Trig Substitution - Calculus 2

To evaluate the integral $\int_{4}^{6} \frac{dx}{\sqrt{x^2-8x+25}}$ , which of the following is the most appropriate first step?
Answer: Complete the square on the quadratic expression in the denominator. The integrand contains a quadratic expression under a radical. Before a trigonometric substitution can be applied, the quadratic $x^2-8x+25$ must be rewritten to match one of the standard forms. This is done by completing the square: $x^2-8x+25 = (x^2-8x+16) - 16 + 25 = (x-4)^2+9$ . After this step, a substitution like $x-4=3\tan\theta$ can be made. The other options are incorrect because they either misidentify the form of the integral or suggest an inappropriate technique.
An attempt to evaluate $\int \frac{dx}{\sqrt{x^2-16}}$ using the substitution $x=4\sin\theta$ would fail. What is the primary reason for this failure?
Answer: The domain of $x=4\sin\theta$ is $[-4, 4]$ , which does not match the domain of the integrand. The integrand $\frac{1}{\sqrt{x^2-16}}$ is defined for $x^2 > 16$ , i.e., $x>4$ or $x<-4$ . The substitution $x=4\sin\theta$ restricts $x$ to the interval $[-4, 4]$ , since $\sin\theta$ is between -1 and 1. The domain of the substitution and the domain of the function to be integrated do not align. Furthermore, using this substitution would lead to $\sqrt{16\sin^2\theta-16} = \sqrt{-16\cos^2\theta}$ , which is not a real number for real $\theta$ . Option B is related but C is the more fundamental reason related to the domains of the functions.
The integral $\int \frac{dx}{x\sqrt{x^2-16}}$ where $x > 4$ requires a trigonometric substitution. After applying the appropriate substitution and simplifying, which expression represents the antiderivative?
Answer: $\frac{1}{4}\sec^{-1}\left(\frac{x}{4}\right) + C$ For $\sqrt{x^2-16}$ with $x > 4$ , use $x = 4\sec\theta$ where $0 < \theta < \frac{\pi}{2}$ . Then $dx = 4\sec\theta\tan\theta \, d\theta$ and $\sqrt{x^2-16} = 4\tan\theta$ . The integral becomes $\int \frac{4\sec\theta\tan\theta \, d\theta}{4\sec\theta \cdot 4\tan\theta} = \int \frac{d\theta}{4} = \frac{\theta}{4} + C$ . Since $\sec\theta = \frac{x}{4}$ , we have $\theta = \sec^{-1}\left(\frac{x}{4}\right)$ . Therefore, the answer is $\frac{1}{4}\sec^{-1}\left(\frac{x}{4}\right) + C$ . Note that $\sec^{-1}\left(\frac{x}{4}\right) = \cos^{-1}\left(\frac{4}{x}\right)$ , but the coefficient differs between choices A and D.
Evaluate the definite integral $\int_{\sqrt{2}}^{2} \frac{dx}{x^3\sqrt{x^2-1}}$ .
Answer: $\frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}$ Let $x=\sec\theta$ , so $dx=\sec\theta\tan\theta \, d\theta$ . Limits: When $x=\sqrt{2}$ , $\sec\theta=\sqrt{2} \implies \theta=\pi/4$ . When $x=2$ , $\sec\theta=2 \implies \theta=\pi/3$ . The integral becomes $\int_{\pi/4}^{\pi/3} \frac{\sec\theta\tan\theta}{\sec^3\theta\sqrt{\sec^2\theta-1}} d\theta = \int_{\pi/4}^{\pi/3} \frac{\sec\theta\tan\theta}{\sec^3\theta\tan\theta} d\theta = \int_{\pi/4}^{\pi/3} \frac{1}{\sec^2\theta} d\theta = \int_{\pi/4}^{\pi/3} \cos^2\theta \, d\theta$ . Use the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$ . The integral is $\frac{1}{2}[\theta+\frac{1}{2}\sin(2\theta)]_{\pi/4}^{\pi/3}$ . Evaluate at the limits: At $\theta=\pi/3$ : $\sin(2\pi/3) = \sqrt{3}/2$ . At $\theta=\pi/4$ : $\sin(\pi/2) = 1$ . So we get $\frac{1}{2}[(\frac{\pi}{3}+\frac{\sqrt{3}}{4}) - (\frac{\pi}{4}+\frac{1}{2})] = \frac{1}{2}[\frac{\pi}{12} + \frac{\sqrt{3}}{4} - \frac{1}{2}] = \frac{\pi}{24} + \frac{\sqrt{3}}{8} - \frac{1}{4}$ .
