📝 Special substitutions for integrals (40 MCQs)

📖 Explanation: The integral contains $x^{1/2}$ and $x^{1/3}$. The substitution $u = x^{1/n}$ works best when $n$ is the least common multiple (LCM) of the denominators of the fractional exponents. Here, the denominators are 2 and 3, whose LCM is 6, so $u = x^{1/6}$ is the correct choice to transform the integrand into a rational function.

📖 Explanation: Special substitutions of the form $u = x^{1/n}$ are primarily used to convert integrands with fractional powers into rational functions. This transformation is valuable because rational functions are generally easier to integrate using methods like partial fractions or simple polynomial division. While changing the limits and simplifying $dx$ are part of the process, the core goal is algebraic simplification into a rational form.

📖 Explanation: With $u = x^{1/6}$, we have $x = u^6$ and $dx = 6u^5 du$. The numerator $\sqrt{x} = (u^6)^{1/2} = u^3$ and $\sqrt[3]{x} = (u^6)^{1/3} = u^2$. The integral becomes $\int \frac{u^3}{1+u^2} \cdot 6u^5 du = 6 \int \frac{u^8}{1+u^2} du$. The correct integrand (before the constant 6) is $\frac{u^8}{1+u^2}$.

📖 Explanation: The Weierstrass substitution $u = \tan(x/2)$ is a standard method for converting trigonometric integrals into rational functions. The key identities are $\sin x = \frac{2u}{1+u^2}$ and $\cos x = \frac{1-u^2}{1+u^2}$, along with $dx = \frac{2}{1+u^2} du$. Option A correctly provides these identities, which are essential for transforming the integral into a rational form in $u$.

📖 Explanation: After integration, the result is in terms of $u = \tan(x/2)$. To express it in terms of $x$, we must perform the back-substitution. Option C demonstrates a common algebraic simplification: $1 - \tan(x/2) = \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2)}$. This is necessary to present the final answer in a standard, simplified form, often leading to logarithmic or arctangent expressions.

📖 Explanation: To eliminate fractional powers, we substitute $u = x^{1/n}$, where $n$ is the least common multiple (LCM) of the denominators of the fractional exponents. The denominators here are 4 and 6. The LCM of 4 and 6 is 12, so $u = x^{1/12}$ is the correct substitution. This ensures that $x^{1/4} = u^3$ and $x^{1/6} = u^2$, which are integer powers of $u$.

📖 Explanation: The primary advantage of the Weierstrass substitution is that it provides a universal, systematic method for integrating any rational function of $\sin x$ and $\cos x$. By converting the trigonometric functions into rational expressions in $u$, it guarantees that the resulting integral can be approached using standard methods for rational functions, such as partial fractions. While it may not always be the most efficient method, its universality is its main strength.

📖 Explanation: Using the Weierstrass substitution, $\cos x = \frac{1-u^2}{1+u^2}$ and $dx = \frac{2 du}{1+u^2}$. The integral becomes $\int \frac{1}{1 + \frac{1-u^2}{1+u^2}} \cdot \frac{2}{1+u^2} du = \int \frac{2}{2} du = \int du = u + C = \tan(x/2) + C$. Wait, this simplifies to $\int du$, giving $u + C = \tan(x/2) + C$. Let's re-evaluate: $1+\cos x = \frac{2}{1+u^2}$, so the integral is $\int \frac{1+u^2}{2} \cdot \frac{2}{1+u^2} du = \int du = u + C$. The derivative of $-\cot(x/2)$ is $\frac{1}{2}\csc^2(x/2)$, not the integrand. The correct answer is $\tan(x/2)+C$.

📖 Explanation: A common error when using any substitution, including $u = \tan(x/2)$, is to forget to change the limits of integration when evaluating a definite integral. The expressions for $\sin x$, $\cos x$, and $dx$ are correctly given in options B, C, and D. The limits of integration must be converted from $x$-values to $u$-values using $u = \tan(x/2)$. Failing to do so will lead to an incorrect numerical result.

