Integration by Substitution - Calculus 2

1. Find $\int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx$
   Answer: $2\sin(\sqrt{x}) + C$ Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$. Also, $\frac{1}{\sqrt{x}} = \frac{1}{u}$. The integral becomes $\int \frac{\cos(u)}{u} \cdot 2u \, du = \int 2\cos(u) \, du = 2\sin(u) + C = 2\sin(\sqrt{x}) + C$. Choice B forgets the factor of 2 from the substitution. Choice C incorrectly integrates cosine as negative cosine instead of sine. Choice D attempts to integrate the original expression term by term without proper substitution.
2. Evaluate the integral $\int \frac{\sin(2x)}{1+\cos^2(x)} dx$.
   Answer: $-\ln(1+\cos^2(x)) + C$ First, use the identity $\sin(2x) = 2\sin(x)\cos(x)$. The integral becomes $\int \frac{2\sin(x)\cos(x)}{1+\cos^2(x)} dx$. Let $u = 1+\cos^2(x)$. Then $du = 2\cos(x)(-\sin(x)) dx = -2\sin(x)\cos(x) dx$. Thus, $-du = 2\sin(x)\cos(x) dx$. The integral transforms to $\int \frac{-du}{u} = -\ln|u| + C$. Substituting back gives $-\ln(1+\cos^2(x)) + C$. The absolute value is not needed as $1+\cos^2(x)$ is always positive.
3. To evaluate the integral $\int \frac{x+2}{\sqrt{x+1}} dx$, a substitution is used. Which of the following is the correct antiderivative?
   Answer: $\frac{2}{3}(x+1)^{3/2} + 2(x+1)^{1/2} + C$ Let $u = x+1$. This implies $du = dx$ and $x = u-1$. The numerator becomes $x+2 = (u-1)+2 = u+1$. The integral transforms to $\int \frac{u+1}{\sqrt{u}} du = \int (u^{1/2} + u^{-1/2}) du$. Integrating term-by-term gives $\frac{u^{3/2}}{3/2} + \frac{u^{1/2}}{1/2} + C = \frac{2}{3}u^{3/2} + 2u^{1/2} + C$. Substituting back $u=x+1$ yields the correct answer.
4. Evaluate the integral $\int \frac{x^2}{\sqrt[3]{x^3+8}} dx$.
   Answer: $\frac{1}{2}(x^3+8)^{2/3} + C$ Let $u = x^3+8$. Then $du = 3x^2 dx$, so $\frac{1}{3}du = x^2 dx$. The integral becomes $\int \frac{1}{\sqrt[3]{u}} \cdot \frac{1}{3}du = \frac{1}{3} \int u^{-1/3} du$. Using the power rule for integration, this is $\frac{1}{3} \frac{u^{2/3}}{2/3} + C = \frac{1}{3} \cdot \frac{3}{2} u^{2/3} + C = \frac{1}{2}u^{2/3} + C$. Substituting back gives $\frac{1}{2}(x^3+8)^{2/3} + C$.
5. Evaluate $\int \frac{1}{x \ln(x^3)} dx$.
   Answer: $\frac{1}{3} \ln|\ln x| + C$ When you encounter an integral with logarithmic functions in both the numerator and denominator, substitution is typically your best approach. The key insight here is recognizing that the derivative of $\ln(x^3)$ appears (almost) in the numerator. Let's use the substitution $u = \ln(x^3)$. First, simplify using logarithm properties: $u = \ln(x^3) = 3\ln(x)$, so $du = 3 \cdot \frac{1}{x} dx$, which means $\frac{1}{x} dx = \frac{1}{3} du$. Substituting into our integral: $\int \frac{1}{x \ln(x^3)} dx = \int \frac{1}{\ln(x^3)} \cdot \frac{1}{x} dx = \int \frac{1}{u} \cdot \frac{1}{3} du = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} \ln|u| + C$ Substituting back: $\frac{1}{3} \ln|\ln(x^3)| + C = \frac{1}{3} \ln|3\ln(x)| + C = \frac{1}{3} \ln|3| + \frac{1}{3} \ln|\ln(x)| + C$ Since $\frac{1}{3} \ln|3|$ is just a constant, we can absorb it into $C$, giving us $\frac{1}{3} \ln|\ln x| + C$. Choice A misses the factor of $\frac{1}{3}$ and incorrectly has coefficient 3. Choice B fails to simplify $\ln(x^3) = 3\ln(x)$ and doesn't account for the substitution factor. Choice C incorrectly attempts integration by parts or treats this as a power function rather than recognizing the logarithmic structure. Strategy tip: When you see nested functions like $f(g(x))$, always check if the derivative of the inner function $g'(x)$ appears elsewhere in the integrand—this signals u-substitution.
