📝 Trig substitution for sqrt(a² + x²) (80 MCQs)

📖 Explanation: For integrands containing $\sqrt{a^2 - x^2}$, the standard substitution is $x = a \sin \theta$. Here, $a=2$, so $x = 2\sin\theta$ is correct. This substitution simplifies the radical to $2\cos\theta$, which cancels with the $dx = 2\cos\theta d\theta$ term. Using $x=2\cos\theta$ would also work with a different restriction, but $\sin\theta$ is the standard choice because it directly maps the interval $[-a,a]$ to $[-\pi/2,\pi/2]$ without sign ambiguity.

📖 Explanation: With $x=3\sin\theta$, we have $dx=3\cos\theta d\theta$ and $\sqrt{9-x^2} = \sqrt{9-9\sin^2\theta} = 3|\cos\theta|$. Restricting $\theta$ to $[-\pi/2,\pi/2]$ makes $\cos\theta \ge 0$, so the radical becomes $3\cos\theta$. The product is $(3\cos\theta)(3\cos\theta d\theta) = 9\cos^2\theta d\theta$. The $\cos^2\theta$ term is then integrated using the half-angle identity $\cos^2\theta = (1+\cos2\theta)/2$.

📖 Explanation: The correct limits: when $x=0$, $0=2\sin\theta \Rightarrow \theta=0$; when $x=2$, $2=2\sin\theta \Rightarrow \theta=\pi/2$. The integral becomes $\int_0^{\pi/2} \frac{2\cos\theta d\theta}{2\cos\theta} = \int_0^{\pi/2} d\theta = \pi/2$. Using limits $0$ to $\pi$ would give $\pi$, which is incorrect because the substitution $x=2\sin\theta$ is not one-to-one on $[0,\pi]$; $\sin\theta$ repeats values, and the restriction $-\pi/2 \le \theta \le \pi/2$ is required for a valid substitution. Using $\theta=\pi$ would also make $\cos\theta$ negative in some regions, altering the simplification.

📖 Explanation: Substitute $x=4\sin\theta$. Then $dx=4\cos\theta d\theta$, $\sqrt{16-x^2}=4\cos\theta$. The integral becomes $\int \frac{4\cos\theta d\theta}{16\sin^2\theta \cdot 4\cos\theta} = \frac{1}{16}\int \csc^2\theta d\theta = -\frac{1}{16}\cot\theta + C$. From $\sin\theta = x/4$, construct a right triangle: opposite = $x$, hypotenuse = 4, adjacent = $\sqrt{16-x^2}$. Thus $\cot\theta = \frac{\sqrt{16-x^2}}{x}$. Substituting gives $-\frac{\sqrt{16-x^2}}{16x} + C$. This result is valid for $x>0$; for $x<0$, the absolute value in the square root requires careful handling, but the general antiderivative form holds.

📖 Explanation: The substitution $x=a\sin\theta$ is specifically designed for integrands containing $\sqrt{a^2 - x^2}$. Option B contains this radical in the denominator, making it the correct choice. Option A requires $x=a\sec\theta$, Option C requires $x=a\tan\theta$, and Option D is a standard arctangent integral that doesn't require trigonometric substitution (though $x=a\tan\theta$ can be used, it's not necessary). Recognizing the correct substitution is a key skill: $\sqrt{a^2-x^2} \rightarrow \sin\theta$, $\sqrt{a^2+x^2} \rightarrow \tan\theta$, $\sqrt{x^2-a^2} \rightarrow \sec\theta$.

📖 Explanation: The integral represents the area of a quarter circle of radius $a$ in the first quadrant. The full circle area is $\pi a^2$, so a quarter of it is $\pi a^2/4$. The student's error likely comes from forgetting the factor of 1/4 or incorrectly applying the formula for the area of a semicircle. Substitution $x=a\sin\theta$ gives $\int_0^{\pi/2} a^2\cos^2\theta d\theta = a^2(\pi/4) = \pi a^2/4$. This is a classic Easy of trigonometric substitution to find the area of a circle without relying on geometry alone.