📖 Explanation: The first step is to let $u = x-2$, which gives $x = u+2$ and $dx = du$. The integral becomes $\int (u+2)\sqrt{u} du = \int (u+2)u^{1/2} du$. This integral is now a simple polynomial in $u$, which can be evaluated directly by expanding: $\int (u^{3/2} + 2u^{1/2}) du$. This is a classic example of a substitution that simplifies the algebraic structure.

📖 Explanation: Let $u = x^{1/6}$, so $x = u^6$, $dx = 6u^5 du$. The integral becomes $\int \frac{6u^5}{u^3 - u^2} du = 6\int \frac{u^3}{u-1} du$. Polynomial long division gives $\frac{u^3}{u-1} = u^2 + u + 1 + \frac{1}{u-1}$. Integrating gives $6(\frac{u^3}{3} + \frac{u^2}{2} + u + \ln|u-1|) + C = 2u^3 + 3u^2 + 6u + 6\ln|u-1| + C$. However, $2u^3+3u^2+6u = 6u^5 - 6u^4 + 6u^3 - 6u^2 + 6u$ when expanded? Let's re-evaluate the long division and the multiplication. $u^3/(u-1) = u^2 + u + 1 + 1/(u-1)$. Integrating gives $u^3/3 + u^2/2 + u + \ln|u-1| + C$. Multiplying by 6 gives $2u^3 + 3u^2 + 6u + 6\ln|u-1| + C$. The expression $2u^3+3u^2+6u$ is not equal to $6u^5 - 6u^4 + 6u^3 - 6u^2 + 6u$ without the constants. Let's check the polynomial division again. $\frac{u^3}{u-1} = u^2 + u + 1 + \frac{1}{u-1}$. The integral is $6 \int (u^2 + u + 1 + \frac{1}{u-1}) du = 2u^3 + 3u^2 + 6u + 6\ln|u-1| + C$. This simplifies to $2x^{1/2} + 3x^{1/3} + 6x^{1/6} + 6\ln|x^{1/6}-1| + C$. None of the options match this exactly, but option B has a $-\ln|1-u|$ which is equivalent to $\ln|u-1|$ if we adjust the sign? Actually, $\ln|u-1| = \ln|1-u|$. So $6\ln|u-1| = 6\ln|1-u|$, but the answer is $2u^3 + 3u^2 + 6u + 6\ln|u-1|$. Since $\ln|u-1| = \ln|1-u|$, the correct choice is the one with the polynomial $2u^3+3u^2+6u$. No option matches. Let's recalculate the polynomial division: $\frac{u^3}{u-1}$ gives quotient $u^2+u+1$ and remainder 1. Integrating $\int \frac{6u^5}{u^2(u-1)} du = \int \frac{6u^3}{u-1} du = 6\int (u^2+u+1 + \frac{1}{u-1}) du = 2u^3+3u^2+6u+6\ln|u-1|+C$. Substituting $u=x^{1/6}$ gives $2x^{1/2} + 3x^{1/3} + 6x^{1/6} + 6\ln|x^{1/6}-1| + C$. Option A is $6u^5 - 6u^4 + 6u^3 - 6u^2 + 6u - 6\ln|1+u| + C$. Option B is $6u^5 - 6u^4 + 6u^3 - 6u^2 + 6u - 6\ln|1-u| + C$. None match. The correct answer is not among the options. Let's re-evaluate the problem setup. $\int \frac{dx}{x^{1/2} - x^{1/3}} = \int \frac{6u^5}{u^3 - u^2} du = 6\int \frac{u^3}{u-1} du$. The division of $u^3$ by $u-1$ is $u^2+u+1$ with remainder 1. Integral is $6(u^3/3 + u^2/2 + u + \ln|u-1|) = 2u^3+3u^2+6u+6\ln|u-1|$. Wait, option A has $6u^5 - 6u^4 + 6u^3 - 6u^2 + 6u$. This is the expansion of $6\frac{u^3}{u-1}$? No. Let's check: $6u^5 - 6u^4 + 6u^3 - 6u^2 + 6u = 6u(u^4 - u^3 + u^2 - u + 1)$. This is not the integral. The correct answer should be $2u^3+3u^2+6u+6\ln|u-1|+C$. Since none match, the question is flawed. The intended correct option might be a misprint. The closest is option B but with the wrong polynomial. Let's correct the problem and say the correct transformation leads to $2x^{1/2} + 3x^{1/3} + 6x^{1/6} + 6\ln|x^{1/6}-1| + C$.

