📝 Trigonometric substitution integration (35 MCQs)

📖 Explanation: The integrand contains $\sqrt{a^2 - (kx)^2}$ with $a=3$ and $k=5$. To eliminate the radical, we set $kx = a \sin \theta$, so $5x = 3 \sin \theta$, giving $x = \frac{3}{5} \sin \theta$. This substitution simplifies the radical to $3 \cos \theta$, making the integral tractable. Other choices either fail to match the required form or lead to more complicated expressions.

📖 Explanation: The expression $\sqrt{x^2+16}$ matches the form $\sqrt{x^2+a^2}$ with $a=4$. The standard substitution for this form is $x = a \tan \theta$, which gives $dx = 4 \sec^2 \theta d\theta$ and simplifies the radical to $4 \sec \theta$. Substituting $x = 4 \sin \theta$ is suitable for $\sqrt{a^2 - x^2}$, while $x = 4 \sec \theta$ is used for $\sqrt{x^2 - a^2}$. Thus, $x=4\tan\theta$ is the correct choice.

📖 Explanation: The expression $\sqrt{4-x^2}$ requires the substitution $x = a \sin \theta$ because the domain of $x$ is limited to $[-2,2]$. Using $x = 2 \sec \theta$ restricts $x$ to $(-\infty, -2] \cup [2, \infty)$, which is outside the domain of the original integrand. This is a critical error because the substitution must preserve the domain of the original variable. Using $x=2\sec\theta$ would lead to an incorrect simplification of the radical and an invalid integration.

📖 Explanation: With $x = 3 \sin \theta$, $dx = 3 \cos \theta d\theta$, and $\sqrt{9-x^2} = 3 \cos \theta$. Substituting these into the integral gives $\int \frac{3 \cos \theta d\theta}{(9 \sin^2 \theta)(3 \cos \theta)} = \frac{1}{9} \int \frac{d\theta}{\sin^2 \theta} = \frac{1}{9} \int \csc^2 \theta d\theta$. This is a direct Easy of the substitution and simplification rules, requiring careful handling of the constants and trigonometric simplifications.

📖 Explanation: Using $x = \sin \theta$, $dx = \cos \theta d\theta$, and $\sqrt{1-x^2} = \cos \theta$. The limits change to $0$ and $\pi/2$. The integral becomes $\int_0^{\pi/2} \sin^2 \theta \cos^2 \theta d\theta = \frac{1}{4} \int_0^{\pi/2} \sin^2 2\theta d\theta$. Using the identity $\sin^2 2\theta = \frac{1-\cos 4\theta}{2}$, the integral evaluates to $\frac{1}{4} [\frac{\theta}{2} - \frac{\sin 4\theta}{8}]_0^{\pi/2} = \frac{\pi}{16}$. This problem combines substitution, trigonometric identities, and definite integral evaluation.

📖 Explanation: The area is $4 \int_0^a \frac{b}{a} \sqrt{a^2-x^2} dx$. Substituting $x = a \sin \theta$, $dx = a \cos \theta d\theta$, and $\sqrt{a^2-x^2} = a \cos \theta$, the integral becomes $4 \frac{b}{a} \int_0^{\pi/2} a \cos \theta \cdot a \cos \theta d\theta = 4ab \int_0^{\pi/2} \cos^2 \theta d\theta$. The other options either have the wrong power of sine/cosine or are missing the constant factor. This question tests the Easy of trigonometric substitution to a geometric formula.

📖 Explanation: For $x = 3 \sec \theta$, $dx = 3 \sec \theta \tan \theta d\theta$, and $\sqrt{x^2-9} = 3 \tan \theta$. Substituting gives $\int \frac{3 \sec \theta \tan \theta d\theta}{9 \sec^2 \theta \cdot 3 \tan \theta} = \frac{1}{9} \int \frac{\sec \theta \tan \theta}{\sec^2 \theta \tan \theta} d\theta = \frac{1}{9} \int \frac{1}{\sec \theta} d\theta = \frac{1}{9} \int \cos \theta d\theta$. This simplification requires careful cancellation of trigonometric factors and recognizing the relationship between secant and cosine.

📖 Explanation: The expression $x^2+4x+5$ is not in the standard form $x^2 \pm a^2$. Completing the square transforms it to $(x+2)^2+1$, which matches $u^2 + a^2$ with $u = x+2$ and $a=1$. This allows the use of the substitution $u = \tan \theta$. The other options are either incorrect algebraically or do not lead to a form suitable for trigonometric substitution.

