u-Substitution - Calculus 1

1. Find the antiderivative of $f(x) = \frac{\sin(\sqrt{x})}{\sqrt{x}}$.
   Answer: $-2\cos(\sqrt{x}) + C$ When you encounter an antiderivative involving a composite function like $\frac{\sin(\sqrt{x})}{\sqrt{x}}$, substitution is your go-to technique. The key insight is recognizing that the denominator $\sqrt{x}$ is related to the derivative of the inner function $\sqrt{x}$ in the sine term. Let's use substitution with $u = \sqrt{x}$. Then $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$, which means $dx = 2\sqrt{x} \, du = 2u \, du$. Substituting into our integral: $\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx = \int \frac{\sin(u)}{u} \cdot 2u \, du = \int 2\sin(u) \, du = -2\cos(u) + C$ Converting back to $x$: $-2\cos(\sqrt{x}) + C$. Now let's examine why the other answers are incorrect. Choice A gives $-\frac{1}{2}\cos(\sqrt{x}) + C$, which results from incorrectly handling the substitution factor of 2. Choice B, $\frac{1}{2}\cos(\sqrt{x}) + C$, makes the same substitution error plus incorrectly integrates sine (forgetting the negative sign). Choice C, $2\cos(\sqrt{x}) + C$, correctly handles the factor of 2 but misses the negative sign that comes from integrating sine. The correct answer is D: $-2\cos(\sqrt{x}) + C$. Study tip: When dealing with composite functions in integration, always check if u-substitution applies. Look for the derivative of the inner function appearing elsewhere in the integrand—it's often disguised but present, as $\frac{1}{\sqrt{x}}$ was here.
2. Evaluate $\int \frac{\cos(x)}{\sin^2(x)} \, dx$.
   Answer: $-\csc(x) + C$ When you encounter an integral with trigonometric functions where one appears in the numerator and a power of another in the denominator, look for substitution opportunities or recognize standard antiderivative patterns. For $\int \frac{\cos(x)}{\sin^2(x)} \, dx$, notice that the derivative of $\sin(x)$ is $\cos(x)$, which appears in the numerator. This suggests using substitution with $u = \sin(x)$, so $du = \cos(x) \, dx$. The integral becomes $\int \frac{1}{u^2} \, du = \int u^{-2} \, du = -u^{-1} + C = -\frac{1}{\sin(x)} + C = -\csc(x) + C$. Alternatively, you can recognize this directly as the derivative pattern: since $\frac{d}{dx}[-\csc(x)] = \csc(x)\cot(x) = \frac{1}{\sin(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{\cos(x)}{\sin^2(x)}$, the antiderivative is $-\csc(x) + C$. Choice A, $\ln(\sin^2(x)) + C$, would have derivative $\frac{2\cos(x)}{\sin(x)}$, not our integrand. Choice C, $\cot(x) + C$, has derivative $-\csc^2(x) = -\frac{1}{\sin^2(x)}$, missing the $\cos(x)$ factor. Choice D, $-\frac{1}{3\sin^3(x)} + C$, would come from integrating $\frac{\cos(x)}{\sin^4(x)}$, not $\sin^2(x)$. Study tip: When you see $\cos(x)$ in the numerator with powers of $\sin(x)$ in the denominator, immediately consider $u$-substitution with $u = \sin(x)$. Also memorize that $\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)$.
3. A particle moves along a line with velocity $v(t) = \frac{t}{t^2+1}$ meters per second at time $t \ge 0$. What is the total distance traveled by the particle from $t=0$ to $t=3$ seconds?
