Basic Trig Integrals - Calculus 1

1. Evaluate $\int \cos(x) \sec^2(\sin x) dx$.
   Answer: $\tan(\sin x) + C$ 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 $\sec^2(u)$ is $\tan(u)$, 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.
2. Evaluate $\int_{0}^{\sqrt{\pi}} x \sin(x^2) \,dx$
   Answer: $1$ 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.
3. A particle's velocity is given by $v(t) = \sec(t)\tan(t)$. If its position at $t=0$ is $s(0)=5$, what is its position at $t=\pi/4$?
   Answer: $4 + \sqrt{2}$ Position $s(t)$ is the antiderivative of velocity $v(t)$. So, $s(t) = \int \sec(t)\tan(t) dt = \sec(t) + C$. We use the initial condition $s(0)=5$ to find the constant $C$. $s(0) = \sec(0) + C = 1 + C$. Setting this equal to 5 gives $1+C=5$, so $C=4$. The position function is $s(t) = \sec(t) + 4$. Evaluating at $t=\pi/4$ gives $s(\pi/4) = \sec(\pi/4) + 4 = \sqrt{2} + 4$.
4. Find the indefinite integral $\int \frac{\csc^2(x)}{1 + \cot(x)} \,dx$
   Answer: $- \ln|1 + \cot(x)| + C$ 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 $\cot(x)$ is $-\csc^2(x)$. 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.
5. Evaluate $\int_{-\pi/3}^{\pi/3} (\sin(x) + \sec^2(x)) dx$.
   Answer: $2\sqrt{3}$ We can split the integral into two parts: $\int_{-\pi/3}^{\pi/3} \sin(x) dx + \int_{-\pi/3}^{\pi/3} \sec^2(x) dx$. Since $\sin(x)$ is an odd function and it is being integrated over a symmetric interval $[-a, a]$, the value of the first integral is 0. The function $\sec^2(x)$ is an even function, so $\int_{-\pi/3}^{\pi/3} \sec^2(x) dx = 2 \int_0^{\pi/3} \sec^2(x) dx$. This evaluates to $2[\tan(x)]_0^{\pi/3} = 2(\tan(\pi/3) - \tan(0)) = 2(\sqrt{3} - 0) = 2\sqrt{3}$. The total value is $0 + 2\sqrt{3} = 2\sqrt{3}$.
6. What is the value of the definite integral $\int_{0}^{3\pi/2} |\cos(x)| \,dx$?
   Answer: $3$ 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 $0$ to $\pi$.
7. Evaluate $\int \frac{\sin(2x)}{\sin(x)} dx$.
   Answer: $2\sin(x) + C$ 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 $2\sin(x) + C$.
8. Evaluate $\int_0^{\pi/4} \frac{dx}{\cos^2(x)}$.
   Answer: 1 The integrand $\frac{1}{\cos^2(x)}$ is equivalent to $\sec^2(x)$. The problem is to evaluate $\int_0^{\pi/4} \sec^2(x) dx$. The antiderivative of $\sec^2(x)$ is $\tan(x)$. Evaluating the definite integral gives $[\tan(x)]_0^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1$.
9. Evaluate $\int_{\pi/4}^{\pi/2} \csc(x)\cot(x) dx$.
   Answer: $\sqrt{2}-1$ The antiderivative of $\csc(x)\cot(x)$ is $-\csc(x)$. Evaluating the definite integral gives $[-\csc(x)]_{\pi/4}^{\pi/2} = -\csc(\pi/2) - (-\csc(\pi/4))$. Since $\csc(\pi/2) = 1$ and $\csc(\pi/4) = \sqrt{2}$, the result is $-1 - (-\sqrt{2})) = \sqrt{2}-1$.
