Integration by Substitution - Calculus 1

1. Which of the following is the indefinite integral of $\int x(x+1)^5 \,dx$?
   Answer: $\frac{(x+1)^7}{7} - \frac{(x+1)^6}{6} + C$ To solve $\int x(x+1)^5 \,dx$, let $u = x+1$. This implies $x = u-1$ and $du = dx$. Substituting these into the integral gives: $\int (u-1)u^5 \,du = \int (u^6 - u^5) \,du$ Now, integrate term by term: $\frac{u^7}{7} - \frac{u^6}{6} + C$ Finally, substitute back $u = x+1$: $\frac{(x+1)^7}{7} - \frac{(x+1)^6}{6} + C$ B: This answer incorrectly attempts to integrate the factors separately, which is not a valid integration rule. C: This is the result of applying integration by parts incorrectly. D: This answer has a sign error in the second term, resulting from an error in distributing $(u-1)u^5$.
2. Evaluate $\int \tan^3(x) \sec^2(x) \,dx$
   Answer: $\frac{1}{4}\tan^4(x) + C$ For this integral, a good choice for substitution is $u = \tan(x)$. Then $du = \sec^2(x) \,dx$. The integral becomes a simple power rule problem: $\int u^3 \,du$ Integrating this gives $\frac{u^4}{4} + C$. Substituting back $u = \tan(x)$ gives: $\frac{1}{4}\tan^4(x) + C$ B: This answer might result from attempting to break down $\tan^3(x)$ into $\tan(x)(\sec^2(x)-1)$ and making integration errors. C: This answer would be correct if the substitution was $u=\sec(x)$, but the integrand was $\tan(x)\sec^3(x)$. D: This answer seems to be the result of incorrectly applying the product rule for differentiation instead of integrating.
3. Evaluate $\int \frac{3x+3}{\sqrt{x^2+2x+5}} \,dx$
   Answer: $3\sqrt{x^2+2x+5} + C$ Let $u = x^2+2x+5$. Then $du = (2x+2) \,dx = 2(x+1) \,dx$. The numerator of the integrand is $3x+3 = 3(x+1)$. We can rewrite the integral as: $\int \frac{3(x+1)}{\sqrt{x^2+2x+5}} \,dx$ We see that $(x+1)dx = \frac{1}{2}du$. So we can substitute: $\int \frac{3}{\sqrt{u}} \cdot \frac{1}{2} \,du = \frac{3}{2} \int u^{-1/2} \,du$ Integrating gives: $\frac{3}{2} (2u^{1/2}) + C = 3u^{1/2} + C = 3\sqrt{u} + C$ Substituting back $u = x^2+2x+5$ gives: $3\sqrt{x^2+2x+5} + C$ B: This answer incorrectly omits the factor of 2 from the integration of $u^{-1/2}$. C: This answer results from applying an incorrect integration rule, treating the form as $1/u$. D: This answer results from an incorrect application of the power rule for integration.
4. The integral $\int \frac{x+2}{x^2+4} \,dx$ can be split into two separate integrals. What is the resulting antiderivative?
   Answer: $\frac{1}{2}\ln(x^2+4) + \arctan(\frac{x}{2}) + C$ First, split the integral into two parts: $\int \frac{x+2}{x^2+4} \,dx = \int \frac{x}{x^2+4} \,dx + \int \frac{2}{x^2+4} \,dx$ For the first integral, let $u = x^2+4$, so $du = 2x \,dx$ and $\frac{1}{2}du = x \,dx$. $\int \frac{x}{x^2+4} \,dx = \frac{1}{2} \int \frac{1}{u} \,du = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+4)$ For the second integral, we use the arctangent formula $\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\arctan(\frac{x}{a})$. Here $a=2$. $\int \frac{2}{x^2+4} \,dx = 2 \int \frac{1}{x^2+2^2} \,dx = 2 \left(\frac{1}{2}\arctan\left(\frac{x}{2}\right)\right) = \arctan\left(\frac{x}{2}\right)$ Combining the results gives: $\frac{1}{2}\ln(x^2+4) + \arctan(\frac{x}{2}) + C$. B: This answer misses the $1/2$ factor on the logarithm and uses the incorrect form for the arctangent. C: This answer incorrectly integrates the second part as if it were $\int \frac{1}{x^2+1}dx$. D: This answer incorrectly handles the constant in the second integral, resulting in $2\arctan(x/2)$ instead of $\arctan(x/2)$.