An ellipse is described by the equation $9x^2+16y^2=144$ . What is the area of the region in the first quadrant bounded by this ellipse?
Answer: $3\pi$ The equation of the ellipse can be written in standard form as $\frac{x^2}{16} + \frac{y^2}{9} = 1$ . This shows the semi-axes are $a=4$ and $b=3$ . The total area of the ellipse is $\pi ab = \pi(4)(3) = 12\pi$ . The question asks for the area in the first quadrant, which is one-fourth of the total area. Therefore, the area is $\frac{12\pi}{4} = 3\pi$ . This can be verified by setting up the integral $A = \int_0^4 y \, dx = \int_0^4 3\sqrt{1-\frac{x^2}{16}} dx = \frac{3}{4}\int_0^4 \sqrt{16-x^2} dx$ . The integral $\int_0^4 \sqrt{16-x^2} dx$ represents the area of a quarter circle of radius 4, which is $\frac{1}{4}\pi(4^2) = 4\pi$ . So the area is $\frac{3}{4}(4\pi) = 3\pi$ . The trigonometric substitution $x=4\sin\theta$ would be used to evaluate this integral formally.
When evaluating $\int \frac{dx}{\sqrt{x^2+a^2}}$ using a trigonometric substitution, the final result is $\ln|\sqrt{x^2+a^2}+x| + C$ . Which intermediate integral in terms of $\theta$ produces this result?
Answer: $\int \sec\theta \, d\theta$ When you encounter an integral with the form $\int \frac{dx}{\sqrt{x^2+a^2}}$ , this signals a trigonometric substitution problem. The presence of $x^2+a^2$ under the square root tells you to use the substitution $x = a\tan\theta$ , which transforms the radical into a simpler trigonometric expression. With $x = a\tan\theta$ , we have $dx = a\sec^2\theta \, d\theta$ and $\sqrt{x^2+a^2} = \sqrt{a^2\tan^2\theta + a^2} = a\sqrt{\tan^2\theta + 1} = a\sec\theta$ . Substituting these into the original integral: $\int \frac{dx}{\sqrt{x^2+a^2}} = \int \frac{a\sec^2\theta \, d\theta}{a\sec\theta} = \int \sec\theta \, d\theta$ This is answer choice D. The integral $\int \sec\theta \, d\theta = \ln|\sec\theta + \tan\theta| + C$ , which when converted back to $x$ gives $\ln|\sqrt{x^2+a^2}+x| + C$ as stated. Choice A, $\int \sec\theta\tan\theta \, d\theta = \sec\theta + C$ , would come from a different substitution setup. Choice B, $\int \csc\theta \, d\theta$ , appears in integrals involving $\sqrt{a^2-x^2}$ with $x = a\sin\theta$ . Choice C, $\int \sec^2\theta \, d\theta = \tan\theta + C$ , is the derivative of tangent but doesn't match our substitution result. Study tip: Memorize the three standard trig substitution patterns: $x^2+a^2$ uses $x = a\tan\theta$ , $a^2-x^2$ uses $x = a\sin\theta$ , and $x^2-a^2$ uses $x = a\sec\theta$ .
If the substitution $x-2 = \sqrt{5}\tan\theta$ is used to evaluate an integral, which of the following expressions is equivalent to $\sqrt{x^2-4x+9}$ ?