10. What is the average value of the function $f(x) = \sec^2(x)$ on the interval $[0, \pi/4]$?
    Answer: $\frac{4}{\pi}$ The average value of a function $f(x)$ on an interval $[a, b]$ is given by $\frac{1}{b-a} \int_a^b f(x) dx$. Here, $a=0$ and $b=\pi/4$. The average value is $\frac{1}{\pi/4 - 0} \int_0^{\pi/4} \sec^2(x) dx = \frac{4}{\pi} \int_0^{\pi/4} \sec^2(x) dx$. The antiderivative of $\sec^2(x)$ is $\tan(x)$. So, we have $\frac{4}{\pi} [\tan(x)]_0^{\pi/4} = \frac{4}{\pi}(\tan(\pi/4) - \tan(0)) = \frac{4}{\pi}(1-0) = \frac{4}{\pi}$.
11. Let $F(x) = \int_{x}^{\pi/4} \sec^2(t) \,dt$. What is the value of $F'(\pi/4)$?
    Answer: $-2$ 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.
12. The acceleration of a particle is $a(t) = -\sin(t)$. If the particle's initial velocity is $v(0)=1$ and initial position is $s(0)=3$, what is its position at $t=\pi$?
    Answer: 3 First, find the velocity function $v(t)$ by integrating the acceleration function $a(t)$: $v(t) = \int -\sin(t) dt = \cos(t) + C_1$. Use the initial velocity $v(0)=1$ to find $C_1$: $1 = \cos(0) + C_1 \implies 1 = 1 + C_1 \implies C_1 = 0$. So, $v(t) = \cos(t)$. Next, find the position function $s(t)$ by integrating the velocity function $v(t)$: $s(t) = \int \cos(t) dt = \sin(t) + C_2$. Use the initial position $s(0)=3$ to find $C_2$: $3 = \sin(0) + C_2 \implies 3 = 0 + C_2 \implies C_2 = 3$. So, $s(t) = \sin(t) + 3$. Finally, evaluate the position at $t=\pi$: $s(\pi) = \sin(\pi) + 3 = 0 + 3 = 3$.
13. What is the average value of the function $f(x) = \sin(x)$ on the closed interval $[0, \pi]$?
    Answer: $2/\pi$ 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 $f(x) = \sin(x)$ on $[0,\pi]$, 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 $\sin(x)$ is $-\cos(x)$, 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.
14. A function is defined by $f(x) = \int_x^{\pi/4} \sec(t)\tan(t) dt$. What is the value of $f(0)$?
    Answer: $\sqrt{2}-1$ When you encounter a definite integral that defines a function, you need to use the Fundamental Theorem of Calculus to evaluate it. Here, $f(x) = \int_x^{\pi/4} \sec(t)\tan(t) dt$, so you're looking for $f(0) = \int_0^{\pi/4} \sec(t)\tan(t) dt$. To evaluate this integral, recognize that $\sec(t)\tan(t)$ is the derivative of $\sec(t)$. You can verify this: $\frac{d}{dt}[\sec(t)] = \sec(t)\tan(t)$. Therefore: $\int_0^{\pi/4} \sec(t)\tan(t) dt = \sec(t)\Big|_0^{\pi/4} = \sec(\pi/4) - \sec(0)$ Since $\sec(\pi/4) = \frac{1}{\cos(\pi/4)} = \frac{1}{\sqrt{2}/2} = \sqrt{2}$ and $\sec(0) = \frac{1}{\cos(0)} = 1$: $f(0) = \sqrt{2} - 1$ Looking at the wrong answers: Choice A gives -1, which would result from incorrectly computing $\sec(0) - \sec(\pi/4)$ (reversing the order). Choice B gives $1 - \sqrt{2}$, which happens if you mistakenly use $\sec(0) - \sec(\pi/4)$ instead of the correct order. Choice C gives 1, which occurs if you forget to subtract $\sec(0)$ entirely and only use $\sec(\pi/4) - 1$ incorrectly. Study tip: When evaluating definite integrals, always remember that the Fundamental Theorem gives you $F(b) - F(a)$ for $\int_a^b f(t) dt$, where the upper limit comes first. Also, memorize that $\frac{d}{dt}[\sec(t)] = \sec(t)\tan(t)$ — it's a frequently tested derivative-antiderivative pair.