5. What is the value of the definite integral $$\int_{0}^{1} (x+1)e^{x^2+2x} \,dx$$ ?
   Answer: $\frac{e^3 - 1}{2}$ When you encounter a definite integral with a composite function in the exponent, look for opportunities to use substitution. The key insight here is recognizing that the integrand $(x+1)e^{x^2+2x}$ has a structure where the coefficient matches the derivative of the exponent. Notice that if $u = x^2 + 2x$, then $\frac{du}{dx} = 2x + 2 = 2(x+1)$. This means $(x+1)dx = \frac{1}{2}du$, which perfectly matches our integrand structure. Using the substitution $u = x^2 + 2x$: - When $x = 0$: $u = 0^2 + 2(0) = 0$ - When $x = 1$: $u = 1^2 + 2(1) = 3$ The integral becomes: $\int_0^3 e^u \cdot \frac{1}{2} du = \frac{1}{2}\int_0^3 e^u du = \frac{1}{2}[e^u]_0^3 = \frac{1}{2}(e^3 - e^0) = \frac{e^3 - 1}{2}$ Answer choice A gives $e^3$, which would result from forgetting the $\frac{1}{2}$ factor and incorrectly evaluating the bounds. Answer choice B gives $e^3 - 1$, which happens when you correctly find $e^3 - 1$ but forget the $\frac{1}{2}$ coefficient from the substitution. Answer choice C gives $\frac{e^3}{2}$, which occurs when you remember the $\frac{1}{2}$ factor but forget to subtract 1 when evaluating $e^3 - e^0$. Study tip: In substitution problems, always check that the derivative relationship works out completely, including any constant factors that arise from $du$.
6. Which of the following integrals results from making the substitution $u = x^2$ in $\int_{1}^{2} x e^{x^2} \,dx$?
   Answer: $\frac{1}{2} \int_{1}^{4} e^u \,du$ When you encounter a substitution problem, you need to systematically replace three things: the variable, the differential, and the limits of integration. Starting with $u = x^2$, you first find the differential by taking the derivative: $\frac{du}{dx} = 2x$, so $du = 2x \, dx$. This means $x \, dx = \frac{1}{2} du$. Next, update the limits of integration. When $x = 1$: $u = 1^2 = 1$. When $x = 2$: $u = 2^2 = 4$. So your new limits are from 1 to 4. Now substitute everything into the original integral: $\int_{1}^{2} x e^{x^2} \,dx = \int_{1}^{4} e^u \cdot \frac{1}{2} \,du = \frac{1}{2} \int_{1}^{4} e^u \,du$ This matches answer choice D. Looking at the wrong answers: A keeps the original limits (1 to 2) instead of updating them to reflect the substitution. B has the correct new limits but missing the $\frac{1}{2}$ factor that comes from $x \, dx = \frac{1}{2} du$. C has the wrong coefficient (2 instead of $\frac{1}{2}$) and keeps the original limits. Study tip: For u-substitution, always follow the three-step checklist: substitute the function, substitute the differential (don't forget coefficients!), and update the limits using your substitution equation. Missing any of these steps leads to the common wrong answers you see here.
7. Evaluate the indefinite integral $\int \frac{5^{x}}{1+5^{x}} \,dx$
   Answer: $\frac{1}{\ln 5} \ln(1+5^x) + C$ When you encounter an integral with exponential functions in both numerator and denominator, look for substitution opportunities that can simplify the expression. For $\int \frac{5^{x}}{1+5^{x}} \,dx$, let's use the substitution $u = 5^x$. Then $du = 5^x \ln 5 \, dx$, which means $dx = \frac{du}{5^x \ln 5} = \frac{du}{u \ln 5}$. Substituting into our integral: $\int \frac{5^{x}}{1+5^{x}} \,dx = \int \frac{u}{1+u} \cdot \frac{du}{u \ln 5} = \frac{1}{\ln 5} \int \frac{1}{1+u} \,du$ This simplifies to $\frac{1}{\ln 5} \ln(1+u) + C = \frac{1}{\ln 5} \ln(1+5^x) + C$, confirming answer D. Let's examine why the other choices are incorrect: Choice A, $\frac{5^x}{\ln 5} + x + C$, can be eliminated by differentiation - this would give $5^x + 1$, not our original integrand. Choice B, $\ln(1+5^x) + C$, is missing the crucial factor of $\frac{1}{\ln 5}$ that comes from the substitution process. Choice C, $x - \frac{\ln(1+5^x)}{\ln 5} + C$, incorrectly includes an $x$ term and has the wrong sign on the logarithmic term. Study tip: When integrating rational functions involving exponentials, try substituting $u$ equal to the exponential expression. The derivative will often provide exactly the factor needed to simplify the integral, but don't forget to account for constants like $\ln 5$ that appear in exponential derivatives.