Answer: $\sqrt{5}\sec\theta$ When you encounter trigonometric substitution problems, the key is recognizing which substitution works for different radical forms. The expression $\sqrt{x^2-4x+9}$ needs to be rewritten using the given substitution $x-2 = \sqrt{5}\tan\theta$ . First, solve for $x$ : $x = 2 + \sqrt{5}\tan\theta$ . Now substitute this into the original expression: $\sqrt{x^2-4x+9} = \sqrt{(2 + \sqrt{5}\tan\theta)^2 - 4(2 + \sqrt{5}\tan\theta) + 9}$ Expanding the squared term: $(2 + \sqrt{5}\tan\theta)^2 = 4 + 4\sqrt{5}\tan\theta + 5\tan^2\theta$ Substituting everything: $\sqrt{4 + 4\sqrt{5}\tan\theta + 5\tan^2\theta - 8 - 4\sqrt{5}\tan\theta + 9}$ The middle terms cancel: $\sqrt{5 + 5\tan^2\theta} = \sqrt{5(1 + \tan^2\theta)}$ Using the fundamental trigonometric identity $1 + \tan^2\theta = \sec^2\theta$ : $\sqrt{5\sec^2\theta} = \sqrt{5}|\sec\theta| = \sqrt{5}\sec\theta$ (We drop the absolute value since $\sec\theta > 0$ in the typical domain for this substitution.) Choice A ( $\sqrt{5}\tan\theta$ ) ignores the trigonometric identity entirely. Choice B ( $5\sec\theta$ ) incorrectly brings the 5 outside the square root without taking its square root. Choice D ( $5\tan\theta$ ) makes both errors—wrong trigonometric function and incorrect handling of the constant. Study tip: For trigonometric substitutions, always complete the square first, then apply the fundamental trigonometric identities. Remember that $\sqrt{a^2} = |a|$ , and consider the appropriate sign based on your domain.
To evaluate the integral $\int \frac{dx}{(x^2+4x+13)^2}$ , one must first complete the square in the denominator. After doing so and making the appropriate trigonometric substitution, which of the following integrals in $\theta$ is obtained?
Answer: $\frac{1}{27} \int \cos^2\theta \, d\theta$ First, complete the square for the quadratic in the denominator: $x^2+4x+13 = (x^2+4x+4)+9 = (x+2)^2+9$ . The integral becomes $\int \frac{dx}{((x+2)^2+9)^2}$ . Let $u=x+2$ , so $du=dx$ . The integral is $\int \frac{du}{(u^2+9)^2}$ . This suggests the substitution $u=3\tan\theta$ , so $du=3\sec^2\theta \, d\theta$ . The denominator becomes $( (3\tan\theta)^2+9)^2 = (9\tan^2\theta+9)^2 = (9\sec^2\theta)^2 = 81\sec^4\theta$ . Substituting into the integral gives $\int \frac{3\sec^2\theta}{81\sec^4\theta} d\theta = \frac{3}{81} \int \frac{1}{\sec^2\theta} d\theta = \frac{1}{27} \int \cos^2\theta \, d\theta$ .
Find the volume of the solid generated by revolving the region bounded by $y=\frac{1}{\sqrt{x^2+1}}$ , $y=0$ , $x=0$ , and $x=1$ about the x-axis.