📖 Explanation: Rewrite the integral as $\int \frac{dx}{\sqrt{25-4x^2}} = \int \frac{dx}{\sqrt{25(1 - (4x^2/25))}} = \int \frac{dx}{5\sqrt{1 - (2x/5)^2}}$. Let $u = 2x/5$, then $du = (2/5)dx \Rightarrow dx = (5/2)du$. The integral becomes $\int \frac{(5/2)du}{5\sqrt{1-u^2}} = \frac{1}{2}\int \frac{du}{\sqrt{1-u^2}} = \frac{1}{2}\sin^{-1}u + C = \frac{1}{2}\sin^{-1}(2x/5) + C$. The factor 1/2 is crucial; forgetting it is a common error when the coefficient of $x^2$ is not 1.

📖 Explanation: The integral represents the area under the curve $y=\sqrt{9-x^2}$ from $x=0$ to $x=3$. This is exactly the area of a quarter circle with radius 3 in the first quadrant. The area of a full circle with radius 3 is $\pi(3)^2 = 9\pi$, so a quarter of that is $9\pi/4$. This geometric interpretation avoids the need for trigonometric substitution and provides a quick check for answers. The substitution $x=3\sin\theta$ would give the same result: $\int_0^{\pi/2} 9\cos^2\theta d\theta = 9\pi/4$.

📖 Explanation: The integrand contains $\sqrt{1-x^2}$, so the standard substitution is $x = \sin\theta$, with $dx = \cos\theta d\theta$. This transforms $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos\theta$, and the integral becomes $\int \sin^2\theta d\theta$, which can be evaluated using the half-angle identity. While $x=\cos\theta$ would also work (giving $-\int \cos^2\theta d\theta$), $x=\sin\theta$ is more standard because it directly matches the form $\sqrt{a^2-x^2}$ with $a=1$.

📖 Explanation: When $x=0$, $0=\sin\theta \Rightarrow \theta=0$. When $x=1$, $1=\sin\theta \Rightarrow \theta=\pi/2$ (using the principal value). The correct limits are $0$ to $\pi/2$. Failing to change the limits and integrating with respect to $\theta$ from 0 to 1 would be incorrect because $\theta$ is not the same as $x$. This is a common mistake; always remember to convert the limits when performing a substitution in a definite integral.

📖 Explanation: This is a standard formula derived by substituting $x=a\sin\theta$. The integral becomes $a^2\int \cos^2\theta d\theta = a^2(\theta/2 + \sin2\theta/4) = a^2\theta/2 + a^2\sin\theta\cos\theta/2$. Substituting back $\theta=\sin^{-1}(x/a)$, $\sin\theta=x/a$, and $\cos\theta=\sqrt{a^2-x^2}/a$ gives the result. Option A is correct. Option B uses $\cos^{-1}$, which is related but has a different constant. The formula is a combination of the algebraic and inverse trigonometric terms, reflecting the geometry of the circle.

📖 Explanation: With $u=4-x^2$, $du=-2x dx$, so $dx = du/(-2x)$. The integral becomes $\int \frac{1}{x\sqrt{u}} \cdot \frac{du}{-2x} = -\frac{1}{2}\int \frac{du}{x^2\sqrt{u}}$. But $x^2 = 4-u$, giving $-\frac{1}{2}\int \frac{du}{(4-u)\sqrt{u}}$, which still contains $u$ in the denominator. This integral can be solved with another substitution $u=t^2$, but it's more complicated than the trigonometric substitution $x=2\sin\theta$. The trigonometric substitution simplifies the radical directly and is generally preferred for this form. This illustrates the importance of choosing the right method.

📖 Explanation: The substitution gives $\sqrt{a^2 - a^2\sin^2\theta} = \sqrt{a^2(1-\sin^2\theta)} = \sqrt{a^2\cos^2\theta} = a|\cos\theta|$. Because $\cos\theta$ can be negative, the absolute value is mathematically correct. However, we restrict $\theta$ to $[-\pi/2,\pi/2]$ so that $\cos\theta \ge 0$, allowing us to drop the absolute value and write $a\cos\theta$. This restriction is essential for the substitution to be one-to-one and for the simplification to be valid without absolute values. Understanding this nuance is important for rigorous evaluation.