8. Evaluate $\int_0^{\ln(2)} \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$.
   Answer: $\ln\left(\frac{5}{4}\right)$ When you encounter an integral with exponential expressions like $\frac{e^x - e^{-x}}{e^x + e^{-x}}$, look for substitution opportunities. This particular form should remind you of hyperbolic functions or suggest a u-substitution. Let's use substitution with $u = e^x + e^{-x}$. Then $du = (e^x - e^{-x})dx$, which is exactly our numerator! This transforms our integral into $\int \frac{1}{u} du = \ln|u| + C$. Substituting back: $\int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx = \ln|e^x + e^{-x}| + C$ Now evaluate the definite integral from 0 to $\ln(2)$: At $x = \ln(2)$: $e^{\ln(2)} + e^{-\ln(2)} = 2 + \frac{1}{2} = \frac{5}{2}$ At $x = 0$: $e^0 + e^0 = 1 + 1 = 2$ Therefore: $\ln\left(\frac{5}{2}\right) - \ln(2) = \ln\left(\frac{5/2}{2}\right) = \ln\left(\frac{5}{4}\right)$ Choice A gives $\ln\left(\frac{3}{4}\right)$, likely from incorrectly calculating the bounds. Choice B gives $\ln\left(\frac{3}{2}\right)$, possibly from an error in the upper bound calculation. Choice C gives $\ln\left(\frac{5}{2}\right)$, which occurs if you forget to subtract the lower bound evaluation—a common mistake in definite integrals. The answer is D. Key strategy: When you see rational expressions involving $e^x$ and $e^{-x}$, check if the numerator is the derivative of the denominator. This often leads to a clean logarithmic antiderivative.
9. Evaluate $\int_{\pi/6}^{\pi/2} \frac{\cos(x)}{\sin^3(x)} dx$.
   Answer: $\frac{3}{2}$ Let $u = \sin(x)$. Then $du = \cos(x) dx$. We must change the limits of integration. When $x = \pi/6$, $u = \sin(\pi/6) = 1/2$. When $x = \pi/2$, $u = \sin(\pi/2) = 1$. The integral transforms to: $\int_{1/2}^1 \frac{1}{u^3} du = \int_{1/2}^1 u^{-3} du$ Now, integrate with respect to $u$: $\left[ \frac{u^{-2}}{-2} \right]_{1/2}^1 = \left[ -\frac{1}{2u^2} \right]_{1/2}^1 = \left(-\frac{1}{2(1)^2}\right) - \left(-\frac{1}{2(1/2)^2}\right) = -\frac{1}{2} - \left(-\frac{1}{2(1/4)}\right) = -\frac{1}{2} - (-2) = \frac{3}{2}$ Distractor B comes from an incorrect power rule, integrating $u^{-3}$ to $u^{-4}/(-4)$. Distractor C arises from forgetting the factor of $1/(-2)$ from the power rule. Distractor D is the result of a sign error when evaluating the definite integral, calculating $-1/2 - 2$.
10. Evaluate the definite integral $$\int_{1}^{e^2} \frac{1}{x\sqrt{\ln x}} \,dx$$
    Answer: $2\sqrt{2}$ Let $u = \ln x$. Then $du = \frac{1}{x} \,dx$. We must change the limits of integration. When $x = 1$, $u = \ln(1) = 0$. When $x = e^2$, $u = \ln(e^2) = 2$. The integral becomes: $$\int_{0}^{2} \frac{1}{\sqrt{u}} \,du = \int_{0}^{2} u^{-1/2} \,du$$ Evaluating this integral: $$[2u^{1/2}]_0^2 = [2\sqrt{u}]_0^2 = 2\sqrt{2} - 2\sqrt{0} = 2\sqrt{2}$$ A: This answer would result if the upper limit was $e^{\sqrt{2}}$ instead of $e^2$, leading to $\int_0^{\sqrt{2}} u^{-1/2} du$. B: This answer may result from forgetting the coefficient 2 from the power rule for integration, yielding $[\sqrt{u}]_0^2$. D: This answer would be correct if the upper limit of integration was $e$ instead of $e^2$, which would make the transformed upper limit $u=1$.