Answer: $\frac{\pi^2}{4}$ When you encounter a problem asking for the volume of a solid of revolution about the x-axis, you'll use the disk method. The volume formula is $V = \pi \int_a^b [f(x)]^2 dx$ , where the region is bounded by $y = f(x)$ and the x-axis. Here, you're revolving the region bounded by $y = \frac{1}{\sqrt{x^2+1}}$ , $y = 0$ , $x = 0$ , and $x = 1$ about the x-axis. Setting up the integral: $V = \pi \int_0^1 \left(\frac{1}{\sqrt{x^2+1}}\right)^2 dx = \pi \int_0^1 \frac{1}{x^2+1} dx$ The antiderivative of $\frac{1}{x^2+1}$ is $\arctan(x)$ , so: $V = \pi[\arctan(x)]_0^1 = \pi(\arctan(1) - \arctan(0)) = \pi\left(\frac{\pi}{4} - 0\right) = \frac{\pi^2}{4}$ This confirms answer choice D is correct. Choice A ( $\frac{\pi^2}{2}$ ) likely results from incorrectly evaluating $\arctan(1)$ as $\frac{\pi}{2}$ instead of $\frac{\pi}{4}$ . Choice B ( $\frac{\pi}{2}$ ) comes from forgetting to square the function in the disk method formula. Choice C ( $\pi$ ) represents the error of both forgetting to square the function AND incorrectly evaluating the arctangent. Remember: always square the function in the disk method, and memorize that $\arctan(1) = \frac{\pi}{4}$ . These are two of the most common mistakes on volume of revolution problems.
Evaluate $\int \frac{dx}{(1+x^2)^{3/2}}$ .
Answer: $\frac{x}{\sqrt{1+x^2}} + C$ Use the trigonometric substitution $x=\tan\theta$ , so $dx=\sec^2\theta \, d\theta$ . The denominator becomes $(1+\tan^2\theta)^{3/2} = (\sec^2\theta)^{3/2} = \sec^3\theta$ . The integral transforms to $\int \frac{\sec^2\theta}{\sec^3\theta} d\theta = \int \frac{1}{\sec\theta} d\theta = \int \cos\theta \, d\theta = \sin\theta + C$ . To convert back to $x$ , we use the relation $x=\tan\theta$ . In a right triangle, $\tan\theta=x/1$ , so the opposite side is $x$ , the adjacent side is 1, and the hypotenuse is $\sqrt{x^2+1}$ . Therefore, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2+1}}$ . The final answer is $\frac{x}{\sqrt{1+x^2}} + C$ .
A student attempts to evaluate $\int \frac{x}{\sqrt{16+x^2}} \, dx$ using the trigonometric substitution $x = 4\tan\theta$ . However, this integral can be solved more efficiently using a different method. What is the correct antiderivative using the most efficient approach?
Answer: $\sqrt{16+x^2} + C$ While trigonometric substitution would work, this integral is much more efficiently solved using ordinary substitution. Let $u = 16 + x^2$ , then $du = 2x \, dx$ , so $x \, dx = \frac{1}{2} du$ . The integral becomes $\int \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^{-1/2} \, du = \frac{1}{2} \cdot 2u^{1/2} + C = \sqrt{u} + C = \sqrt{16+x^2} + C$ . Choice B represents the result if the student incorrectly tried to use trigonometric substitution on this problem (which would be unnecessarily complicated). Choice C looks like the result for $\int \sqrt{16+x^2} \, dx$ , and Choice D would be incorrect due to the square root in the denominator.
Consider the integral $\int \frac{dx}{(x^2 + 6x + 13)^{3/2}}$ . After completing the square and applying the appropriate trigonometric substitution, the integral can be expressed in which of the following forms?
Answer: $\frac{x+3}{4\sqrt{x^2+6x+13}} + C$ First, complete the square: $x^2 + 6x + 13 = (x+3)^2 + 4$ . Let $u = x+3$ , so the integral becomes $\int \frac{du}{(u^2 + 4)^{3/2}}$ . Using the substitution $u = 2\tan\theta$ , we have $du = 2\sec^2\theta \, d\theta$ and $u^2 + 4 = 4\tan^2\theta + 4 = 4\sec^2\theta$ . The integral becomes $\int \frac{2\sec^2\theta \, d\theta}{(4\sec^2\theta)^{3/2}} = \int \frac{2\sec^2\theta \, d\theta}{8\sec^3\theta} = \frac{1}{4}\int \cos\theta \, d\theta = \frac{1}{4}\sin\theta + C$ . Since $\tan\theta = \frac{u}{2} = \frac{x+3}{2}$ , we have $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{u/2}{\sqrt{1+(u/2)^2}} = \frac{u/2}{\sqrt{(u^2+4)/4}} = \frac{u}{\sqrt{u^2+4}} = \frac{x+3}{\sqrt{x^2+6x+13}}$ . Therefore, the answer is $\frac{1}{4} \cdot \frac{x+3}{\sqrt{x^2+6x+13}} + C = \frac{x+3}{4\sqrt{x^2+6x+13}} + C$ .