15. Which of the following definite integrals has a value of 0?
    Answer: $\int_0^{\pi} \cos(x) dx$ We evaluate each integral: A) $\int_0^{\pi} \sin(x) dx = [-\cos(x)]_0^{\pi} = -(-1) - (-1) = 2$. B) $\int_0^{\pi} \cos(x) dx = [\sin(x)]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0$. C) $\int_{-\pi/4}^{\pi/4} \sec^2(x) dx = [\tan(x)]_{-\pi/4}^{\pi/4} = 1 - (-1) = 2$. D) $\int_0^{\pi/2} \sin(x) dx = [-\cos(x)]_0^{\pi/2} = 0 - (-1) = 1$. The only integral that evaluates to 0 is B.
16. If $F(x) = \int_{\pi/4}^{x^2} \sec^2(t) dt$, what is $F'(\sqrt{\pi/3})$?
    Answer: $8\sqrt{\frac{\pi}{3}}$ By the Fundamental Theorem of Calculus Part 1, combined with the Chain Rule, if $F(x) = \int_a^{g(x)} f(t) dt$, then $F'(x) = f(g(x)) \cdot g'(x)$. Here, $f(t) = \sec^2(t)$ and $g(x) = x^2$, so $g'(x) = 2x$. Therefore, $F'(x) = \sec^2(x^2) \cdot 2x$. We need to evaluate this at $x = \sqrt{\pi/3}$. $F'(\sqrt{\pi/3}) = \sec^2((\sqrt{\pi/3})^2) \cdot 2\sqrt{\pi/3} = \sec^2(\pi/3) \cdot 2\sqrt{\pi/3}$. Since $\cos(\pi/3) = 1/2$, $\sec(\pi/3) = 2$, and $\sec^2(\pi/3) = 4$. The result is $4 \cdot 2\sqrt{\pi/3} = 8\sqrt{\frac{\pi}{3}}$.
17. If $\int_{\pi/2}^c \cos(x) dx = -1$ and $c > \pi/2$, which of the following could be a value for $c$?
    Answer: $\pi$ First, evaluate the integral in terms of $c$: $\int_{\pi/2}^c \cos(x) dx = [\sin(x)]_{\pi/2}^c = \sin(c) - \sin(\pi/2) = \sin(c) - 1$. We are given that this equals -1, so $\sin(c) - 1 = -1$, which implies $\sin(c) = 0$. Among the answer choices with $c > \pi/2$, we need $\sin(c) = 0$. Checking: $\sin(\pi) = 0$. Verification: $\int_{\pi/2}^\pi \cos(x)dx = [\sin(x)]_{\pi/2}^\pi = \sin(\pi) - \sin(\pi/2) = 0 - 1 = -1$. ✓
18. What is the value of $\int_0^{2\pi} |\sin(x)| dx$ in terms of A?
    Answer: 4A 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 $4A$.
19. What is the value of $\int_{\pi/4}^{\pi/3} (1+\cot^2(x)) dx$?
    Answer: $1 - \frac{\sqrt{3}}{3}$ First, use the Pythagorean identity $1+\cot^2(x) = \csc^2(x)$. The integral simplifies to $\int_{\pi/4}^{\pi/3} \csc^2(x) dx$. The antiderivative of $\csc^2(x)$ is $-\cot(x)$. Now, evaluate the definite integral: $[-\cot(x)]_{\pi/4}^{\pi/3} = -\cot(\pi/3) - (-\cot(\pi/4)) = -\frac{1}{\sqrt{3}} - (-1) = 1 - \frac{1}{\sqrt{3}} = 1 - \frac{\sqrt{3}}{3}$.
20. Which of the following is an antiderivative of $f(\theta) = 2\tan^2(\theta)$?
    Answer: $2\tan(\theta) - 2\theta + C$ 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 $2\sec^2(\theta)$ is $2\tan(\theta)$, and the integral of $2$ is $2\theta$. 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.

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