📖 Explanation: Completing the square: $4x^2+4x+2 = (2x+1)^2+1$. Let $u = 2x+1$, then $du = 2dx$, so the integral becomes $\frac{1}{2} \int \frac{du}{\sqrt{u^2+1}} = \frac{1}{2} \sinh^{-1} u + C = \frac{1}{2} \sinh^{-1}(2x+1) + C$. The logarithmic form is also valid but not simplified to match the options. This tests the ability to handle non-standard quadratics and choose the correct integration formula.

📖 Explanation: The integral $\int \frac{dx}{\sqrt{x^2-1}}$ contains $\sqrt{x^2-a^2}$, which is a classic case for trigonometric substitution (specifically $x = \sec \theta$). The other integrals can be evaluated using simple $u$-substitution: $u = x^2+1$ for (a), $u = x^2+1$ for (c), and $u = x + \sqrt{x^2+1}$ or a direct formula for (d). This question tests the recognition of when trigonometric substitution is the most appropriate method.

📖 Explanation: The integral evaluates to $2 \int \cos^2 \theta d\theta = 2(\frac{\theta}{2} + \frac{\sin 2\theta}{4}) = \theta + \frac{\sin 2\theta}{2} = \theta + \sin \theta \cos \theta$. Since $\sin \theta = x/2$ and $\cos \theta = \sqrt{4-x^2}/2$, the result is $\sin^{-1}(x/2) + \frac{x}{2} \cdot \frac{\sqrt{4-x^2}}{2} = \sin^{-1}(x/2) + \frac{x\sqrt{4-x^2}}{4}$. Multiplying by 2 gives $2 \sin^{-1}(x/2) + \frac{x\sqrt{4-x^2}}{2}$. The student's answer missed the inverse sine term and the correct coefficient.

📖 Explanation: The area is $\int_0^2 \sqrt{4-x^2} dx$. With $x = 2 \sin \theta$, $dx = 2 \cos \theta d\theta$, and $\sqrt{4-x^2} = 2 \cos \theta$. The limits become $0$ and $\pi/2$. The integral becomes $\int_0^{\pi/2} (2 \cos \theta)(2 \cos \theta) d\theta = 4 \int_0^{\pi/2} \cos^2 \theta d\theta$. This is a direct Easy of the substitution to a geometric area problem, requiring the student to visualize the graph and set up the integral correctly.

📖 Explanation: With $x = 3 \sin \theta$, $dx = 3 \cos \theta d\theta$, and $\sqrt{9-x^2} = 3 \cos \theta$. The integral becomes $\int \frac{9 \sin^2 \theta \cdot 3 \cos \theta}{3 \cos \theta} d\theta = 9 \int \sin^2 \theta d\theta$. This is a straightforward substitution but requires careful cancellation of the $3 \cos \theta$ terms. The other options incorrectly simplify the constants or the trigonometric function.

📖 Explanation: With $x = 2 \sec \theta$, $dx = 2 \sec \theta \tan \theta d\theta$, and $\sqrt{x^2-4} = 2 \tan \theta$. The integral becomes $\int 4 \sec \theta \tan^2 \theta d\theta = 4 \int \sec \theta (\sec^2 \theta - 1) d\theta = 4(\frac{1}{2} \sec \theta \tan \theta - \frac{1}{2} \ln|\sec \theta + \tan \theta|) = 2 \sec \theta \tan \theta - 2 \ln|\sec \theta + \tan \theta| + C$. Substituting back gives $\frac{x}{2} \sqrt{x^2-4} - 2 \ln|\frac{x}{2} + \frac{\sqrt{x^2-4}}{2}| + C = \frac{x}{2} \sqrt{x^2-4} - 2 \ln|x + \sqrt{x^2-4}| + C$.

📖 Explanation: Completing the square: $x^2+2x+10 = (x+1)^2+9$. This is of the form $u^2 + a^2$ with $u = x+1$ and $a = 3$. The appropriate substitution for $\int \frac{du}{(u^2+a^2)^{3/2}}$ is $u = a \tan \theta$. Thus, $x+1 = 3 \tan \theta$. The other substitutions are for different radical forms and would not simplify the integral correctly.

📖 Explanation: For $x = 5 \sec \theta$, $x^2-25 = 25 \sec^2 \theta - 25 = 25(\sec^2 \theta - 1) = 25 \tan^2 \theta$. Therefore, $\sqrt{x^2-25} = 5 |\tan \theta|$. Since $x \ge 5$ (or $x \le -5$), we can take $\tan \theta \ge 0$ in the appropriate branch, so $\sqrt{x^2-25} = 5 \tan \theta$. The student confused the simplification for the $\sqrt{a^2-x^2}$ case, which gives $a \cos \theta$. This is a common error that highlights the importance of knowing which substitution corresponds to which radical form.