Which trigonometric substitution is most appropriate to begin evaluating the integral $\int \sqrt{7-2x^2} dx$ ?
Answer: $x = \sqrt{7/2} \sin\theta$ The integral contains a term of the form $\sqrt{a^2-u^2}$ . We first rewrite the expression as $\sqrt{7-2x^2} = \sqrt{2(\frac{7}{2}-x^2)}$ - this is incorrect. We must factor out the coefficient of $x^2$ from the whole expression inside the radical. Or, better, identify $u^2$ and $a^2$ . The expression is $\sqrt{(\sqrt{7})^2 - (\sqrt{2}x)^2}$ . This is of the form $\sqrt{a^2-u^2}$ with $a=\sqrt{7}$ and $u=\sqrt{2}x$ . The appropriate substitution is $u=a\sin\theta$ , which translates to $\sqrt{2}x = \sqrt{7}\sin\theta$ . Solving for $x$ gives $x = \frac{\sqrt{7}}{\sqrt{2}}\sin\theta = \sqrt{7/2} \sin\theta$ .
Consider the integral $\int \frac{dx}{x^2\sqrt{x^2+3}}$ . After making the substitution $x=\sqrt{3}\tan\theta$ , the integral is transformed into an integral with respect to $\theta$ . After integrating, the result is $-\frac{\csc\theta}{3} + C$ . Which of the following is the final answer in terms of $x$ ?
Answer: $-\frac{\sqrt{x^2+3}}{3x} + C$ Given the substitution $x=\sqrt{3}\tan\theta$ , we have $\tan\theta = \frac{x}{\sqrt{3}}$ . We can construct a right triangle where the opposite side is $x$ and the adjacent side is $\sqrt{3}$ . The hypotenuse is then $\sqrt{x^2+(\sqrt{3})^2} = \sqrt{x^2+3}$ . The cosecant function is the ratio of the hypotenuse to the opposite side, so $\csc\theta = \frac{\sqrt{x^2+3}}{x}$ . Substituting this into the given result $-\frac{\csc\theta}{3} + C$ gives $-\frac{\sqrt{x^2+3}}{3x} + C$ .
Evaluate $\int \frac{dx}{(5-4x-x^2)^{3/2}}$ .
Answer: $\frac{x+2}{9\sqrt{5-4x-x^2}} + C$ First, complete the square: $5-4x-x^2 = 5 - (x^2+4x) = 5 - (x^2+4x+4-4) = 5 - (x+2)^2 + 4 = 9 - (x+2)^2$ . The integral is $\int \frac{dx}{(9-(x+2)^2)^{3/2}}$ . Let $u=x+2$ , so $du=dx$ . We have $\int \frac{du}{(9-u^2)^{3/2}}$ . Now substitute $u=3\sin\theta$ , so $du=3\cos\theta\, d\theta$ . The integral becomes $\int \frac{3\cos\theta}{(9-9\sin^2\theta)^{3/2}} d\theta = \int \frac{3\cos\theta}{(9\cos^2\theta)^{3/2}} d\theta = \int \frac{3\cos\theta}{27\cos^3\theta} d\theta = \frac{1}{9} \int \frac{1}{\cos^2\theta} d\theta = \frac{1}{9} \int \sec^2\theta d\theta = \frac{1}{9}\tan\theta + C$ . To convert back, from $u=3\sin\theta$ , we have $\sin\theta=u/3$ . In a right triangle, opposite is $u$ , hypotenuse is 3, so adjacent is $\sqrt{9-u^2}$ . Then $\tan\theta = \frac{u}{\sqrt{9-u^2}}$ . Substituting back $u=x+2$ , we get $\frac{1}{9} \frac{x+2}{\sqrt{9-(x+2)^2}} + C = \frac{x+2}{9\sqrt{5-4x-x^2}} + C$ .
