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AP Calculus AB : Antiderivatives by substitution of variables

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Example Questions

Example Question #1 : Antiderivatives By Substitution Of Variables

Use a change of variable (aka a u-substitution) to evaluate the integral,

$\int x\sqrt{1+x^2}dx$

Possible Answers:

$\frac{1}{3}x^{\frac{3}{2}}+C$

$\frac{1}{6}(2x+1)^{\frac{2}{3}}+C$

$\frac{3}{4}(x^2+1)^{\frac{3}{2}}+C$

$\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

$\frac{1}{6}(x^2+1)^{\frac{2}{3}}+\sqrt{x}+C$

Correct answer:

$\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

Explanation:

$\int x\sqrt{1+x^2}dx$

Integrals such as this are seen very commonly in introductory calculus courses. It is often useful to look for patterns such as the fact that the polynomial under the radical in our example, 1+x^2 , happens to be one order higher than the factor outside the radical, You know that if you take a derivative of a second order polynomial you will get a first order polynomial, so let's define the variable:

u = 1+x^2 (1)

Now differentiate with respect to to write the differential for ,

du = 2x dx (2)

Looking at equation (2), we can solve for xdx , to obtain $xdx =\frac{1}{2}du$ . Now if we look at the original integral we can rewrite in terms of

$\int x\sqrt{1+x^2}dx = \int\sqrt{1+x^2}\underbrace{xdx} = \int\sqrt{u}\left(\frac{1}{2}du \right )$

Now proceed with the integration with respect to .

$\int\sqrt{u}\left(\frac{1}{2}du \right ) = \frac{1}{2}\int u^{\frac{1}{2}}du$

$= \frac{1}{2}\left[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1} \right ]+C$

$=\frac{1}{2} \left(\frac{2}{3}u^{\frac{3}{2}} \right )+C$

$=\frac{1}{3}u^{\frac{3}{2}}+C$

Now write the result in terms of using equation (1), we conclude,

$\int x\sqrt{1+x^2}dx=\frac{1}{3}(x^2+1)^{\frac{3}{2}}+C$

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Example Question #2 : Antiderivatives By Substitution Of Variables

Use u-subtitution to fine

$\int \sin^2(x)\cos (x)dx$

Possible Answers:

$\frac{1}{6}sin^3(x)cos^2(x)$

$\frac{1}{4}sin^4(x)$

$\frac{-1}{6}cos^3(x)sin^2(x)$

$\frac{1}{6}sin^2(x)$

$\frac{1}{3}sin^3(x)$

Correct answer:

$\frac{1}{3}sin^3(x)$

Explanation:

Let u=sin(x)

Then du=cos(x)dx

Now we can subtitute

$\int sin^2(x) \cos(x) dx=\int u^2du= \frac{1}{3}u^3$

Now we substitute back

$\frac{1}{3}u^3=\frac{1}{3} \sin^3(x)$

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Example Question #3 : Antiderivatives By Substitution Of Variables

Evaluate $\int_0^1 \frac{6x}{1+x^2} dx$

Possible Answers:

$3\ln2$

$\ln3$

$2\ln3$

$\ln2$

Correct answer:

$3\ln2$

Explanation:

We can use substitution for this integral.

Let u = 1+x^2 ,

then du = 2x dx .

Multiplying this last equation by , we get 3du = 6x dx .

Now we can make our substitutions

$\int_0^1 \frac{6x}{1+x^2} dx$ . Start

$=\int_1^2 \frac{3}{u}du$ . Swap out 1+x^2 with , and 6x dx with 3du . Make sure you also plug the bounds on the integral into u = 1+x^2 for to get the new bounds.

$=3\int_1^2 \frac{1}{u}du$ . Factor out the .

$=3 \ln(u)|_1^2$ . Integrate (absolute value signs are not needed since $1\le u \le2$ .)

$=3(\ln(2)-\ln(1))$ . Evaluate

$=3\ln2 -0 = 3\ln2$ .