11. Evaluate the definite integral $\int_{0}^{\ln(\sqrt{3})} \frac{e^x}{1+e^{2x}} \,dx$
    Answer: $\frac{\pi}{12}$ Use the substitution $u = e^x$. Then $du = e^x \,dx$. We change the limits of integration: when $x=0$, $u=e^0=1$. When $x=\ln(\sqrt{3})$, $u=e^{\ln(\sqrt{3})}=\sqrt{3}$. The integral becomes: $\int_{1}^{\sqrt{3}} \frac{1}{1+u^2} \,du = [\arctan(u)]_{1}^{\sqrt{3}} = \arctan(\sqrt{3}) - \arctan(1) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$ B: This answer corresponds to $\arctan(\sqrt{3}) - \arctan(0)$, which would arise if the lower limit $u=1$ was incorrectly evaluated as $u=0$. C: This answer results from incorrectly integrating $1/(1+u^2)$ to $\ln(1+u^2)$, a common error confusing it with $u/(1+u^2)$. D: This answer corresponds to just $\arctan(\sqrt{3})$, forgetting to subtract the value at the lower limit.
12. If $\int_{0}^{k} \frac{x}{x^2+1} \,dx = \ln(3)$, what is the value of $k$? Assume $k>0$.
    Answer: $\sqrt{8}$ First, evaluate the integral. Let $u = x^2+1$, so $du = 2x \,dx$ or $x \,dx = \frac{1}{2}du$. Change the limits: when $x=0, u=1$; when $x=k, u=k^2+1$. $$\int_{1}^{k^2+1} \frac{1}{u} \cdot \frac{1}{2}du = \frac{1}{2} [\ln|u|]_1^{k^2+1} = \frac{1}{2}(\ln(k^2+1) - \ln(1)) = \frac{1}{2}\ln(k^2+1)$$ We are given this equals $\ln(3)$. $$\frac{1}{2}\ln(k^2+1) = \ln(3) \implies \ln(k^2+1) = 2\ln(3) \implies \ln(k^2+1) = \ln(3^2) = \ln(9)$$ Therefore, $k^2+1 = 9$, which means $k^2 = 8$. Since $k>0$, $k = \sqrt{8} = 2\sqrt{2}$. A: This answer results from setting $k^2=8$ but then forgetting the square root, or from setting $k+1=9$. B: This answer incorrectly equates $\frac{1}{2}\ln(k^2+1)$ with $\ln(3)$ by simply setting $k^2+1=3$ and ignoring the $1/2$ coefficient and solving $k^2=2$. D: This answer might arise from incorrectly solving the equation $k^2+1=3$.
13. Evaluate the definite integral $\int_0^1 x^3 \sqrt{1-x^2} \, dx$.
    Answer: $\frac{2}{15}$ To evaluate this integral, use the substitution $u = 1-x^2$. This implies $du = -2x \, dx$, so $x \, dx = -\frac{1}{2} du$. We also need to express $x^2$ in terms of $u$, which is $x^2 = 1-u$. The limits of integration must be changed: when $x=0$, $u=1-0^2=1$, and when $x=1$, $u=1-1^2=0$. The integral can be rewritten as $\int_0^1 x^2 \sqrt{1-x^2} (x \, dx)$. Substituting gives: $\int_1^0 (1-u)\sqrt{u} \left(-\frac{1}{2} du\right) = \frac{1}{2} \int_0^1 (1-u)u^{1/2} du = \frac{1}{2} \int_0^1 (u^{1/2} - u^{3/2}) du$ Evaluating the integral: $\frac{1}{2} \left[ \frac{2}{3}u^{3/2} - \frac{2}{5}u^{5/2} \right]_0^1 = \frac{1}{2} \left( (\frac{2}{3} - \frac{2}{5}) - 0 \right) = \frac{1}{2} \left( \frac{10-6}{15} \right) = \frac{1}{2} \left( \frac{4}{15} \right) = \frac{2}{15}$
14. The substitution $u = 3x+1$ is used to find $\int \frac{x}{(3x+1)^3} \,dx$. Which of the following is the correct antiderivative?