📖 Explanation: Substitute $x=2\sin\theta$, $dx=2\cos\theta d\theta$, $\sqrt{4-x^2}=2\cos\theta$. Limits: $x=0 \Rightarrow \theta=0$, $x=2 \Rightarrow \theta=\pi/2$. The integral becomes $\int_0^{\pi/2} \frac{4\sin^2\theta}{2\cos\theta} \cdot 2\cos\theta d\theta = 4\int_0^{\pi/2} \sin^2\theta d\theta = 4(\pi/4) = \pi$. This is a classic example where the trigonometric substitution simplifies the integral to a standard form. The result can also be interpreted geometrically as the volume of a solid of revolution, but the direct integration is straightforward.

📖 Explanation: The expression is of the form $\sqrt{a^2 - u^2}$ with $u = x-2$ and $a=3$. The standard substitution is $u = a\sin\theta$, so $x-2 = 3\sin\theta$. This is a shift in the variable, which is common when the quadratic is not centered at the origin. Completing the square often reveals this form. The substitution $x = 3\sin\theta$ would incorrectly treat $x$ as the variable, ignoring the shift. Always identify the 'inside' function $u$ before applying the substitution.

📖 Explanation: The restriction $-\pi/2 \le \theta \le \pi/2$ ensures that $\sin\theta$ is one-to-one on the interval and maps onto $[-1,1]$. This is the principal value range for $\sin^{-1}$. It also ensures $\cos\theta \ge 0$, so $\sqrt{a^2-x^2} = a\cos\theta$ without absolute values. Other ranges like $[0,\pi]$ would make $\sin\theta$ one-to-one, but $\cos\theta$ changes sign, complicating the simplification. The standard choice is the principal range for its simplicity and consistency with the inverse sine function.

📖 Explanation: Let's verify: Substitute $u = \sqrt{1-x^2}$ or $x=\sin\theta$. Using $x=\sin\theta$, $dx=\cos\theta d\theta$, $\sqrt{1-x^2}=\cos\theta$. The integral becomes $\int \sin^3\theta d\theta = \int (1-\cos^2\theta)\sin\theta d\theta = -\cos\theta + \cos^3\theta/3 + C$. Substituting back $\cos\theta = \sqrt{1-x^2}$ gives $-\sqrt{1-x^2} + \frac{(1-x^2)^{3/2}}{3} + C$. The student's answer is correct. This integral can also be solved with a simple $u$-substitution, demonstrating that trigonometric substitution is not always necessary but can be a useful tool.

📖 Explanation: The equation $x^2 + y^2 = a^2$ describes a circle. Solving for $y$ gives $y = \pm\sqrt{a^2 - x^2}$. The positive root gives the upper semicircle. This is the geometric interpretation of the integral $\int \sqrt{a^2 - x^2} dx$ as the area under a semicircle. Recognizing this can help in quickly evaluating definite integrals from 0 to $a$ as the area of a quarter circle, and from $-a$ to $a$ as the area of a semicircle.

📖 Explanation: The integral represents the area under the semicircle $y=\sqrt{4-x^2}$ from $x=-2$ to $x=2$. This is exactly the area of a semicircle with radius 2. The area of a full circle is $\pi(2)^2 = 4\pi$, so the area of the semicircle is $2\pi$. Using the substitution $x=2\sin\theta$, limits $-\pi/2$ to $\pi/2$, the integral becomes $\int_{-\pi/2}^{\pi/2} 4\cos^2\theta d\theta = 4(\pi/2) = 2\pi$. This is a good example of using both geometry and calculus to solve the same problem.

📖 Explanation: With $x=a\sin\theta$, $dx=a\cos\theta d\theta$, and $(a^2-x^2)^{3/2} = (a^2\cos^2\theta)^{3/2} = a^3\cos^3\theta$. The integral becomes $\int \frac{a\cos\theta d\theta}{a^3\cos^3\theta} = \frac{1}{a^2}\int \sec^2\theta d\theta = \frac{1}{a^2}\tan\theta + C$. Substituting back $\tan\theta = \frac{x}{\sqrt{a^2-x^2}}$ gives $\frac{x}{a^2\sqrt{a^2-x^2}} + C$. This is a useful formula to remember. The power of the denominator is 3/2, which leads to a $\sec^2\theta$ term after simplification, unlike the $1/\sqrt{a^2-x^2}$ case which gives $\cos\theta$.