   Answer: $\frac{1}{2} \ln(10)$ When you see a velocity function and need to find total distance traveled, you're working with the fundamental relationship between velocity, displacement, and distance. The key insight is that distance is always positive, while displacement can be negative if the particle changes direction. To find total distance, you need to integrate the absolute value of velocity: $\int_0^3 |v(t)| \, dt$. However, first check if the velocity changes sign in the interval. Since $v(t) = \frac{t}{t^2+1}$ and both $t \geq 0$ and $t^2+1 > 0$ for $t \in [0,3]$, the velocity is always non-negative. This means $|v(t)| = v(t)$, so you can integrate directly. Calculate $\int_0^3 \frac{t}{t^2+1} \, dt$. Using substitution with $u = t^2+1$, so $du = 2t \, dt$ and $t \, dt = \frac{1}{2} du$. When $t=0,u=1$; when $t=3,u=10$. The integral becomes $\frac{1}{2} \int_1^{10} \frac{1}{u} \, du = \frac{1}{2}[\ln(u)]_1^{10} = \frac{1}{2}(\ln(10) - \ln(1)) = \frac{1}{2}\ln(10)$. Choice A) $\ln(10)$ results from forgetting the $\frac{1}{2}$ factor in the substitution. Choice C) $\arctan(3)$ comes from incorrectly integrating $\frac{1}{t^2+1}$ instead of $\frac{t}{t^2+1}$. Choice D) $\frac{9}{10}$ suggests a computational error or wrong approach entirely. Remember: always check if velocity changes sign when finding distance traveled, and be careful with substitution factors during integration.
4. Which of the following is the indefinite integral of $f(x) = \sin(x) \cos(\cos(x))$?
   Answer: $-\sin(\cos(x)) + C$ To find the integral $\int \sin(x) \cos(\cos(x)) dx$, we use u-substitution. Let the inner function be $u = \cos(x)$. Then, its differential is $du = -\sin(x) dx$. This means $\sin(x) dx = -du$. Substituting these into the integral gives: $\int \cos(\cos(x)) (\sin(x) dx) = \int \cos(u) (-du) = -\int \cos(u) du$ The integral of $\cos(u)$ is $\sin(u)$. Therefore: $-\sin(u) + C$ Substituting back for $u = \cos(x)$, we get the final answer: $-\sin(\cos(x)) + C$
5. If an integral is transformed by the substitution $u=\sin(x)$ into $\int_{1/2}^1 \frac{1}{u} \, du$, what was the original integral?
   Answer: $\int_{\pi/6}^{\pi/2} \cot(x) \, dx$ When you encounter reverse substitution problems, you need to work backwards from the transformed integral to find the original. The key is understanding how substitution changes both the integrand and the limits of integration. Given that $u = \sin(x)$ transforms an integral into $\int_{1/2}^1 \frac{1}{u} \, du$, let's reconstruct the original. Since $u = \sin(x)$, we have $du = \cos(x) \, dx$. The transformed integrand $\frac{1}{u}$ becomes $\frac{1}{\sin(x)}$, and when we substitute back $du = \cos(x) \, dx$, we get $\frac{1}{\sin(x)} \cdot \cos(x) \, dx = \frac{\cos(x)}{\sin(x)} \, dx = \cot(x) \, dx$. For the limits: when $u = 1/2$, we need $\sin(x) = 1/2$, so $x = \pi/6$. When $u = 1$, we need $\sin(x) = 1$, so $x = \pi/2$. Therefore, the original integral is $\int_{\pi/6}^{\pi/2} \cot(x) \, dx$, which is answer C. Answer A has the correct integrand $\frac{\cos(x)}{\sin(x)}$ but wrong limits $[1/2, 1]$ instead of $[\pi/6, \pi/2]$. Answer B uses $\tan(x)$ instead of $\cot(x)$ - this would give $\frac{\sin(x)}{\cos(x)}$, not $\frac{1}{u}$ after substitution. Answer D has the correct integrand but wrong lower limit $\pi/3$ (where $\sin(x) = \sqrt{3}/2$) instead of $\pi/6$. Remember: in reverse substitution problems, always check both the integrand transformation AND the limit conversions. Both must match the given transformed integral.
