, and double-check your signs when applying limits.

Question 17

Find the indefinite integral $\int \frac{\sin(x) - 1}{\cos^2(x)} \,dx$

$\sec(x) + \tan(x) + C$
$\sec(x) - \tan(x) + C$ (correct answer)
$\tan(x) - \sec(x) + C$
$\ln(\cos^2(x)) + \sec(x) + C$

Explanation: When you encounter a rational function with trigonometric expressions, look for ways to split the fraction and use standard derivative patterns. The key insight here is recognizing that you can separate this integral into two simpler parts. Start by splitting the fraction:

\int \frac{\sin(x) - 1}{\cos^2(x)} \,dx = \int \frac{\sin(x)}{\cos^2(x)} \,dx - \int \frac{1}{\cos^2(x)} \,dx

For the first integral, rewrite

\frac{\sin(x)}{\cos^2(x)} = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)} = \tan(x) \sec(x)

. Since

\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)

, we have

\int \tan(x)\sec(x) \,dx = \sec(x)

. For the second integral, recognize that

\frac{1}{\cos^2(x)} = \sec^2(x)

, and since

\frac{d}{dx}[\tan(x)] = \sec^2(x)

, we get

\int \sec^2(x) \,dx = \tan(x)

. Therefore:

\int \frac{\sin(x) - 1}{\cos^2(x)} \,dx = \sec(x) - \tan(x) + C

This confirms answer B is correct. A has the wrong sign between terms. C reverses both terms incorrectly. D introduces a logarithmic term that doesn't arise from this integration and would require a completely different approach. Study tip: When integrating rational trigonometric functions, always try splitting fractions first, then look for standard derivative patterns like

\sec(x)\tan(x)

and

\sec^2(x)

. Memorizing the derivatives of all six trig functions will make these problems much faster.

Question 18

Evaluate $\int_{-\pi/4}^{\pi/4} (x^3 + \tan(x) + \sec^2(x)) \,dx$

$1$
$0$
$2$ (correct answer)
$2 + \ln(2)$

Explanation: When evaluating definite integrals, look for opportunities to use symmetry properties and recognize standard antiderivatives. This problem combines both strategies effectively. Break this integral into three parts:

\int_{-\pi/4}^{\pi/4} x^3 \,dx + \int_{-\pi/4}^{\pi/4} \tan(x) \,dx + \int_{-\pi/4}^{\pi/4} \sec^2(x) \,dx

For the first integral, notice that

x^3

is an odd function (

f(-x) = -f(x)

), and you're integrating over a symmetric interval

[-\pi/4, \pi/4]

. When an odd function is integrated over a symmetric interval, the result is always zero because the positive and negative areas cancel out. For the second integral,

\tan(x)

is also an odd function, so this integral also equals zero by the same reasoning. For the third integral, you need the antiderivative of

\sec^2(x)

, which is

\tan(x)

. Evaluating:

[\tan(x)]_{-\pi/4}^{\pi/4} = \tan(\pi/4) - \tan(-\pi/4) = 1 - (-1) = 2

Therefore, the total integral equals

0 + 0 + 2 = 2

. Choice A gives 1, which might result from forgetting the symmetric cancellation and incorrectly evaluating one of the odd function integrals. Choice B gives 0, which you'd get if you mistakenly applied odd function symmetry to the entire integrand without separating

\sec^2(x)

. Choice D gives

2 + \ln(2)

, which suggests confusion with antiderivatives—perhaps mixing up derivatives of trigonometric and logarithmic functions. Study tip: Always check for odd/even function symmetry first when you see symmetric integration limits—it can eliminate terms immediately and simplify your work significantly.

Question 19

Find the indefinite integral $\int \frac{\csc^2(x)}{1 + \cot(x)} \,dx$

$\frac{-1}{1 + \cot(x)} + C$
$\ln|1 + \cot(x)| + C$
$- \ln|1 + \cot(x)| + C$ (correct answer)
$\frac{-\cot(x)}{x - \csc^2(x)} + C$

Explanation: When you encounter an integral with trigonometric functions like

\csc^2(x)

and

\cot(x)

, look for substitution opportunities. The key insight here is recognizing that the derivative of

\cot(x)

-\csc^2(x)

. Let's use substitution with

u = 1 + \cot(x)

. Then

du = -\csc^2(x) \, dx

, which means

\csc^2(x) \, dx = -du

. Substituting into our integral:

\int \frac{\csc^2(x)}{1 + \cot(x)} \,dx = \int \frac{1}{u} \cdot (-du) = -\int \frac{1}{u} \,du = -\ln|u| + C

Substituting back:

-\ln|1 + \cot(x)| + C

This confirms answer C is correct. Let's examine why the other options fail: A suggests

\frac{-1}{1 + \cot(x)} + C

. If you differentiate this, you get

\frac{\csc^2(x)}{(1 + \cot(x))^2}

, which doesn't match our integrand. B gives

\ln|1 + \cot(x)| + C

. This is missing the negative sign that comes from our substitution

du = -\csc^2(x) \, dx

. D presents an unnecessarily complex expression that doesn't relate to the standard antiderivative forms we expect from this substitution. Study tip: When you see

\csc^2(x)

in an integral, immediately think about

\cot(x)

since

\frac{d}{dx}[\cot(x)] = -\csc^2(x)

. This derivative relationship is your gateway to effective substitution in trigonometric integrals.

Question 20

Which of the following is an antiderivative of $f(\theta) = 2\tan^2(\theta)$ ?

$2\tan(\theta) - 2\theta + C$ (correct answer)
$\frac{2}{3}\tan^3(\theta) + C$
$2\tan(\theta) + C$
$2\sec^2(\theta) - 2 + C$

Explanation: To find the integral of

2\tan^2(\theta)

, we first use the Pythagorean identity

\tan^2(\theta) = \sec^2(\theta) - 1

. So, the integral becomes:

\int 2\tan^2(\theta) \,d\theta = \int 2(\sec^2(\theta) - 1) \,d\theta = \int (2\sec^2(\theta) - 2) \,d\theta

We can integrate term by term. The integral of

2\sec^2(\theta)

2\tan(\theta)

, and the integral of

2

2\theta

. Therefore, the antiderivative is

2\tan(\theta) - 2\theta + C

. Distractor B arises from incorrectly applying the power rule to a trigonometric function. Distractor C results from forgetting to integrate the

-1

term from the identity. Distractor D is the derivative of the original function's rewritten form, not its antiderivative.