📖 Explanation: Using the Weierstrass substitution, $\sin x = \frac{2u}{1+u^2}$ and $dx = \frac{2 du}{1+u^2}$. The integral becomes $\int \frac{1}{1 + \frac{2u}{1+u^2}} \cdot \frac{2}{1+u^2} du = \int \frac{2}{(1+u)^2} du = -\frac{2}{1+u} + C$. Substituting $u = \tan(x/2)$ gives $-\frac{2}{1+\tan(x/2)} + C$.

📖 Explanation: The Weierstrass substitution is a general method for rational functions of $\sin x$ and $\cos x$. However, it is often inefficient for integrals of the form $\int \sin^m x \cos^n x dx$ where $m$ and $n$ are integers, especially if one of them is odd. For such integrals, a simpler $u$-substitution (like $u = \sin x$ or $u = \cos x$) is more appropriate. Option C is a classic example where a direct substitution is far simpler than the Weierstrass approach.

📖 Explanation: The integral $\int \frac{dx}{x\sqrt{x^2-1}}$ is a classic example for trigonometric substitution $x = \sec \theta$, which simplifies the radical to $\tan \theta$. The substitution $u = x^2 - 1$ gives $du = 2x dx$, which does not match the $x$ in the denominator $x\sqrt{x^2-1}$. The substitution $u = \sqrt{x^2-1}$ would work, but the student chose a poor special substitution, highlighting the importance of selecting a method that aligns with the algebraic structure of the integrand.

📖 Explanation: Completing the square: $x^2 + 4x + 5 = (x+2)^2 + 1$. The substitution $u = x+2$ simplifies the expression to $\sqrt{u^2+1}$, which is a standard form for a trigonometric substitution (e.g., $u = \tan \theta$). This is a classic multi-step approach where completing the square is followed by a special substitution to rationalize or simplify the integrand.

📖 Explanation: As shown in a previous question, $1+\cos x = \frac{2}{1+u^2}$. The integral becomes $\int \frac{1+u^2}{2} \cdot \frac{2}{1+u^2} du = \int du = u + C$. Back-substituting $u = \tan(x/2)$ yields $\tan(x/2) + C$. This demonstrates the power of the Weierstrass substitution in simplifying trigonometric integrals into rational forms.

📖 Explanation: The fractional exponents are $3/4$ and $5/6$. The denominators are 4 and 6. The least common multiple (LCM) of 4 and 6 is 12. Therefore, a substitution like $u = x^{1/12}$ would rationalize both terms, as $x^{3/4} = u^{9}$ and $x^{5/6} = u^{10}$, which are integer powers of $u$. This is a fundamental step in applying the special substitution technique.

📖 Explanation: Let $u = x^{1/6}$. Then $x = u^6$, $dx = 6u^5 du$. The integral becomes $\int \frac{6u^5}{u^3 + u^2} du = 6\int \frac{u^3}{u+1} du$. Polynomial division: $u^3/(u+1) = u^2 - u + 1 - \frac{1}{u+1}$. The integral is $6(\frac{u^3}{3} - \frac{u^2}{2} + u - \ln|u+1|) + C = 2u^3 - 3u^2 + 6u - 6\ln|u+1| + C$. Substituting back: $2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6\ln|x^{1/6}+1| + C$. This matches option A if we reorder the terms.

📖 Explanation: The integral involves a sum of radicals $\sqrt{1+x} + \sqrt{1-x}$. While the Weierstrass substitution $u = \tan(x/2)$ is powerful for rational functions of $\sin$ and $\cos$, it is not relevant here because the integrand is algebraic, not trigonometric. Substitutions like $x = \sin \theta$ or $x = \cos \theta$ are appropriate because they directly eliminate the radicals, and $u = \sqrt{1+x}$ is also a valid approach to rationalize the expression.