📖 Explanation: $x=a\sec\theta$ is used for integrals containing $\sqrt{x^2-a^2}$, not $\sqrt{a^2-x^2}$. With $x=a\sec\theta$, $\sqrt{a^2-x^2} = \sqrt{a^2 - a^2\sec^2\theta} = a\sqrt{1-\sec^2\theta} = a\sqrt{-\tan^2\theta}$, which is not real for most $\theta$. So it's invalid. Both $x=a\sin\theta$ and $x=a\cos\theta$ are valid (with appropriate restrictions), and $x=a\tanh u$ is a hyperbolic substitution that also works because $1-\tanh^2u = \sech^2u$. Recognizing invalid substitutions is as important as knowing the correct ones.

📖 Explanation: The curve is a semicircle of radius 2. The path from $x=0$ to $x=2$ is a quarter of the circle. The circumference of the full circle is $2\pi(2) = 4\pi$, so a quarter of that is $\pi$. Using the arc length formula: $L = \int_0^2 \sqrt{1 + (dy/dx)^2} dx$, with $dy/dx = -x/\sqrt{4-x^2}$, gives $L = \int_0^2 \frac{2}{\sqrt{4-x^2}} dx = 2[\sin^{-1}(x/2)]_0^2 = 2(\pi/2) = \pi$. This connects geometry, arc length, and trigonometric substitution.

📖 Explanation: Substitute $x=\sin\theta$, $dx=\cos\theta d\theta$, $\sqrt{1-x^2}=\cos\theta$. Limits: $x=0 \Rightarrow \theta=0$, $x=1 \Rightarrow \theta=\pi/2$. The integral becomes $\int_0^{\pi/2} \sin^3\theta d\theta$. Using the reduction formula or $\sin^3\theta = (1-\cos^2\theta)\sin\theta$, we get $[-\cos\theta + \cos^3\theta/3]_0^{\pi/2} = (0+0) - (-1+1/3) = 2/3$. This is a good example of how trigonometric substitution simplifies powers of $x$ combined with the square root. The integral can also be solved with a $u$-substitution $u=1-x^2$, but the trigonometric method is systematic.

📖 Explanation: The curve is a semicircle of radius 5 centered at the origin. The line $y=3$ intersects at $x=4$ (since $3 = \sqrt{25-x^2} \Rightarrow x=4$). The area is $\int_{-4}^{4} (\sqrt{25-x^2} - 3) dx$. By symmetry, $2\int_0^4 \sqrt{25-x^2} dx - 24$. Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(x/a)$, we get $2[ \frac{x}{2}\sqrt{25-x^2} + \frac{25}{2}\sin^{-1}(x/5) ]_0^4 - 24 = 2[ \frac{4}{2}(3) + \frac{25}{2}\sin^{-1}(4/5) - 0 ] - 24 = 12 + 25\sin^{-1}(4/5) - 24 = 25\sin^{-1}(4/5) - 12$. This is a Hard Easy combining geometry, definite integrals, and inverse trigonometric functions.

📖 Explanation: The derivative of $\sin^{-1}(x/a)$ is $\frac{1}{\sqrt{a^2-x^2}}$. The derivative of $\cos^{-1}(x/a)$ is $-\frac{1}{\sqrt{a^2-x^2}}$. So \int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C = -\cos^{-1}(x/a) + C&#039;. The two antiderivatives differ by a constant $\pi/2$, which can be absorbed into the constant of integration. So the student's claim is technically correct but not in the standard form; the standard form uses $\sin^{-1}$.

📖 Explanation: The area of an ellipse is $\pi ab$. This is derived by integrating the upper half $y = \frac{b}{a}\sqrt{a^2-x^2}$ from $-a$ to $a$ and multiplying by 2. The integral of $\sqrt{a^2-x^2}$ from $-a$ to $a$ is $\pi a^2/2$ (area of a semicircle). So the area is $2 \cdot \frac{b}{a} \cdot \frac{\pi a^2}{2} = \pi ab$. This is a classic Easy of trigonometric substitution to derive a fundamental geometric formula. The special case $a=b=r$ gives the circle area $\pi r^2$.