To compute the integral $\int \frac{e^{2t}}{\sqrt{e^{2t}+1}} dt$ , a student first uses the substitution $x=e^t$ . Which trigonometric substitution should be applied next?
Answer: $x=\tan\theta$ When you encounter integrals involving expressions like $\sqrt{a^2 + x^2}$ , $\sqrt{a^2 - x^2}$ , or $\sqrt{x^2 - a^2}$ , trigonometric substitution is your go-to technique. The key is recognizing which trigonometric identity matches your expression's form. After the substitution $x = e^t$ , your integral becomes $\int \frac{x^2}{\sqrt{x^2 + 1}} dx$ . This expression contains $\sqrt{x^2 + 1}$ , which has the form $\sqrt{x^2 + a^2}$ where $a = 1$ . For expressions of the form $\sqrt{x^2 + a^2}$ , you should use the substitution $x = a\tan\theta$ . Since $a = 1$ here, the correct substitution is $x = \tan\theta$ . This works because of the identity $1 + \tan^2\theta = \sec^2\theta$ , which transforms $\sqrt{x^2 + 1} = \sqrt{\tan^2\theta + 1} = \sqrt{\sec^2\theta} = \sec\theta$ . Looking at the wrong answers: Choice A) $x^2 = \tan\theta$ gives you $x = \sqrt{\tan\theta}$ , which doesn't simplify your square root expression. Choice B) $x = \sin\theta$ is used for expressions like $\sqrt{a^2 - x^2}$ , not $\sqrt{x^2 + a^2}$ . Choice C) $x = \sec\theta$ is used for expressions like $\sqrt{x^2 - a^2}$ . Remember this pattern: $\sqrt{x^2 + a^2} \rightarrow x = a\tan\theta$ , $\sqrt{a^2 - x^2} \rightarrow x = a\sin\theta$ , and $\sqrt{x^2 - a^2} \rightarrow x = a\sec\theta$ . Matching the form of your expression to the correct trigonometric identity is essential for successful trigonometric substitution.
The integral $\int \frac{1}{x^2-a^2} dx$ is typically solved using partial fractions. However, it can also be evaluated using the trigonometric substitution $x=a\sec\theta$ . If this method is used, what is the resulting expression in terms of $x$ ?
Answer: $\frac{1}{2a} \ln|\frac{x-a}{x+a}| + C$ Let $x=a\sec\theta$ , so $dx = a\sec\theta\tan\theta \, d\theta$ . The denominator is $x^2-a^2 = a^2\sec^2\theta-a^2 = a^2\tan^2\theta$ . The integral becomes $\int \frac{a\sec\theta\tan\theta}{a^2\tan^2\theta} d\theta = \frac{1}{a}\int \frac{\sec\theta}{\tan\theta} d\theta = \frac{1}{a}\int \csc\theta \, d\theta$ . The antiderivative is $\frac{1}{a} \ln|\csc\theta - \cot\theta| + C$ . Now, we must convert back to $x$ . From $x=a\sec\theta$ , we have $\sec\theta = x/a$ . The triangle has hypotenuse $x$ and adjacent side $a$ , so the opposite side is $\sqrt{x^2-a^2}$ . Then $\csc\theta = \frac{x}{\sqrt{x^2-a^2}}$ and $\cot\theta = \frac{a}{\sqrt{x^2-a^2}}$ . The expression is $\frac{1}{a}\ln|\frac{x-a}{\sqrt{x^2-a^2}}| + C = \frac{1}{a}\ln|\frac{x-a}{\sqrt{(x-a)(x+a)}}| + C = \frac{1}{a}\ln|\sqrt{\frac{x-a}{x+a}}| + C = \frac{1}{2a}\ln|\frac{x-a}{x+a}| + C$ . This matches the result from partial fractions.