6. Find the indefinite integral $\int (x + \frac{1}{x})^3 (1 - \frac{1}{x^2}) \, dx$.
   Answer: $\frac{1}{4}(x + \frac{1}{x})^4 + C$ When you encounter an integral with a complicated expression raised to a power multiplied by another expression, look for substitution opportunities. The key insight here is recognizing that the derivative of $(x + \frac{1}{x})$ appears elsewhere in the integrand. Let's use substitution with $u = x + \frac{1}{x}$. Taking the derivative: $\frac{du}{dx} = 1 - \frac{1}{x^2}$, which means $du = (1 - \frac{1}{x^2})dx$. This is exactly the second factor in our integrand! Substituting into the original integral: $\int (x + \frac{1}{x})^3 (1 - \frac{1}{x^2}) \, dx = \int u^3 \, du$ This simplifies to: $\int u^3 \, du = \frac{u^4}{4} + C = \frac{1}{4}(x + \frac{1}{x})^4 + C$ Answer choice A, $\frac{1}{4}(1 - \frac{1}{x^2})^4 + C$, incorrectly treats $(1 - \frac{1}{x^2})$ as the substitution variable instead of recognizing it as the derivative. Answer choice B, $(x + \frac{1}{x})^4 + C$, forgets the coefficient $\frac{1}{4}$ that comes from integrating $u^3$. Answer choice C, $3(x + \frac{1}{x})^2 + C$, appears to confuse integration with differentiation, giving something closer to a derivative form. Study tip: When you see a composite function raised to a power multiplied by another expression, immediately check if that other expression might be the derivative of the inner function. This pattern appears frequently in calculus and makes substitution problems much more manageable.
7. If $\int g(x) \, dx = \cos(x^2) + C$, then $\int x g(x^2) \, dx =$ ?
   Answer: $\frac{1}{2}\cos(x^4) + C$ When you're given information about an antiderivative and asked to find a related integral, substitution is your key tool. Here, you know that $\int g(x) \, dx = \cos(x^2) + C$, which means $g(x) = \frac{d}{dx}[\cos(x^2)]$. Let's find $g(x)$ first. Using the chain rule: $g(x) = \frac{d}{dx}[\cos(x^2)] = -\sin(x^2) \cdot 2x = -2x\sin(x^2)$. Now for $\int x g(x^2) \, dx$, substitute what we know about $g$: $g(x^2) = -2x^2\sin(x^4)$, so the integral becomes $\int x \cdot (-2x^2\sin(x^4)) \, dx = \int -2x^3\sin(x^4) \, dx$. Use substitution with $u = x^4$, so $du = 4x^3 \, dx$, which means $x^3 \, dx = \frac{1}{4} du$. The integral transforms to: $\int -2\sin(u) \cdot \frac{1}{4} \, du = -\frac{1}{2}\int \sin(u) \, du = \frac{1}{2}\cos(u) + C = \frac{1}{2}\cos(x^4) + C$. Choice A incorrectly multiplies by $x$ instead of using proper substitution. Choice C misses the factor of $\frac{1}{2}$ that comes from the substitution process. Choice D incorrectly includes an extra $x^2$ term and misapplies the chain rule. The correct answer is B: $\frac{1}{2}\cos(x^4) + C$. Study tip: When working with composite functions in integration, always carefully track the derivatives that arise from substitution—the chain rule often introduces coefficients that become crucial factors in your final answer.
8. The integral $\int e^{2x} \sqrt{e^x+1} \, dx$ is equivalent to which of the following indefinite integrals after a suitable u-substitution?
   Answer: $\int (u-1)u^{1/2} \, du$ To solve $\int e^{2x} \sqrt{e^x+1} \, dx$, we first rewrite $e^{2x}$ as $e^x \cdot e^x$. The integral becomes $\int e^x \sqrt{e^x+1} \cdot e^x \, dx$. Let's choose the substitution $u = e^x+1$. This means $du = e^x dx$. We also need to express the remaining $e^x$ in the integrand in terms of $u$. From our substitution, we have $e^x = u-1$. Substituting all parts into the integral, we get: $\int (u-1) \sqrt{u} \, du$ which can also be written as $\int (u-1)u^{1/2} \, du$. This integral is now a simple polynomial in $u$ that can be solved with the power rule. The other choices represent common errors in substitution.
9. Evaluate $\int_{-1}^1 x^2 \sin(x^3) \, dx$.
   Answer: 0 When you encounter an integral with both polynomial and trigonometric functions, always check if the function has special symmetry properties before diving into complex integration techniques. The key insight here is recognizing that $f(x) = x^2 \sin(x^3)$ is an odd function. To verify this, check $f(-x) = (-x)^2 \sin((-x)^3) = x^2 \sin(-x^3) = -x^2 \sin(x^3) = -f(x)$. Since $f(-x) = -f(x)$, the function is odd. When you integrate an odd function over a symmetric interval like $[-1, 1]$, the result is always zero. This happens because the negative area from $x = -1$ to $x = 0$ exactly cancels the positive area from $x = 0$ to $x = 1$. Therefore, $\int_{-1}^1 x^2 \sin(x^3) \, dx = 0$. Choice A, $2(1-\cos(1))/3$, likely comes from attempting substitution integration and making sign errors. Choice B, $2\cos(1)/3$, suggests a similar substitution approach but with different computational mistakes. Choice D, $1-\cos(1)$, might result from forgetting the factor of 2 that appears when evaluating the antiderivative or from integration errors. Study tip: Before attempting complex integration techniques like substitution or integration by parts, always check if your function is odd or even over a symmetric interval. Recognizing $f(-x) = -f(x)$ (odd) or $f(-x) = f(x)$ (even) can save you significant time and prevent computational errors.
