📝 Integration by Substitution u-Substitution (31 MCQs)

📖 Explanation: This is a direct Easy of the u-substitution method. Recognizing that the derivative of $x^2$ is $2x$, which is present in the integrand, is the key. Setting $u = x^2$ simplifies the integral to a basic cosine integral. Other options incorrectly identify $u$ or the resulting integrand.

📖 Explanation: The core of u-substitution is reversing the chain rule. This requires identifying a composite function $f(g(x))$ and recognizing that the derivative g'(x) (or a constant multiple of it) is also present in the integrand. This allows the integral to be transformed into a simpler form. The other options describe special cases or are incorrect generalizations.

📖 Explanation: The student correctly performed the u-substitution: $u = x^2, du = 2x dx$, so the integral becomes $\frac{1}{2} \int \sin(u) du$. The error is in the final integration. The antiderivative of $\sin(u)$ is $-\cos(u)$, not $\cos(u)$. Therefore, the correct final answer is $-\frac{1}{2} \cos(x^2) + C$. This question tests the ability to identify common sign errors in integration.

📖 Explanation: This requires applying u-substitution to a definite integral. Let $u = x^2$, so $du = 2x dx$. The limits change from $x=0$ to $u=0$, and from $x=\sqrt{\pi}$ to $u=\pi$. The integral becomes $\int_{0}^{\pi} \cos(u) \, du = \sin(u) \big|_{0}^{\pi} = \sin(\pi) - \sin(0) = 0$. However, the correct answer is 0, but option C is '1'. Let's re-evaluate. $\sin(\pi)=0$ and $\sin(0)=0$, so the integral is 0. It seems there is a mistake in the options. The correct answer should be 0. Let's adjust the options. Assuming the question is $\int_{0}^{\sqrt{\pi/2}} 2x \cos(x^2) \, dx$, then upper limit would be $u = \pi/2$, and the integral would be $\sin(\pi/2) - 0 = 1$. Let's use that revised question to make a good HOTS question. The question has been revised to $\int_{0}^{\sqrt{\pi/2}} 2x \cos(x^2) \, dx$.

📖 Explanation: Applying u-substitution, we let $u = x^2$, so $du = 2x dx$. The new limits of integration are $u(0) = 0$ and $u(\sqrt{\pi/2}) = \pi/2$. The integral becomes $\int_{0}^{\pi/2} \cos(u) du = \sin(u) \big|_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$. This demonstrates the importance of correctly changing the limits of integration when evaluating a definite integral using substitution.

📖 Explanation: For $u = \ln x$, we have $du = \frac{1}{x} dx$. The integral in option B, $\int \frac{\ln x}{x} dx$, can be directly rewritten as $\int u \, du$, which is straightforward. Option A suggests $u = e^x$, Option C suggests $u = x^2$, and Option D suggests a trigonometric substitution. Recognizing which substitution simplifies the integral is a key conceptual skill.

📖 Explanation: Let $u = \sqrt{x}$. Then $du = \frac{1}{2\sqrt{x}} dx$, which means $\frac{dx}{\sqrt{x}} = 2 du$. The integral becomes $\int \sin(u) \cdot 2 du = \int 2 \sin(u) du$. This is a classic example where the substitution simplifies the integrand by eliminating the composite function and its derivative. The other options would lead to more complex or incorrect integrals.

📖 Explanation: This question checks if a student can differentiate a composite function correctly. If $u = e^{x^3}$, then by the chain rule, $\frac{du}{dx} = e^{x^3} \cdot 3x^2$. Therefore, $du = 3x^2 e^{x^3} dx$. The other options are incorrect derivatives. This highlights the importance of accurately computing $du$ in the u-substitution process.

📖 Explanation: The presence of $x^2$ in $f(x) = x^2 \cdot h(x^3)$ strongly suggests the substitution $u = x^3$, because its derivative, $3x^2$, is present (up to a constant). The integral becomes $\frac{1}{3} \int h(u) du$. This is a conceptual question that requires understanding the structure of an integrand to choose an appropriate substitution, even when the exact form of $h$ is unknown. The graph is a distractor to see if students focus on the symbolic form.

📖 Explanation: The integral $\int \tan(x) dx = \int \frac{\sin(x)}{\cos(x)} dx$. Letting $u = \cos(x)$, we get $du = -\sin(x) dx$. The integral becomes $-\int \frac{1}{u} du = -\ln|u| + C = -\ln|\cos(x)| + C = \ln|\sec(x)| + C$. This is a standard but essential Easy of u-substitution. Option A leads to a dead end, and Option C requires a more complex approach.

📖 Explanation: Let $u = e^x - e^{-x}$. Then $du = (e^x + e^{-x}) dx$. The numerator is exactly the derivative of the denominator, making the integral $\int \frac{1}{u} du = \ln|u| + C$. This is a prime example of recognizing that the integrand is of the form f'(x)/f(x). The other options do not simplify the integrand as effectively.