6. Evaluate the definite integral $\int_{0}^{1} (1-x)^9 dx$.
   Answer: $\frac{1}{10}$ Let $u = 1-x$. Then $du = -dx$, or $-du = dx$. The limits change: when $x=0$, $u=1$; when $x=1$, $u=0$. The integral becomes $\int_1^0 u^9 (-du) = \int_0^1 u^9 du$. This evaluates to $[\frac{u^{10}}{10}]_0^1 = \frac{1^{10}}{10} - \frac{0^{10}}{10} = \frac{1}{10}$.
7. Find the antiderivative of $f(x) = x^2 \sin(x^3+1)$.
   Answer: $-\frac{1}{3}\cos(x^3+1) + C$ To evaluate $\int x^2 \sin(x^3+1) dx$, let $u = x^3+1$. Then $du = 3x^2 dx$, which means $\frac{1}{3}du = x^2 dx$. The integral becomes $\int \sin(u) \cdot \frac{1}{3} du = \frac{1}{3} \int \sin(u) du$. This evaluates to $\frac{1}{3}(-\cos(u)) + C = -\frac{1}{3}\cos(x^3+1) + C$.
8. The substitution $u=\sqrt{x}-3$ transforms a definite integral $\int_a^b f(x) dx$ into $\int_{-2}^{-1} 2(u+3)e^u du$. What was the original integral?
   Answer: $\int_1^4 e^{\sqrt{x}-3} dx$ From $u = \sqrt{x}-3$, we find $\sqrt{x} = u+3$ and $x=(u+3)^2$. Then $dx = 2(u+3)du$. The transformed integral is $\int_{-2}^{-1} e^u \cdot 2(u+3)du$. The part $2(u+3)du$ corresponds to $dx$, so the original function $f(x)$ must correspond to $e^u$. Substituting $u=\sqrt{x}-3$ gives $f(x) = e^{\sqrt{x}-3}$. For the limits: if $u=-2$, then $-2=\sqrt{a}-3 \implies \sqrt{a}=1 \implies a=1$. If $u=-1$, then $-1=\sqrt{b}-3 \implies \sqrt{b}=2 \implies b=4$. Thus, the original integral was $\int_1^4 e^{\sqrt{x}-3} dx$.
9. Evaluate $\int \frac{x+1}{x^2+1} dx$.
   Answer: $\frac{1}{2}\ln(x^2+1) + \arctan(x) + C$ This integral should be split into two parts: $\int \frac{x}{x^2+1} dx + \int \frac{1}{x^2+1} dx$. The second integral is the standard form for $\arctan(x)$. For the first integral, use the substitution $u=x^2+1$, so $du=2xdx$ or $\frac{1}{2}du=xdx$. This integral becomes $\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+1)$. Combining the two parts gives $\frac{1}{2}\ln(x^2+1) + \arctan(x) + C$.
10. To evaluate $\int x^3 \sqrt{x^2+1} \, dx$, a helpful substitution is $u=x^2+1$. Which of the following is the correct antiderivative?