10. Evaluate the integral $\int_0^{\pi/6} \sin^4(3x) \cos(3x) dx$
    Answer: $\frac{1}{15}$ When you see an integral with a power of one trigonometric function multiplied by another trig function, think substitution. The key insight is recognizing that $\cos(3x)$ is the derivative of $\sin(3x)$ (up to a constant factor). Let's use the substitution $u = \sin(3x)$. Then $du = 3\cos(3x)dx$, which means $\cos(3x)dx = \frac{1}{3}du$. Our integral becomes: \int_0^{\pi/6} \sin^4(3x) \cos(3x) dx = \int \u^4 \cdot \frac{1}{3} du = \frac{1}{3} \int u^4 du Now we need the new limits. When $x = 0$: $u = \sin(0) = 0$. When $x = \pi/6$: $u = \sin(\pi/2) = 1$. So: $\frac{1}{3} \int_0^1 u^4 du = \frac{1}{3} \left[\frac{u^5}{5}\right]_0^1 = \frac{1}{3} \cdot \frac{1}{5} = \frac{1}{15}$ A ($0$) would result from incorrectly thinking this is an odd function over a symmetric interval, but our interval isn't symmetric about zero. B ($\frac{1}{5}$) comes from forgetting the factor of $\frac{1}{3}$ from the substitution. C ($\frac{1}{12}$) might arise from computational errors with the chain rule factor or limits. Study tip: For integrals of the form $\int (\text{trig function})^n \cdot (\text{derivative of that function}) dx$, always try $u$-substitution first. The derivative relationship is your biggest clue that substitution will work cleanly.
11. Find the indefinite integral $\int x \sqrt{x+3} \, dx$.
    Answer: $\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C$ Let $u = x+3$. Then $du = dx$ and $x = u-3$. The integral becomes $\int (u-3)\sqrt{u} \, du = \int (u^{3/2} - 3u^{1/2}) \, du$. Integrating with respect to $u$ gives $\frac{2}{5}u^{5/2} - 3\left(\frac{2}{3}u^{3/2}\right) + C = \frac{2}{5}u^{5/2} - 2u^{3/2} + C$. Substituting back $u=x+3$ yields $\frac{2}{5}(x+3)^{5/2} - 2(x+3)^{3/2} + C$.
12. Let $F(x) = \int_2^x \frac{t}{\sqrt{t^2+5}} \, dt$. What is the value of $F(2\sqrt{5})$?
    Answer: $2$ We need to evaluate $\int_2^{2\sqrt{5}} \frac{t}{\sqrt{t^2+5}} \, dt$. Let $u = t^2+5$. Then $du = 2t \, dt$, so $t \, dt = \frac{1}{2}du$. Change the limits: when $t=2$, $u=2^2+5=9$. When $t=2\sqrt{5}$, $u=(2\sqrt{5})^2+5 = 4(5)+5=25$. The integral becomes $\frac{1}{2} \int_9^{25} \frac{1}{\sqrt{u}} \, du = \frac{1}{2} \int_9^{25} u^{-1/2} \, du = \frac{1}{2}[2u^{1/2}]_9^{25} = [\sqrt{u}]_9^{25} = \sqrt{25} - \sqrt{9} = 5-3=2$.