📖 Explanation: The primary goal of the Weierstrass substitution is to convert a trigonometric integral into a rational function. The most powerful and systematic method for integrating rational functions is the method of partial fractions. Therefore, after applying $u = \tan(x/2)$, the next logical step is to decompose the resulting rational function using partial fractions, if necessary.

📖 Explanation: Trigonometric substitutions are a type of special substitution used to eliminate radicals by substituting trigonometric functions. The process of eliminating radicals from an integrand is known as rationalization, even though the result might not be a rational function. This is a core technique in integral calculus, distinct from algebraic manipulation or integration by parts.

📖 Explanation: The integral contains $x^{1/4}$. To eliminate the fractional exponent, we let $u = x^{1/4}$, which means $x = u^4$ and $dx = 4u^3 du$. The integrand becomes $\frac{4u^3}{u + u^4} = \frac{4u^2}{1+u^3}$, which is a rational function. This substitution directly removes the radical, showcasing the purpose of the special substitution technique.

📖 Explanation: The integrand $\frac{1}{\sqrt{x} + \sqrt[4]{x}}$ is an algebraic function, not a trigonometric one. The Weierstrass substitution $u = \tan(x/2)$ is specifically designed for rational functions of $\sin x$ and $\cos x$. Applying it to an algebraic integrand is a category error. The correct approach is to use a special substitution like $u = x^{1/4}$ to rationalize the expression.

📖 Explanation: The identity $\cosh^2 u - \sinh^2 u = 1$ means that $\sqrt{1+\sinh^2 u} = \cosh u$. This directly eliminates the radical, leading to $\int du = u + C = \sinh^{-1} x + C$. While a trigonometric substitution also works, the hyperbolic substitution uses the fundamental property of hyperbolic functions to simplify the radical in a single step, making it a practical Easy of special substitutions.

📖 Explanation: For $\sqrt{x^2 - 4}$, the standard trigonometric substitution is $x = 2\sec \theta$. Then $dx = 2\sec \theta \tan \theta d\theta$, and $\sqrt{x^2 - 4} = \sqrt{4(\sec^2 \theta - 1)} = 2\tan \theta$. The integral becomes $\int \frac{2\sec \theta \tan \theta}{(2\sec \theta)(2\tan \theta)} d\theta = \int \frac{1}{2} d\theta = \frac{1}{2}\theta + C = \frac{1}{2}\sec^{-1}(x/2) + C$. The correct resulting integral is $\frac{1}{2}\int d\theta$, not $\sec \theta$.

📖 Explanation: In integration by parts, when calculating $v = \int dv$, we usually omit the constant of integration because it cancels out. However, sometimes choosing a specific constant $C_1$ can simplify the resulting integral $\int v du$. This is a nuanced technique related to integration by parts, not directly a special substitution. The connection is that both involve strategic choices to simplify the process of integration, highlighting the importance of algebraic manipulation and insight in calculus.

📖 Explanation: The integrand is algebraic, not trigonometric. The Weierstrass substitution $u = \tan(x/2)$ is for rational functions of $\sin$ and $\cos$, so it is inappropriate. Substitutions $u = x^{1/6}$, $u = \sqrt{x}$, or $u = x^{1/3}$ would only partially rationalize the integrand, but $u = x^{1/6}$ is the most effective because 6 is the LCM of 2 and 3. The question asks for which is NOT appropriate.

📖 Explanation: With $u = x^{1/6}$, we have $x = u^6$, $dx = 6u^5 du$. The denominator $x^{2/3} + x^{1/2} = (u^6)^{2/3} + (u^6)^{1/2} = u^4 + u^3$. The numerator in the integrand is $dx = 6u^5 du$. So the integrand becomes $\frac{6u^5}{u^4 + u^3} du$. Thus, the exponents in the numerator and denominator are 5, 4, and 3, respectively.

📖 Explanation: The first student's answer $-\tan(x/2) - 1 + C$ and the second's $-\frac{2}{1+\tan(x/2)} + C$ are equivalent. This can be shown using the identity $\tan(x/2) = \frac{1-\cos x}{\sin x}$ or by differentiating both results. The constants of integration absorb the difference. This demonstrates that antiderivatives can appear different but still be correct.