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Example Question #4 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{x^3(2+x^4)^5}dx$

Possible Answers:

$\frac{1}{6}(2+x^4)^6+c$

$\frac{1}{2}x^4+\frac{1}{24}x^{24}+c$

$2x+x^{20}+c$

$2x^3+x^{23}+c$

$\frac{1}{24}(2+x^4)^6+c$

Correct answer:

$\frac{1}{24}(2+x^4)^6+c$

Explanation:

To solve the integral, we have to simplify it by using a variable u to substitute for a variable of x.

For this problem, we will let u replace the expression 2+x^4 .

Next, we must take the derivative of u. Its derivative is du=4x^3dx .

Next, solve this equation for dx so that we may replace it in the integral.

Plug in place of (2+x^4) and $\frac{du}{4x^3}$ in place of into the original integral and simplify.

The x^3 in the denominator cancels out the remaining x^3 in the integral, leaving behind a $\frac{1}{4}$ . We can pull the $\frac{1}{4}$ out front of the integral. Next, take the anti-derivative of the integrand and replace u with the original expression, adding the constant to the answer.

The specific steps are as follows:

1. $\int{x^3(2+x^4)^5dx}$

2. u=2+x^4

3. du=4x^3dx

4. $dx=\frac{du}{4x^3}$

5. $\int{x^3u^5(\frac{du}{4x^3})}$

6. $\frac{1}{4}\int{u^5}du$

7. $\frac{1}{4}(\frac{1}{6}u^6)$

8. $\frac{1}{4}(\frac{1}{6}(2+x^4)^6$

9. $\frac{1}{24}(2+x^4)^6+c$

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Example Question #5 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{x^2\sqrt{x^3+1}}dx$

Possible Answers:

$\frac{2}{3}(x^3+1)^9+c$

$x^{\frac{7}{2}}+x^2+c$

$\frac{1}{3}(x^3+1)^{\frac{5}{2}}+c$

$\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

$\sqrt[3]{(x^3+1)}(\frac{1}{x^2})+c$

Correct answer:

$\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^3+1 . Next, we must take the derivative of u. Its derivative is 3x^2dx . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^3+1 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The x^2 in the denominator cancels out the remaining x^2 in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{x^2\sqrt{x^3+1}}dx$

2. = x^3+1

3. du=3x^2dx

4. $\int{x^2\sqrt{u}\frac{du}{3x^2}}$

5. $\frac{1}{3}\int{\sqrt{u}du}$

6. $\frac{1}{3}(\frac{2}{3}u^{\frac{3}{2}})+c$

7. $\frac{2}{9}u^{\frac{3}{2}}+c$

8. $\frac{2}{9}(x^3+1)^{\frac{3}{2}}+c$

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Example Question #6 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{cos^3xsinxdx}$

Possible Answers:

$\frac{-cos^4x}{4}+c$

$\frac{-1}{3}(cosx)^3+c$

$\frac{sin^4x}{4}+c$

cotxcos^2x+c

$\frac{1}{3}(cosx)^3+c$

Correct answer:

$\frac{-cos^4x}{4}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression cosx . Next, we must take the derivative of . Its derivative is -sinxdx . Next, solve this equation for so that we may replace it in the integral. Plug in place of cosx and $\frac{du}{-sinx}$ in place of into the original integral and simplify. The sinx in the denominator cancels out the remaining sinx in the integral, leaving behind a . Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{cos^3xsinxdx}$

2. u=cosx

3. du=-sinxdx

4. $\int{u^3sinx(\frac{du}{-sinx})}$

5. $\int-u^3du$

6. $\frac{-1}{4}u^4+c$

7. $\frac{-cos^4x}{4}+c$

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Example Question #7 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{xsin(x^2)dx}$

Possible Answers:

2cos(x^2)+c

$\frac{1}{2}sin(x^2)+c$

$\frac{-1}{2}sin(x^2)+c$

$\frac{1}{2}cos(x^2)+c$

$\frac{-1}{2}cos(x^2)+c$

Correct answer:

$\frac{-1}{2}cos(x^2)+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^2 . Next, we must take the derivative of . Its derivative is . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^2 and $\frac{du}{2x}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{2}$ . We can pull the $\frac{1}{2}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{xsin(x^2)dx}$

2. u=x^2

3. du=2xdx

4. $\int{x(sinu)\frac{du}{2x}}$

5. $\frac{1}{2}\int{sinudu}$

6. $\frac{-1}{2}cosu+c$

7. $\frac{-1}{2}cos(x^2)+c$

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Example Question #2 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{x^2e^{x^3}dx}$

Possible Answers:

$\frac{1}{2}e^{x2}+c$

$\frac{3}{2}e^{x^2}+c$

$\frac{1}{3}e^{x^3}+c$

$\frac{3}{2}e^{x^3}+c$

$2xe^{x^3}+c$

Correct answer:

$\frac{1}{3}e^{x^3}+c$

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression x^3 . Next, we must take the derivative of . Its derivative is 3x^2 . Next, solve this equation for so that we may replace it in the integral. Plug in place of x^3 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The x^2 in the denominator cancels out the remaining x^2 in the integral, leaving behind a $\frac{1}{3}$ . We can pull the $\frac{1}{3}$ out front of the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{x^2e^{x^3}dx}$

2. u=x^3

3. du=3x^2

4. $\int{x^2e^u(\frac{du}{3x^2})}$

5. $\frac{1}{3}\int{e^udu}$

6. $\frac{1}{3}e^u+c$

7. $\frac{1}{3}e^{x^3}+c$

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Example Question #9 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{e^xcos(e^x)dx}$

Possible Answers:

-cos(e^x)+c

-sin(e^x)+c

sin(e^x)+c

$e^{cosx}+c$

$e^{sinx}+c$

Correct answer:

sin(e^x)+c

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression e^x . Next, we must take the derivative of . Its derivative is e^x . Next, solve this equation for so that we may replace it in the integral. Plug in place of e^x and $\frac{du}{e^x}$ in place of into the original integral and simplify. The e^x in the denominator cancels out the remaining e^x in the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{e^xcos(e^x)dx}$

2. u=e^x

3. du=e^xdx

4. $\int{e^xcosu(\frac{du}{e^x})}$

5. $\int{cosudu}$

6. sinu+c

7. sin(e^x)+c

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Example Question #9 : Antiderivatives By Substitution Of Variables

Solve the following integral using substitution:

$\int{\frac{sin(lnx)}{x}dx}$

Possible Answers:

$sinx(\frac{1}{x})+c$

-cos(lnx)+c

cos(lnx)+c

$\frac{1}{\sqrt{1-lnx}}+c$

$-sinx(\frac{1}{x})+c$

Correct answer:

-cos(lnx)+c

Explanation:

To solve the integral, we have to simplify it by using a variable to substitute for a variable of . For this problem, we will let u replace the expression lnx . Next, we must take the derivative of . Its derivative is $\frac{1}{x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of lnx and dux in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral. Next, take the anti-derivative of the integrand and replace with the original expression, adding the constant to the answer. The specific steps are as follows:

1. $\int{\frac{sin(lnx)}{x}dx}$

2. u=lnx

3. $du=(\frac{1}{x}dx)$

4. dux=dx

5. $\int{\frac{sinu}{x}dux}$

6. $\int{sinudu}$

7. -cosu+c

8. -cos(lnx)+c

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AP Calculus AB : Antiderivatives by substitution of variables

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Antiderivatives By Substitution Of Variables

Example Question #2 : Antiderivatives By Substitution Of Variables

Example Question #3 : Antiderivatives By Substitution Of Variables

Example Question #4 : Antiderivatives By Substitution Of Variables

Example Question #5 : Antiderivatives By Substitution Of Variables

Example Question #6 : Antiderivatives By Substitution Of Variables

Example Question #7 : Antiderivatives By Substitution Of Variables

Example Question #2 : Antiderivatives By Substitution Of Variables

Example Question #9 : Antiderivatives By Substitution Of Variables

Example Question #9 : Antiderivatives By Substitution Of Variables

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