13. Evaluate $\int_0^{\pi/4} \sec^2(x) \tan^2(x) \, dx$.
    Answer: 1/3 When you encounter an integral with products of trigonometric functions like $\sec^2(x) \tan^2(x)$, look for substitution opportunities involving trigonometric derivatives. Since the derivative of $\tan(x)$ is $\sec^2(x)$, this integral is perfectly set up for u-substitution. Let $u = \tan(x)$, so $du = \sec^2(x) \, dx$. Notice that $\tan^2(x) = u^2$, and our integral becomes: $\int_0^{\pi/4} \sec^2(x) \tan^2(x) \, dx = \int_0^{\pi/4} u^2 \, du$ We need to change the limits of integration. When $x = 0$, $u = \tan(0) = 0$. When $x = \pi/4$, $u = \tan(\pi/4) = 1$. So: $\int_0^1 u^2 \, du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$ This confirms answer choice D is correct. Answer choice A (3) likely comes from forgetting the fraction in the antiderivative $\frac{u^3}{3}$. Answer choice B (1) results from incorrectly evaluating the definite integral as just the upper limit value. Answer choice C (-1/3) might occur from sign errors during substitution or evaluation. Study tip: When you see $\sec^2(x)$ paired with functions of $\tan(x)$, immediately consider $u = \tan(x)$ substitution since $\frac{d}{dx}[\tan(x)] = \sec^2(x)$. This pattern appears frequently in calculus and makes otherwise complex trigonometric integrals straightforward.
14. To evaluate the integral $\int \frac{x+1}{x^2+2x+5} \, dx$, the most effective first step is the substitution $u = x^2+2x+5$. Which of the following is the resulting integral in terms of $u$?
    Answer: $\int \frac{1}{2u} \, du$ Let $u = x^2+2x+5$. Then the derivative is $\frac{du}{dx} = 2x+2 = 2(x+1)$. Rearranging gives $du = 2(x+1)\,dx$, or $(x+1)\,dx = \frac{1}{2}du$. Substituting both $u$ and $\frac{1}{2}du$ into the original integral gives $\int \frac{1}{u} \cdot \frac{1}{2}du = \int \frac{1}{2u} \, du$.
15. Which of the following is an antiderivative of $f(x) = \frac{e^{3x}}{1+e^{6x}}$?
    Answer: $\frac{1}{3}\arctan(e^{3x}) + C$ We need to evaluate $\int \frac{e^{3x}}{1+e^{6x}} \, dx$. Notice that $e^{6x} = (e^{3x})^2$. Let $u = e^{3x}$. Then $du = 3e^{3x} \, dx$, so $e^{3x} \, dx = \frac{1}{3}du$. The integral becomes $\int \frac{1}{1+u^2} \cdot \frac{1}{3}du = \frac{1}{3} \int \frac{1}{1+u^2} \, du = \frac{1}{3}\arctan(u) + C$. Substituting back $u=e^{3x}$ gives $\frac{1}{3}\arctan(e^{3x}) + C$.
16. What is the average value of $f(x) = \frac{(\ln x)^2}{x}$ on the interval $[1, e]$?
    Answer: $\frac{1}{3(e-1)}$ When you encounter a question asking for the average value of a function on an interval, you need to apply the average value formula: $\text{Average} = \frac{1}{b-a}\int_a^b f(x)\,dx$ For $f(x) = \frac{(\ln x)^2}{x}$ on $[1,e]$, this becomes $\frac{1}{e-1}\int_1^e \frac{(\ln x)^2}{x}\,dx$. To evaluate the integral, use substitution with $u = \ln x$, so $du = \frac{1}{x}dx$. When $x = 1$, $u = \ln 1 = 0$; when $x = e$, $u = \ln e = 1$. The integral transforms to: $\int_0^1 u^2\,du = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3} - 0 = \frac{1}{3}$ Therefore, the average value is $\frac{1}{e-1} \cdot \frac{1}{3} = \frac{1}{3(e-1)}$, which is answer D. Answer A ($1$) ignores the coefficient $\frac{1}{e-1}$ in the average value formula—you can't just evaluate the integral without dividing by the interval length. Answer B ($\frac{1}{3}$) is what you get if you evaluate the integral correctly but forget to multiply by $\frac{1}{e-1}$. Answer C ($\frac{e^3-1}{3(e-1)}$) results from incorrectly handling the substitution limits or making an algebraic error when evaluating $u^3$ at the bounds. Remember: average value problems always require both computing the definite integral AND dividing by the interval length $(b-a)$. Practice u-substitution with logarithmic functions, as this combination appears frequently on calculus exams.