    Answer: $-\frac{1}{9(3x+1)} + \frac{1}{18(3x+1)^2} + C$ When you encounter an integral with a polynomial in the numerator and a more complex expression in the denominator, substitution often helps simplify the problem. The key insight here is that after substitution, you'll need to express everything in terms of the new variable. Let's work through this substitution systematically. With $u = 3x + 1$, we have $du = 3dx$, so $dx = \frac{du}{3}$. Also, $x = \frac{u-1}{3}$. Substituting into our integral: $\int \frac{x}{(3x+1)^3} dx = \int \frac{\frac{u-1}{3}}{u^3} \cdot \frac{du}{3} = \frac{1}{9} \int \frac{u-1}{u^3} du$ This splits into: $\frac{1}{9} \int (u^{-2} - u^{-3}) du = \frac{1}{9}(-u^{-1} + \frac{u^{-2}}{2}) + C$ Converting back to $x$: $-\frac{1}{9(3x+1)} + \frac{1}{18(3x+1)^2} + C$ Choice A incorrectly applies integration rules without proper substitution. Choice B has the wrong coefficients—the denominators don't match our calculation. Choice C mistakenly introduces a logarithm, which would only appear if we had $u^{-1}$ in our integral, but we have $u^{-2}$ and $u^{-3}$ terms. Choice D matches our result exactly. Study tip: When using substitution with rational functions, always split complex fractions into simpler terms before integrating. Double-check your coefficients by differentiating your final answer—it should give you the original integrand.
15. Evaluate $$\int \sin(2x)\cos^2(2x) \,dx$$
    Answer: $-\frac{1}{6}\cos^3(2x) + C$ Let $u = \cos(2x)$. Then $du = -\sin(2x) \cdot 2 \,dx = -2\sin(2x) \,dx$. Therefore, $\sin(2x) \,dx = -\frac{1}{2} \,du$. Substituting into the integral: $$\int u^2 \left(-\frac{1}{2} \,du\right) = -\frac{1}{2} \int u^2 \,du$$ Integrating with respect to $u$: $$-\frac{1}{2} \left(\frac{u^3}{3}\right) + C = -\frac{u^3}{6} + C$$ Substituting back $u = \cos(2x)$: $$-\frac{1}{6}\cos^3(2x) + C$$ B: This answer has a sign error, likely from forgetting the negative sign in the derivative of cosine. C: This answer omits the factor of $\frac{1}{2}$ that comes from the chain rule in the differential (i.e., from the '2' in $\cos(2x)$). D: This answer results from incorrectly integrating a product of functions by integrating each factor separately.
16. What is the value of the definite integral $$\int_{-2}^{2} \frac{x^3}{x^4+x^2+1} \,dx$$ ?
    Answer: 0 The integral can be evaluated by observing the properties of the integrand. Let $f(x) = \frac{x^3}{x^4+x^2+1}$. Let's check if $f(x)$ is an odd function, i.e., if $f(-x) = -f(x)$. $$f(-x) = \frac{(-x)^3}{(-x)^4+(-x)^2+1} = \frac{-x^3}{x^4+x^2+1} = -f(x)$$ Since the integrand is an odd function and the interval of integration $[-2, 2]$ is symmetric about $x=0$, the value of the definite integral is 0. A student could also attempt a u-substitution with $u=x^2$, but recognizing the symmetry is the most direct method. B: This expression would be correct if the integrand were an even function, not an odd function. C: This distractor might arise from an attempt to integrate using a substitution like $u=x^4+x^2+1$, which is not effective and leads to an incorrect form. D: This is a generic incorrect value that a student might guess if unable to solve the problem.
17. Evaluate the indefinite integral $\int \frac{\sin(\ln x)}{x \cos^2(\ln x)} dx$.
    Answer: $\sec(\ln x) + C$ When you encounter an integral with $\ln x$ inside trigonometric functions, substitution is your key strategy. The presence of $\frac{1}{x}$ in the integrand is a strong hint that $u = \ln x$ will work perfectly. Let $u = \ln x$, so $du = \frac{1}{x}dx$. This transforms our integral into: $\int \frac{\sin u}{\cos^2 u} du$ Now you can recognize this as $\int \sin u \cdot \sec^2 u \, du$. To evaluate this, use integration by parts or recognize that this equals $\int \tan u \sec u \, du$, which has the antiderivative $\sec u + C$. Substituting back: $\sec(\ln x) + C$, which is answer D. Let's examine why the other options are wrong. Answer A, $\ln|\cos(\ln x)| + C$, would result from integrating $\int \frac{\sin(\ln x)}{x\cos(\ln x)}dx$ (notice the missing squared term in the denominator). Answer B, $\tan(\ln x) + C$, comes from incorrectly thinking that $\int \frac{\sin u}{\cos^2 u}du = \tan u$, but the derivative of $\tan u$ is $\sec^2 u$, not $\frac{\sin u}{\cos^2 u}$. Answer C has the wrong sign—you might get this if you made an error with the integration by parts or misremembered the antiderivative. Remember: when you see $\ln x$ in the integrand along with $\frac{1}{x}$, immediately consider $u = \ln x$ substitution. This pattern appears frequently and dramatically simplifies these problems.