📖 Explanation: The denominator $u^2 + 2u + 2 = (u+1)^2 + 1$. This suggests a substitution $v = u+1$ and using the arctangent formula. The integral is $\int \frac{2}{(u+1)^2+1} du = 2\tan^{-1}(u+1) + C$. This is a case where a rational function integrates to an inverse trigonometric function, which is a common outcome of the Weierstrass substitution.

📖 Explanation: The integrand is $\frac{1}{\sin x + \tan x} = \frac{1}{\sin x + \frac{\sin x}{\cos x}} = \frac{\cos x}{\sin x(1+\cos x)}$. This is a rational function of $\sin x$ and $\cos x$. The Weierstrass substitution $u = \tan(x/2)$ is a universal method for such integrals, guaranteeing a rational function in $u$. While other substitutions might simplify the expression, this is the most systematic and reliable approach for a mixed-concept problem involving multiple trigonometric functions.

📖 Explanation: While the Weierstrass substitution is universal for rational functions of $\sin$ and $\cos$, it often results in complex rational functions with high-degree polynomials. This can make the subsequent integration (e.g., by partial fractions) very tedious and time-consuming. The limitation is not in the method's applicability but in its computational efficiency, as it may not be the most practical choice for hand computations.

📖 Explanation: The Weierstrass substitution is a powerful tool because it provides a general, foolproof algorithm for integrating any rational function of $\sin x$ and $\cos x$. By converting the integrand into a rational function of $u$, it opens the door to using the well-developed theory of partial fractions. While it may not always be the most efficient, its universality is its key strength.

📖 Explanation: A very common error in all substitution methods, including special substitutions, is forgetting to change the limits of integration when evaluating a definite integral. If $u = x^{1/n}$, then when $x$ goes from $a$ to $b$, $u$ goes from $a^{1/n}$ to $b^{1/n}$. Failing to make this adjustment invalidates the result.

📖 Explanation: The denominators of the fractional exponents are 2 and 3. The least common multiple of 2 and 3 is 6. Therefore, $u = x^{1/6}$ is the appropriate substitution. This will transform the integral into $\int \frac{6u^5}{u^3 + u^2} du$, a rational function.

📖 Explanation: Let $u = x^{1/4}$. Then $x = u^4$, $dx = 4u^3 du$. The integral becomes $\int \frac{4u^3}{u^3 + u^2} du = 4\int \frac{u}{u+1} du = 4\int (1 - \frac{1}{u+1}) du = 4u - 4\ln|u+1| + C = 4x^{1/4} - 4\ln|x^{1/4}+1| + C$. Wait, the integral is $\int \frac{dx}{x^{3/4}+x^{1/2}}$. The substitution is $u=x^{1/4}$, so $x^{3/4}=u^3$, $x^{1/2}=u^2$, $dx=4u^3 du$. The integral is $\int \frac{4u^3}{u^3+u^2} du = 4\int \frac{u}{u+1} du = 4\int (1 - \frac{1}{u+1}) du = 4u - 4\ln|u+1| + C$. The options are $4\ln|x^{1/4}+1| + C$ and $\ln|x^{1/4}+1| + C$. The correct answer includes the $4u$ term. Let's re-evaluate: $\int \frac{4u}{u+1} du = \int 4 - \frac{4}{u+1} du = 4u - 4\ln|u+1| + C$. So the answer is $4x^{1/4} - 4\ln|x^{1/4}+1| + C$. Since this is not an option, let's check the problem statement again. The integral is $\int \frac{dx}{x^{3/4}+x^{1/2}}$. If we factor $x^{1/2}$ from the denominator, we get $\int \frac{dx}{x^{1/2}(x^{1/4}+1)}$. This is not a simple log. Let's try $u=x^{1/4}$. Then the integral is $4\int \frac{u}{u+1} du$, which is $4u - 4\ln|u+1|+C$. This is not a simple logarithm, so options A and B are incomplete. The question may be flawed. The correct answer should be $4x^{1/4} - 4\ln|x^{1/4}+1| + C$.