📖 Explanation: When evaluating a definite integral using substitution, the limits of integration are transformed along with the integrand. The final result is a number, and it is the same whether you compute it by re-substituting and evaluating at the original limits, or by evaluating the transformed integral at the new limits. This is a crucial concept that distinguishes definite from indefinite integrals. Option A is false, Option C is irrelevant, and Option D is incorrect as it describes the procedure for indefinite integrals.

📖 Explanation: This is a direct Easy of the Fundamental Theorem of Calculus (Part 2) combined with understanding the integrand. The derivative of a definite integral from a constant to $x$ is the integrand evaluated at $x$. Here, the integrand is $t \cos(t^2)$. Substituting $t = x$ gives $x \cos(x^2)$. This is not a pure u-substitution question, but it tests the ability to recognize that the integrand is a derivative of a composite function.

📖 Explanation: This is a more complex Easy. The denominator $x^2 + 4x + 5$ can be completed to $(x+2)^2 + 1$. The substitution $u = x+2$ simplifies the integral to $\int \frac{1}{u^2+1} du = \tan^{-1}(u) + C$. This question tests the ability to combine algebraic manipulation (completing the square) with u-substitution. The other options do not simplify the integral into a recognizable form.

📖 Explanation: This is a classic error. If a student uses u-substitution but does not change the limits of integration, they will evaluate the antiderivative in terms of $u$ at the original $x$ limits, which is incorrect. Let's check: $u = 1 - x^2$, when $x=0$, $u=1$, and when $x=1$, $u=0$. The correct integral is $\int_{1}^{0} -\frac{1}{2} \sqrt{u} du = \frac{1}{2} \int_{0}^{1} u^{1/2} du = \frac{1}{3}$. If the limits are not changed, evaluating at $x=0$ and $x=1$ gives $-\frac{1}{3}(1-1)^{3/2} + \frac{1}{3}(1-0)^{3/2} = 1/3$, which is actually the correct answer. The student's answer of 0 suggests a different error, like mis-evaluating the antiderivative. Let's refine the question. A student correctly finds the antiderivative to be $-\frac{1}{3}(1-x^2)^{3/2}$ and evaluates it from 0 to 1, getting $-\frac{1}{3}(0)^{3/2} - (-\frac{1}{3}(1)^{3/2}) = 1/3$. If they got 0, they might have evaluated $1-x^2$ at both limits as 0, which means they didn't understand the substitution. Let's reword to focus on the limits. The student's mistake is evaluating the antiderivative at the original $x$ limits after substituting, but then incorrectly substituting the $x$ values back into the $u$-expression. The core error is not correctly applying the changed limits.

📖 Explanation: This question tests the deep understanding of why u-substitution works. The factor $\frac{1}{x}$ is not just any term; it is the exact derivative of $\ln(x)$. This allows us to set $u = \ln(x)$, $du = \frac{1}{x} dx$, and the integral transforms to $\int \frac{1}{u} du$. Recognizing this pattern is the central insight of the method. The other options demonstrate a superficial understanding of the integrand's structure.

📖 Explanation: This combines u-substitution with solving an initial value problem. $\int \frac{x}{\sqrt{1+x^2}} dx$ requires $u = 1+x^2$, $du = 2x dx$, resulting in $\frac{1}{2} \int u^{-1/2} du = \sqrt{1+x^2} + C$. With $y(0)=2$, $2 = \sqrt{1} + C$, so $C=1$. Thus $y(x) = \sqrt{1+x^2} + 1$. Then $y(1) = \sqrt{2} + 1$. There's a calculation error in the options. Let's set $y(1) = \sqrt{2} + 1$. The question now asks for a value that matches this result. The correct answer is $\sqrt{2} + 1$.

📖 Explanation: The u-substitution method is fundamentally the reverse chain rule. It requires the derivative of the inner function (or a constant multiple of it) to be present in the integrand. If this derivative is not present, a simple u-substitution will not work. The other options are false: u-substitution works for trigonometric and exponential functions, and it can evaluate definite integrals (with limit change). This is a conceptual check on the boundaries of the method's applicability.

📖 Explanation: This is a straightforward Easy. Let $u = \sin(x)$, then $du = \cos(x) dx$. The integral becomes $\int u^3 du = \frac{u^4}{4} + C = \frac{\sin^4(x)}{4} + C$. The other options are common mistakes, such as mis-identifying the derivative or integrating incorrectly.

📖 Explanation: The two answers are indeed equivalent. Since $\sin^2(x) + \cos^2(x) = 1$, we have $\frac{1}{2} \sin^2(x) = \frac{1}{2}(1 - \cos^2(x)) = -\frac{1}{2} \cos^2(x) + \frac{1}{2}$. The difference between the two answers is the constant $\frac{1}{2}$, which is absorbed into the constant of integration $C$. This question tests the understanding that antiderivatives are unique only up to an additive constant.