18. To evaluate $\int \frac{1}{e^x + e^{-x}} \,dx$, a common first step is to multiply the numerator and denominator by $e^x$. After this step and a subsequent u-substitution, what is the resulting antiderivative?
    Answer: $\arctan(e^x) + C$ First, multiply the numerator and denominator by $e^x$: $$\int \frac{1}{e^x + e^{-x}} \cdot \frac{e^x}{e^x} \,dx = \int \frac{e^x}{e^{2x} + 1} \,dx$$ Now, perform a u-substitution. Let $u = e^x$. Then $du = e^x \,dx$. The integral becomes: $$\int \frac{1}{u^2 + 1} \,du$$ This is the standard integral for the inverse tangent function: $$\arctan(u) + C$$ Substituting back $u = e^x$ gives the final answer: $\arctan(e^x) + C$. This function is also $\tanh(x)$, whose integral is not a standard Calc 1 result, but the method described leads to the correct answer. A: This would be the result if the numerator were $e^x - e^{-x}$. C: This comes from the integral of $\cosh^2(x)$, suggesting a misremembered hyperbolic identity. D: This is a plausible but incorrect result from misapplying the power rule.
19. Evaluate the definite integral $\int_0^1 \frac{e^{\arctan(x)}}{1+x^2} dx$.
    Answer: $e^{\pi/4} - 1$ When you encounter a definite integral with a composite function in the numerator and an expression related to that function's derivative in the denominator, think substitution. This is a classic u-substitution setup. Let's use the substitution $u = \arctan(x)$. Then $du = \frac{1}{1+x^2} dx$, which matches exactly what we have in the denominator. When $x = 0$, $u = \arctan(0) = 0$. When $x = 1$, $u = \arctan(1) = \frac{\pi}{4}$. The integral transforms to: $\int_0^{\pi/4} e^u \, du$ This evaluates to $e^u \Big|_0^{\pi/4} = e^{\pi/4} - e^0 = e^{\pi/4} - 1$, confirming answer C. Looking at the wrong answers: Answer A ($e - 1$) would result if you mistakenly thought $\arctan(1) = 1$ instead of $\frac{\pi}{4}$. Answer B ($e^{\pi/4}$) represents forgetting to subtract the lower bound when evaluating the definite integral - you'd get this if you only computed $e^{\pi/4}$ and stopped there. Answer D ($e$) might come from incorrectly assuming the upper limit of integration gives $e^1 = e$. Study tip: When you see $\frac{1}{1+x^2}$ in an integral, immediately think of inverse trig functions since this is the derivative of $\arctan(x)$. Master the derivatives of inverse trig functions - they're key for recognizing u-substitution opportunities in integration problems.
20. Evaluate the definite integral $$\int_{0}^{\pi/2} \sin(x) e^{\cos(x)} \,dx$$
    Answer: e - 1 To evaluate this integral, we use u-substitution. Let $u = \cos(x)$. Then $du = -\sin(x) \,dx$, which means $-du = \sin(x) \,dx$. We must also change the limits of integration. When $x = 0$, $u = \cos(0) = 1$. When $x = \pi/2$, $u = \cos(\pi/2) = 0$. Substituting these into the integral gives: $$\int_{1}^{0} e^u (-du) = -\int_{1}^{0} e^u \,du = \int_{0}^{1} e^u \,du$$ Evaluating this integral gives $[e^u]_0^1 = e^1 - e^0 = e - 1$. A: This answer results from a sign error, likely from forgetting the negative sign in $du = -\sin(x) dx$ and evaluating $-\int_0^1 e^u \,du$. C: This answer results from incorrectly evaluating $e^0$ as 0 instead of 1. D: This answer results from incorrectly evaluating the bounds, perhaps calculating $e^1 - e^1$ or $e^0 - e^0$.

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