📖 Explanation: These are two of the most important inverse trigonometric integrals. $\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$, and $\int \frac{dx}{1+x^2} = \tan^{-1}x + C$. They arise from the derivatives of $\sin^{-1}x$ and $\tan^{-1}x$, respectively. Note that $\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}(x/a) + C$, and $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}(x/a) + C$. These are foundational formulas that appear frequently in calculus and its Easys.

📖 Explanation: Substitute $x=4\sin\theta$, $dx=4\cos\theta d\theta$, $\sqrt{16-x^2}=4\cos\theta$. The integral becomes $\int \frac{16\sin^2\theta}{4\cos\theta} \cdot 4\cos\theta d\theta = 16\int \sin^2\theta d\theta$. This simplifies to $16(\theta/2 - \sin2\theta/4) + C = 8\theta - 4\sin2\theta + C$. Substituting back gives $8\sin^{-1}(x/4) - \frac{x\sqrt{16-x^2}}{2} + C$. This is a good example of how the $a^2$ factor from $x^2$ combines with the differential to produce a constant multiple of the standard $\sin^2\theta$ integral.

📖 Explanation: The primary reason for using $x=a\sin\theta$ is the identity $1-\sin^2\theta = \cos^2\theta$, which eliminates the square root: $\sqrt{a^2-x^2} = a\cos\theta$ (with the appropriate restriction). This often transforms the integral into a trigonometric integral that can be evaluated using known formulas. While it may also reduce the power of $x$ or add trigonometric functions, the main goal is to remove the radical, which is the obstacle to direct integration.

📖 Explanation: The region is a semicircle of radius 2. Rotating it about the x-axis produces a sphere of radius 2. The volume of a sphere is $\frac{4}{3}\pi r^3 = \frac{4}{3}\pi(8) = \frac{32\pi}{3}$. Using the disk method: $V = \int_{-2}^{2} \pi (4-x^2) dx = \pi[4x - x^3/3]_{-2}^{2} = \pi[(8 - 8/3) - (-8 + 8/3)] = \pi[16 - 16/3] = \pi(32/3) = 32\pi/3$. This is a classic Easy of the disk method and trigonometric substitution, though the geometric interpretation is quicker. It connects integration to 3D geometry.

📖 Explanation: The integral contains $\sqrt{9 - (x-1)^2}$, which is of the form $\sqrt{a^2 - u^2}$ with $u=x-1$. Substitutions for this form are $u=a\sin\theta$, $u=a\cos\theta$, or $u=a\tanh u$. $u=a\sec\theta$ is for $\sqrt{u^2 - a^2}$, so it would not simplify the expression; it would introduce a square root of a negative or tangent squared. Recognizing the correct form of the substitution is crucial. The correct substitutions are $x-1 = 3\sin\theta$, $x-1 = 3\cos\theta$, or $x-1 = 3\tanh u$.

📖 Explanation: $\int \frac{dx}{\sqrt{4-x^2}} = \sin^{-1}(x/2) + C$. Evaluating from 0 to 1: $\sin^{-1}(1/2) - \sin^{-1}(0) = \pi/6 - 0 = \pi/6$. This is a direct Easy of the standard formula. It's important to know the exact values of inverse trigonometric functions at common points. This integral also represents the area under the curve $y=1/\sqrt{4-x^2}$, which is related to the inverse sine function.

📖 Explanation: The derivative $dy/dx = -x/\sqrt{a^2-x^2}$. The arc length integrand becomes $\sqrt{1 + x^2/(a^2-x^2)} = \sqrt{a^2/(a^2-x^2)} = a/\sqrt{a^2-x^2}$. The integral $\int \frac{a}{\sqrt{a^2-x^2}} dx = a\sin^{-1}(x/a) + C$ is evaluated using the standard inverse sine formula, which is derived from the trigonometric substitution $x=a\sin\theta$. Thus, the arc length formula for a circle involves trigonometric substitution. This connects arc length, geometry, and integration techniques.