Evaluate the definite integral $\int_{0}^{1/2} \frac{x^2}{\sqrt{1-x^2}} dx$ .
Answer: $\frac{\pi}{12} - \frac{\sqrt{3}}{8}$ Use the substitution $x=\sin\theta$ , so $dx=\cos\theta \, d\theta$ . The limits of integration become: when $x=0, \sin\theta=0 \Rightarrow \theta=0$ ; when $x=1/2, \sin\theta=1/2 \Rightarrow \theta=\pi/6$ . The integral becomes $\int_{0}^{\pi/6} \frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}} \cos\theta \, d\theta = \int_{0}^{\pi/6} \frac{\sin^2\theta}{\cos\theta} \cos\theta \, d\theta = \int_{0}^{\pi/6} \sin^2\theta \, d\theta$ . Using the half-angle identity, $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$ . The integral is $\frac{1}{2} \int_{0}^{\pi/6} (1-\cos(2\theta)) d\theta = \frac{1}{2} [\theta - \frac{1}{2}\sin(2\theta)]_{0}^{\pi/6}$ . Evaluating this gives $\frac{1}{2} [(\frac{\pi}{6} - \frac{1}{2}\sin(\frac{\pi}{3})) - (0-0)] = \frac{1}{2}[\frac{\pi}{6} - \frac{1}{2}(\frac{\sqrt{3}}{2})] = \frac{\pi}{12} - \frac{\sqrt{3}}{8}$ .
Consider the integral $\int \frac{x^3}{\sqrt{x^2+4}} dx$ . After making the substitution $x=2\tan\theta$ , the integral is converted to an integral in $\theta$ . Which of the following is the correct antiderivative in terms of $\theta$ ?
Answer: $\frac{8}{3}\sec^3\theta - 8\sec\theta + C$ With $x=2\tan\theta$ , we have $dx=2\sec^2\theta \, d\theta$ and $\sqrt{x^2+4}=\sqrt{4\tan^2\theta+4}=2\sec\theta$ . The integral becomes $\int \frac{(2\tan\theta)^3}{2\sec\theta}(2\sec^2\theta) d\theta = \int \frac{8\tan^3\theta}{2\sec\theta}(2\sec^2\theta) d\theta = 8\int \tan^3\theta \sec\theta \, d\theta$ . To solve this, rewrite as $8\int \tan^2\theta (\sec\theta\tan\theta) \, d\theta = 8\int (\sec^2\theta-1)(\sec\theta\tan\theta) \, d\theta$ . Let $u=\sec\theta$ , so $du=\sec\theta\tan\theta \, d\theta$ . The integral becomes $8\int (u^2-1) du = 8(\frac{u^3}{3}-u) + C = \frac{8}{3}\sec^3\theta - 8\sec\theta + C$ .
The integral $\int \frac{dx}{x^2 \sqrt{4-x^2}}$ is evaluated using the substitution $x=2\sin\theta$ . The resulting antiderivative in $\theta$ is $-\frac{1}{4}\cot\theta + C$ . What is the antiderivative in terms of $x$ ?
Answer: $-\frac{\sqrt{4-x^2}}{4x} + C$ From the substitution $x=2\sin\theta$ , we have $\sin\theta = x/2$ . We can form a right triangle where the opposite side is $x$ and the hypotenuse is 2. The adjacent side is $\sqrt{2^2-x^2} = \sqrt{4-x^2}$ . The cotangent function is the ratio of the adjacent side to the opposite side, so $\cot\theta = \frac{\sqrt{4-x^2}}{x}$ . Substituting this into the given result $-\frac{1}{4}\cot\theta + C$ yields $-\frac{1}{4} \frac{\sqrt{4-x^2}}{x} + C = -\frac{\sqrt{4-x^2}}{4x} + C$ .

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