📖 Explanation: This is a direct Easy of the chain rule and the Fundamental Theorem of Calculus (Part 2). The derivative is $\frac{1}{\sqrt{1+(x^2)^2}} \cdot \frac{d}{dx}(x^2) = \frac{2x}{\sqrt{1+x^4}}$. This question is Easy because it combines the derivative of the upper limit with the substitution inside the integrand. The other options represent common mistakes, such as forgetting the chain rule or incorrectly simplifying the integrand.

📖 Explanation: Let $u = e^x + 1$, then $du = e^x dx$. The new limits: when $x=0$, $u=2$; when $x=\ln 2$, $u=3$. The integral becomes $\int_{2}^{3} \frac{1}{u} du = \ln|u| \big|_{2}^{3} = \ln(3) - \ln(2) = \ln(3/2)$. This is a straightforward Easy of u-substitution on a definite integral with exponential functions. It tests the ability to change limits and evaluate a logarithmic result.

📖 Explanation: The statement $\int 2x \cos(x^2) dx = \int \cos(u) du$ is actually correct if $u = x^2$. Let's check: $u = x^2$, $du = 2x dx$. The integral becomes $\int \cos(u) du$. So the student has not made a mistake in this line. The question is flawed. Let's create a better Medium question. A student writes $\int x \cos(x^2) dx = \int \cos(u) du$ with $u = x^2$. What is wrong? The student forgot the factor of $1/2$ from $du = 2x dx$. The correct integral should be $\frac{1}{2} \int \cos(u) du$. So the correct answer is that they forgot to account for the constant factor resulting from the substitution.

📖 Explanation: Let $u = x^2+1$, so $du = 2x dx$. The integral becomes $\int \frac{4x}{x^2+1} dx = \int \frac{2}{u} du = 2 \ln|u| + C = 2 \ln(x^2+1) + C$. The absence of absolute value is because $x^2+1$ is always positive. This tests the ability to handle a constant multiple in the substitution.

📖 Explanation: This is a Easy question. Option A is a good start: $u = \ln x$, then $du = \frac{1}{x} dx$, giving $\int \sqrt{1+u} du$. From there, another substitution $w = 1+u$ is needed. Option B combines these steps: $u = 1 + \ln x$, so $du = \frac{1}{x} dx$, giving $\int \sqrt{u} du$. Option C is also valid: $u = \sqrt{1+\ln x}$, then $du = \frac{1}{2\sqrt{1+\ln x}} \cdot \frac{1}{x} dx$, which also simplifies the integral. Therefore, both B and C are valid. This question tests the ability to see that multiple substitutions can lead to the same result.

📖 Explanation: By the chain rule, \frac{d}{dx} [F(g(x))] = F'(g(x)) g'(x) = f(g(x)) g'(x). Therefore, $F(g(x))$ is the antiderivative. This question directly tests the theoretical underpinnings of u-substitution, which is the most important concept. The other options represent a misunderstanding of how the chain rule operates in reverse.

📖 Explanation: This is a classic integral. With $u = \ln(x)$, $du = \frac{1}{x} dx$, the integral becomes $\int \frac{1}{u} du = \ln|u| + C = \ln|\ln(x)| + C$. This is a fundamental example of the \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C pattern. The other options are incorrect Easys of the power rule or other integration rules.

📖 Explanation: The integral in option B involves a quadratic in the denominator. Completing the square gives $x^2+2x+5 = (x+1)^2+4$. Then a substitution like $u = x+1$ (or a trigonometric substitution) can be used. The other options are standard u-substitution problems: A uses $u = x$, C uses $u = \sin(x)$, and D uses $u = e^x$. This question tests the skill of algebraically manipulating the integrand before applying u-substitution.

📖 Explanation: For u-substitution to work, the derivative of the substituted variable must be present in the integrand. In $\int \sin(x^2) dx$, if we let $u = x^2$, then $du = 2x dx$. The $2x$ term is not present, and we cannot create it without complicating the integral (e.g., $x = \sqrt{u}$). This question tests the ability to analyze why a substitution fails, which is an important troubleshooting skill.

📖 Explanation: Let $u = \sin(x)$, so $du = \cos(x) dx$. The new limits are $u(0)=0$ and $u(\pi/2)=1$. The integral becomes $\int_{0}^{1} u^3 du = \left[ \frac{u^4}{4} \right]_{0}^{1} = \frac{1}{4}$. This tests the ability to apply u-substitution to a definite integral with trigonometric functions and correctly change the limits. It's a direct Easy, but a common source of errors.

📖 Explanation: This is an advanced question testing the ability to manipulate trigonometric expressions. The term $\cos^2(2x)$ can be expressed using $\sin^2(2x)$. However, the key is to see if $\frac{1}{4} \cos^2(2x)$ can be absorbed into the constant of integration. Since $\cos^2(2x)$ is not constant, the two expressions are generally not equivalent. The only way for the expressions to be equivalent is if $\frac{1}{4} \cos^2(2x)$ were constant, which it is not. Therefore, the CAS answer is not equivalent to $\frac{1}{2} \sin^2(x) + C$. The correct answer is D.