    Answer: $\frac{1}{15}(3x^2-2)(x^2+1)^{3/2} + C$ Let $u=x^2+1$, so $x^2=u-1$ and $du=2xdx$, which means $\frac{1}{2}du=xdx$. Rewrite the integral as $\int x^2 \sqrt{x^2+1} \cdot x \, dx$. Substitute to get $\int (u-1)\sqrt{u} \cdot \frac{1}{2}du = \frac{1}{2} \int (u^{3/2} - u^{1/2}) du$. Integrating gives $\frac{1}{2} [\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}] + C = \frac{1}{5}u^{5/2} - \frac{1}{3}u^{3/2} + C$. Substitute back $u=x^2+1$ to get $\frac{1}{5}(x^2+1)^{5/2} - \frac{1}{3}(x^2+1)^{3/2} + C$. This can be simplified by factoring out $\frac{1}{15}(x^2+1)^{3/2}$, which gives $\frac{1}{15}(x^2+1)^{3/2} [3(x^2+1) - 5] + C = \frac{1}{15}(x^2+1)^{3/2} [3x^2 - 2] + C$. Choice D is this simplified form.
11. Which of the following is an antiderivative of $\tan(x)$?
    Answer: $\ln|\sec(x)| + C$ To integrate $\tan(x)$, rewrite it as $\int \frac{\sin(x)}{\cos(x)} dx$. Let $u = \cos(x)$, so $du = -\sin(x) dx$. The integral becomes $\int \frac{-du}{u} = -\ln|u| + C = -\ln|\cos(x)| + C$. Using logarithm properties, this is equivalent to $\ln|(\cos(x))^{-1}| + C = \ln|\sec(x)| + C$.
12. Find the value of $\int_1^e \frac{(\ln x)^3}{x} dx$.
    Answer: $\frac{1}{4}$ When you encounter an integral with a composite function like $(\ln x)^3$ divided by $x$, this is a perfect setup for u-substitution. The key insight is recognizing that the derivative of $\ln x$ is $\frac{1}{x}$, which appears in the denominator. Let $u = \ln x$, so $du = \frac{1}{x} dx$. When $x = 1$, $u = \ln 1 = 0$. When $x = e$, $u = \ln e = 1$. The integral becomes: $\int_0^1 u^3 du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$ Choice A ($1$) represents the error of forgetting to divide by 4 when integrating $u^3$. Choice B ($\frac{1}{4}e^4$) occurs if you mistakenly think the upper limit contributes $e^4$ instead of $(\ln e)^4 = 1^4$. Choice C ($\frac{e^4-1}{4}$) results from incorrectly using the original limits of integration (1 to $e$) instead of converting them to the new variable's limits (0 to 1). The correct answer is D ($\frac{1}{4}$). Study tip: When you see $\frac{f'(x) \cdot g(f(x))}{denominators}$ patterns, immediately consider u-substitution with $u = f(x)$. Always remember to convert your limits of integration to match your new variable – this prevents the common error of mixing old and new variables in your final evaluation.
13. Evaluate $\int \frac{2x+3}{x^2+3x+5} dx$.
    Answer: $\ln|x^2+3x+5| + C$ This integral has the form $\int \frac{g'(x)}{g(x)} dx$. Let $u = x^2+3x+5$. Then $du = (2x+3) dx$. The integral becomes $\int \frac{1}{u} du$, which is $\ln|u| + C$. Substituting back gives $\ln|x^2+3x+5| + C$. Since the discriminant of the quadratic is $3^2 - 4(1)(5) = 9-20 < 0$, the quadratic is always positive, so the absolute value is not strictly necessary.
14. Which expression represents the integral $\int \frac{x}{\sqrt{1-x^4}} dx$?
    Answer: $\frac{1}{2}\arcsin(x^2) + C$ The denominator suggests an arcsin form. Let $u = x^2$. Then $du = 2x dx$, so $\frac{1}{2}du = x dx$. The integral becomes $\int \frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2}du = \frac{1}{2} \int \frac{1}{\sqrt{1-u^2}} du$. The antiderivative is $\frac{1}{2} \arcsin(u) + C$. Substituting back for $u$ yields $\frac{1}{2}\arcsin(x^2) + C$.