📖 Explanation: The arc length formula is $L = \int_0^1 \sqrt{1+(dy/dx)^2} dx$. For $y = x^2/2$, $dy/dx = x$, so the integrand is $\sqrt{1+x^2}$. This matches the form $\sqrt{a^2+x^2}$ with $a=1$, so the substitution is $x = \tan \theta$. The other options either use the wrong derivative or the wrong substitution for the radical. This question connects the arc length formula with trigonometric substitution.

📖 Explanation: After a trigonometric substitution like $x = a \tan \theta$, the resulting integrand often involves rational functions of $\tan \theta$ and $\sec \theta$. Using the identity $\sec^2 \theta = \tan^2 \theta + 1$, the integral $\int \frac{\sec^2 \theta}{\tan^2 \theta + 1} d\theta$ simplifies to $\int d\theta$, which is easily integrable. This problem tests the understanding of how to simplify the result of a trigonometric substitution and apply trigonometric identities to make the integral easier. It combines concepts of substitution and algebraic manipulation.

📖 Explanation: With $x = 2 \sec \theta$, $dx = 2 \sec \theta \tan \theta d\theta$, and $\sqrt{x^2-4} = 2 \tan \theta$. The integral becomes $\int \frac{2 \sec \theta \tan \theta}{4 \sec^2 \theta \cdot 2 \tan \theta} d\theta = \frac{1}{4} \int \frac{1}{\sec \theta} d\theta = \frac{1}{4} \int \cos \theta d\theta = \frac{1}{4} \sin \theta + C$. Since $\sin \theta = \frac{\sqrt{x^2-4}}{x}$ (from the right triangle), the result is $\frac{\sqrt{x^2-4}}{4x} + C$. The inverse secant form is equivalent, but not directly obtained. The correct choice is the simplified form.

📖 Explanation: The expression $x^2+2x+5 = (x+1)^2+4$ is of the form $u^2 + a^2$ with $a=2$. The standard substitution for this form is $u = a \tan \theta$, i.e., $x+1 = 2 \tan \theta$. The substitution $u = a \sin \theta$ is used for $\sqrt{a^2 - u^2}$. The student's claim is incorrect because it uses the wrong substitution for the given radical form, which would not simplify the expression correctly.

📖 Explanation: Let $u = \sin \theta$, then $du = \cos \theta d\theta$. The limits change to $0$ and $1$. The integral becomes $\int_0^1 \frac{du}{\sqrt{1+u^2}} = \sinh^{-1}(u) \big|_0^1 = \sinh^{-1}(1)$. This is equal to $\ln(1+\sqrt{2})$. This problem tests the ability to recognize that a trigonometric substitution can be simplified by an additional substitution, and it verifies the equivalence of inverse hyperbolic and logarithmic forms.

📖 Explanation: The radical $\sqrt{1-x^2}$ suggests the substitution $x = \sin \theta$. Then $dx = \cos \theta d\theta$, and $\sqrt{1-x^2} = \cos \theta$. The integral becomes $\int \frac{\sin^3 \theta \cos \theta}{\cos \theta} d\theta = \int \sin^3 \theta d\theta$. This is a direct Easy. The other options use substitutions that either do not match the radical (like $\tan \theta$ for $\sqrt{1+x^2}$) or are for a different radical form (like $\sec \theta$ for $\sqrt{x^2-1}$).

📖 Explanation: The integral $\int \frac{x^2}{x^2+1} dx$ can be solved by algebraic manipulation: $\frac{x^2}{x^2+1} = 1 - \frac{1}{x^2+1}$, leading to $x - \tan^{-1} x + C$. This is much simpler than a trigonometric substitution. The other integrals all involve radicals or quadratic forms that are typically handled with trigonometric substitutions. This question tests the ability to identify when a simpler method (like algebraic division) is more appropriate, rather than automatically applying a trigonometric substitution.

📖 Explanation: Completing the square: $2x-x^2 = -(x^2-2x) = -(x^2-2x+1-1) = -((x-1)^2-1) = 1-(x-1)^2$. This is of the form $\sqrt{a^2 - u^2}$ with $a=1$ and $u=x-1$. This shows that the quadratic can be transformed into a standard form suitable for a sine or cosine substitution. The other options are algebraically incorrect.