17. If $\int_1^k \frac{x}{x^2+1} \, dx = \frac{1}{2}\ln(5)$ and $k>0$, what is the value of $k$?
    Answer: $3$ To evaluate the integral, let $u=x^2+1$, so $du=2x\,dx$ or $x\,dx = \frac{1}{2}du$. The limits become: if $x=1, u=1^2+1=2$; if $x=k, u=k^2+1$. The integral is $\frac{1}{2} \int_2^{k^2+1} \frac{1}{u} \, du = \frac{1}{2} [\ln|u|]_2^{k^2+1} = \frac{1}{2}(\ln(k^2+1) - \ln(2))$. We set this equal to $\frac{1}{2}\ln(5)$. So, $\frac{1}{2}(\ln(\frac{k^2+1}{2})) = \frac{1}{2}\ln(5)$. This implies $\ln(\frac{k^2+1}{2}) = \ln(5)$, so $\frac{k^2+1}{2} = 5$. Then $k^2+1=10$, $k^2=9$, and since $k>0$, $k=3$.
18. Evaluate $\int_0^1 \frac{x+1}{x^2+1} \, dx$.
    Answer: $\frac{1}{2}\ln(2) + \frac{\pi}{4}$ First, split the integral into two parts: $\int_0^1 \frac{x}{x^2+1} \, dx + \int_0^1 \frac{1}{x^2+1} \, dx$. For the first integral, use u-substitution with $u=x^2+1$, $du=2x\,dx$. This gives $\frac{1}{2}\int_1^2 \frac{1}{u} \, du = \frac{1}{2}[\ln|u|]_1^2 = \frac{1}{2}(\ln(2)-\ln(1)) = \frac{1}{2}\ln(2)$. The second integral is a standard form: $[\arctan(x)]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$. The final answer is the sum of the two results: $\frac{1}{2}\ln(2) + \frac{\pi}{4}$.
19. Evaluate $\int_0^4 \frac{1}{(1+\sqrt{x})^2 \sqrt{x}} \, dx$.
    Answer: 4/3 When you encounter an integral with both square roots and complex denominators like this one, substitution is your key strategy. The presence of both $\sqrt{x}$ and $(1+\sqrt{x})^2$ strongly suggests using $u = \sqrt{x}$. Let's work through this substitution: If $u = \sqrt{x}$, then $x = u^2$ and $dx = 2u \, du$. When $x = 0,u = 0$, and when $x = 4,u = 2$. Substituting into our integral: $\int_0^4 \frac{1}{(1+\sqrt{x})^2 \sqrt{x}} \, dx = \int_0^2 \frac{1}{(1+u)^2 \cdot u} \cdot 2u \, du = \int_0^2 \frac{2}{(1+u)^2} \, du$ Now we have a much simpler integral. Since the antiderivative of $\frac{1}{(1+u)^2}$ is $-\frac{1}{1+u}$, we get: $2 \int_0^2 \frac{1}{(1+u)^2} \, du = 2\left[-\frac{1}{1+u}\right]_0^2 = 2\left(-\frac{1}{3} - (-1)\right) = 2 \cdot \frac{2}{3} = \frac{4}{3}$ So answer B is correct. Answer A ($\frac{2}{3}$) likely comes from forgetting the factor of 2 from the substitution. Answer C ($\frac{1}{2}$) might result from incorrect handling of the limits or the antiderivative. Answer D (2) could come from computational errors in evaluating the definite integral. Study tip: When you see nested expressions involving $\sqrt{x}$, always try $u = \sqrt{x}$ first. This substitution often dramatically simplifies the integrand, turning a seemingly complex problem into a standard form you can easily integrate.
20. Evaluate the definite integral $\int_0^{\sqrt{\pi}} x \cos(x^2) dx$.
    Answer: 0 Let $u = x^2$. Then $du = 2x \, dx$, which means $x \, dx = \frac{1}{2} du$. We must also change the limits of integration. When $x=0$, $u=0^2=0$. When $x=\sqrt{\pi}$, $u=(\sqrt{\pi})^2=\pi$. The integral becomes $\frac{1}{2} \int_0^\pi \cos(u) \, du = \frac{1}{2} [\sin(u)]_0^\pi = \frac{1}{2}(\sin(\pi) - \sin(0)) = \frac{1}{2}(0 - 0) = 0$.

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