15. Find $\int \frac{\sec^2(x)}{\tan^2(x) + 4\tan(x) + 3} \, dx$
    Answer: $\frac{1}{2}\ln|\frac{\tan(x) + 1}{\tan(x) + 3}| + C$ Let $u = \tan(x)$, so $du = \sec^2(x) dx$. The integral becomes $\int \frac{1}{u^2 + 4u + 3} du$. Factor the denominator: $u^2 + 4u + 3 = (u + 1)(u + 3)$. Using partial fractions: $\frac{1}{(u+1)(u+3)} = \frac{A}{u+1} + \frac{B}{u+3}$. Solving: $1 = A(u+3) + B(u+1)$. Setting $u = -1$: $1 = 2A$, so $A = \frac{1}{2}$. Setting $u = -3$: $1 = -2B$, so $B = -\frac{1}{2}$. Therefore: $\int (\frac{1/2}{u+1} - \frac{1/2}{u+3}) du = \frac{1}{2}\ln|u+1| - \frac{1}{2}\ln|u+3| + C = \frac{1}{2}\ln|\frac{u+1}{u+3}| + C = \frac{1}{2}\ln|\frac{\tan(x)+1}{\tan(x)+3}| + C$. Choice A omits the $\frac{1}{2}$ coefficient. Choice C has coefficient 1 instead of $\frac{1}{2}$. Choice D shows the individual logarithm terms but is equivalent to choice B.
16. If the substitution $u = 5x+2$ is used to evaluate the definite integral $\int_0^1 \sqrt{5x+2} \, dx$, what are the resulting limits of integration?
    Answer: From 2 to 7 When using u-substitution for definite integrals, the limits of integration must be converted to u-values. The original limits are for $x$. The lower limit is $x=0$, so the new lower limit is $u = 5(0) + 2 = 2$. The upper limit is $x=1$, so the new upper limit is $u = 5(1) + 2 = 7$.
17. After applying the substitution $u = e^x$, which integral is equivalent to $\int \frac{e^x}{1+e^{2x}} dx$?
    Answer: $\int \frac{1}{1+u^2} du$ Let $u = e^x$. Then $du = e^x dx$. Also, note that $e^{2x} = (e^x)^2 = u^2$. Substituting these into the original integral gives $\int \frac{1}{1+u^2} du$.
18. Evaluate the definite integral $\int_0^2 \frac{x}{\sqrt{9-x^2}} dx$.
    Answer: $3 - \sqrt{5}$ Let $u = 9-x^2$. Then $du = -2x dx$, so $-\frac{1}{2}du = x dx$. The limits of integration change: when $x=0$, $u=9-0^2=9$; when $x=2$, $u=9-2^2=5$. The integral becomes $\int_9^5 \frac{1}{\sqrt{u}} (-\frac{1}{2}du) = \frac{1}{2} \int_5^9 u^{-1/2} du$. Integrating gives $\frac{1}{2}[2u^{1/2}]_5^9 = [\sqrt{u}]_5^9 = \sqrt{9}-\sqrt{5} = 3-\sqrt{5}$.
19. Evaluate the definite integral $$\int_0^{\pi/4} \sec^2(x) e^{\tan(x)} dx.$$
    Answer: $e-1$ Let $u = \tan(x)$. Then $du = \sec^2(x) dx$. We must also change the limits of integration. When $x=0$, $u = \tan(0) = 0$. When $x=\pi/4$, $u = \tan(\pi/4) = 1$. The integral becomes $\int_0^1 e^u du$, which evaluates to $[e^u]_0^1 = e^1 - e^0 = e - 1$.
20. Evaluate $\int \frac{x^3}{\sqrt{x^2 + 4}} \, dx$
    Answer: $\frac{1}{3}(x^2 + 4)^{3/2} - 4\sqrt{x^2 + 4} + C$ Using substitution $u = x^2 + 4$, so $du = 2x \, dx$ and $x^2 = u - 4$. The integral becomes $\int \frac{x^2 \cdot x}{\sqrt{x^2 + 4}} \, dx = \frac{1}{2} \int \frac{u - 4}{\sqrt{u}} \, du = \frac{1}{2} \int (u^{1/2} - 4u^{-1/2}) \, du = \frac{1}{2}[\frac{2}{3}u^{3/2} - 8u^{1/2}] + C = \frac{1}{3}(x^2 + 4)^{3/2} - 4\sqrt{x^2 + 4} + C$. Choice B has an incorrect coefficient (-2 instead of -4). Choice C has an incorrect first coefficient (2/3 instead of 1/3). Choice D has an incorrect first coefficient (1/2 instead of 1/3).

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