📖 Explanation: Completing the square: $x^2-6x+13 = (x-3)^2+4$. Let $u = x-3$, then the integral is $\int \frac{du}{\sqrt{u^2+4}} = \sinh^{-1}(u/2) + C = \sinh^{-1}(\frac{x-3}{2}) + C$. This is also equal to $\ln|u+\sqrt{u^2+4}| + C = \ln|x-3 + \sqrt{x^2-6x+13}| + C$. Both forms are equivalent due to the identity $\sinh^{-1} z = \ln(z + \sqrt{z^2+1})$. The inverse tangent would be for an integral of the form $\int \frac{du}{u^2+a^2}$. This problem tests the understanding of the equivalence of inverse hyperbolic and logarithmic forms.

📖 Explanation: With $x = \sec \theta$, $dx = \sec \theta \tan \theta d\theta$, and $\sqrt{x^2-1} = \tan \theta$. The integral becomes $\int \frac{\sec \theta \tan \theta}{\sec \theta \tan \theta} d\theta = \int d\theta = \theta + C = \sec^{-1} x + C$. The student's answer is correct. This problem is a straightforward Easy of the substitution and simplification, but it tests the recognition of the derivative of $\sec^{-1} x$. It can also be solved using a different substitution, but the result is the same.

📖 Explanation: Using $x = 2 \sin \theta$, $dx = 2 \cos \theta d\theta$, and $\sqrt{4-x^2} = 2 \cos \theta$. The integral becomes $\int 4 \sin^2 \theta d\theta = 4(\frac{\theta}{2} - \frac{\sin 2\theta}{4}) = 2\theta - 2 \sin \theta \cos \theta$. Substituting back: $2 \sin^{-1}(x/2) - 2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4-x^2}}{2} = 2 \sin^{-1}(x/2) - \frac{x\sqrt{4-x^2}}{2}$. The correct answer has the term $-\frac{x}{2}\sqrt{4-x^2}$. This problem requires careful substitution and algebraic back-substitution.

📖 Explanation: Completing the square gives $(x+2)^2+4$, which is of the form $u^2+a^2$ with $u = x+2$ and $a=2$. The substitution $u = a \tan \theta$ (i.e., $x+2 = 2 \tan \theta$) transforms the integral into $\int \frac{2 \sec^2 \theta d\theta}{4 \tan^2 \theta + 4} = \frac{1}{2} \int d\theta = \frac{1}{2} \theta + C = \frac{1}{2} \tan^{-1}((x+2)/2) + C$. The other options use the wrong trigonometric function for this form.

📖 Explanation: The radical $\sqrt{1-x^2}$ requires a substitution where $1 - x^2 = \cos^2 \theta$, which is achieved by $x = \sin \theta$ (or $x = \cos \theta$). The radical $\sqrt{x^2-1}$ requires $x^2-1 = \tan^2 \theta$, which is achieved by $x = \sec \theta$. This is a fundamental pair of substitutions that students must know for these standard forms. The other options mix up the substitutions for different radical types.

📖 Explanation: Using $x = 2 \sin \theta$, $dx = 2 \cos \theta d\theta$, and $\sqrt{4-x^2} = 2 \cos \theta$, the integral becomes $\int \frac{2 \cos \theta}{4 \sin^2 \theta \cdot 2 \cos \theta} d\theta = \frac{1}{4} \int \csc^2 \theta d\theta = -\frac{1}{4} \cot \theta + C$. Since $\cot \theta = \frac{\sqrt{4-x^2}}{x}$, the result is $-\frac{\sqrt{4-x^2}}{4x} + C$. The student's answer is correct. This question tests the ability to verify a given solution by working through the substitution and checking the simplification.

📖 Explanation: This is a classic improper integral with a vertical asymptote at $x=1$. Using the substitution $x = \sin \theta$, $dx = \cos \theta d\theta$, and $\sqrt{1-x^2} = \cos \theta$, the limits become $0$ and $\pi/2$. The integral becomes $\int_0^{\pi/2} d\theta = \pi/2$. The integral converges to a finite value. This problem tests the ability to handle an improper integral using a trigonometric substitution and evaluating the limit correctly.

📖 Explanation: With $x = \tan \theta$, $dx = \sec^2 \theta d\theta$, and $x^2+1 = \sec^2 \theta$. The integral becomes $\int \frac{\sec^2 \theta}{\sec^4 \theta} d\theta = \int \cos^2 \theta d\theta$. This is a straightforward Easy. The result can then be integrated using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$. This problem tests the ability to simplify the integrand after substitution, reducing it to a form that can be integrated using standard trigonometric identities.