Techniques of antidifferentiation - AP Calculus AB

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Question

Solve the following integral using substitution:

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

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Answer

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 . 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 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.

3.

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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Question

Integrate:

$\int \frac{x^2}{x^3+2}dx$

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Answer

To integrate, we must make the following substitution:

Rewriting the integral in terms of u and integrating, we get

$\frac{1}{3} \int \frac{du}{u} = \frac{1}{3}\ln \left | u\right |+C$

The integral was performed using the identical rule.

Finally, replace u with the original x term:

$\frac{1}{3}\ln \left | x^3+2\right |+C$

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Question

Calculate the following integral: $\int \frac{1}{\sqrt[3]{1-4x}}dx$

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Answer

$\int \frac{1}{\sqrt[3]{1-4x}}dx$

Use substitution to solve the integral

Rewrite the integrand as a negative exponent: $\frac{1}{\sqrt[3]{1-4x}}dx=(1-4x)^{\frac{-1}{3}}dx$

Therefore: $\int \frac{1}{\sqrt[3]{1-4x}}dx=\int (1-4x)^{\frac{-1}{3}}dx$

Make the following substitution: $-\frac{1}{4}du=dx$

Apply the substitution to the integrand: $\int (1-4x)^{\frac{-1}{3}}dx=\int \frac{-1}{4}u^{\frac{-1}{3}}du$

Solve the integral: $\int \frac{-1}{4}u^{\frac{-1}{3}}du=(\frac{-1}{4})(\frac{3}{2})u^{\frac{2}{3}}+C=\frac{-3}{8}u^{\frac{2}{3}}+C$

Re-substitute the value of u: $\frac{-3}{8}u^{\frac{2}{3}}+C=\frac{-3}{8}(1-4x)^{\frac{2}{3}}+C=\frac{-3}{8}(\sqrt[3]{1-4x})^{2}+C$

Solution: $\int \frac{1}{\sqrt[3]{1-4x}}dx=\frac{-3}{8}(\sqrt[3]{1-4x})^{2}+C$

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Question

Solve the following integral using substitution:

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

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Answer

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 . 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 and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{3}$ . 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.

3.

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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Question

Integrate:

$\int x \sqrt{9x^2+4}dx$

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Answer

To integrate, the following substitution must be made:

Rewriting the integral in terms of u and integrating, we get

$\frac{1}{18}\int \sqrt{u}du= \frac{2}{3}u^{\frac{3}{2}}+C$

The integration was performed using the following rule:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

Finally, we rewrite our answer in terms of x:

$\frac{2}{3}(9x^2+4)^{\frac{3}{2}}+C$

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Question

Calculate the following integral: $\int sin^{5}xcosxdx$

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Answer

$\int sin^{5}xcosxdx$

Solve by u substitution

Make the following substitution:

Apply the substitution the integral: $\int sin^{5}xcosxdx=\int u^{5}du$

Solve the integral: $\int u^{5}du=\frac{1}{6}u^{6}+C$

Re-substitute the value for u: $\frac{1}{6}sin^{6}x+C$

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Question

Calculate the following integral: $\int ((20e^{2-8y}\sqrt{1+e^{2-8y}}+4y^{3}-5\sqrt[3]{y})dy)$

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Answer

$\int ((20e^{2-8y}\sqrt{1+e^{2-8y}}+4y^{3}-5\sqrt[3]{y})dy)$

Use the addition and subtraction properties of integrals to break the integral into 3 separate integrals: $\int ((20e^{2-8y}\sqrt{1+e^{2-8y}}+4y^{3}-5\sqrt[3]{y})dy)=\int 20e^{2-8y}\sqrt{1+e^{2-8y}}dy +\int 4y^{3}dy-\int -5\sqrt[3]{y}dy$ Evaluate the second integral: $\int 4y^{3}dy=y^{4}+m$

Re-write the third integral as follows: $\int 5\sqrt[3]{y}dy=\int 5y^{\frac{1}{3}}dy=5\int y^{\frac{1}{3}}dy$

Evaluate the third integral: $5\int y^{\frac{1}{3}}dy=5(\frac{3}{4}y^{\frac{4}{3}})+l=\frac{15}{4}y^{\frac{4}{3}}+l$

Make the following substitution to evaluate the first integral: $u=1+e^{2-8y}$ $du=-8(e^{2-8y}dy)$ $\frac{-1}{8}du=(e^{2-8y}dy)$

Apply the substitution to the first integral: $\int ((20e^{2-8y}\sqrt{1+e^{2-8y}}=\int 20(\frac{-1}{8}\sqrt{u}du)=-\frac{5}{2}\int \sqrt{u}du$

Evaluate the integral: $-\frac{5}{2}\int \sqrt{u}du=\frac{-5}{2}(\frac{2}{3})u^{\frac{3}{2}}+k=\frac{-5}{3}u^{\frac{3}{2}}+k$

Substitute the value for u back into the equation: $\frac{-5}{3}u^{\frac{3}{2}}+k=\frac{-5}{3}(1+e^{2-8y})^{\frac{3}{2}}+k$

Combine the answers of the three integrals into one equation: $\frac{-5}{3}(1+e^{2-8y})^{\frac{3}{2}}+k+y^{4}+m-\frac{15}{4}y^{\frac{4}{3}}+l=\frac{-5}{3}(1+e^{2-8y})^{\frac{3}{2}}+y^{4}-\frac{15}{4}y^{\frac{4}{3}}+C$

Solution: $\int ((20e^{2-8y}\sqrt{1+e^{2-8y}}+4y^{3}-5\sqrt[3]{y})dy)=\frac{-5}{3}(1+e^{2-8y})^{\frac{3}{2}}+y^{4}-\frac{15}{4}y^{\frac{4}{3}}+C$

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Question

Integrate:

$\int [\cos(2x)+x^3e^{x^4}]dx$

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Answer

To integrate, we can split the integral into the sum of two separate integrals:

$\int [\cos(2x)+x^3e^{x^4}]dx = \int \cos(2x)dx+ \int x^3e^{x^4}dx$

We can solve them independently and add the results together.

To solve the first integral, we make the following substitution:

Rewriting the integral in terms of u and integrating, we get

$\frac{1}{2}\int \cos(u)du=\frac{\sin(u)}{2}+C$

The integral was found using the identical rule.

Rewriting the integral in terms of x, we get

$\frac{\sin(2x)}{2}+C$

We solve the second integral using the following substitution:

Rewriting the integral in terms of u, we get

$\frac{1}{4} \int e^u du= \frac{e^u}{4}+C$

The integral was performed using the identical rule.

Now, rewrite the answer in terms of x:

$\frac{e^{x^4}}{4}+C$

Combining the results from each integral, we get

$\frac{\sin(2x)}{2}+\frac{e^{x^4}}{4}+C$

Note that the two constants of integration combine to make a single constant.

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Question

Calculate the following integral: $\int-3tanxdx$

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Answer

$\int-3tanxdx$

Factor out the constant: $-3\int tanxdx$

Re-write tanx in terms of sine and cosine: $-3\int \frac{sinx}{cosx}dx=-3\int sinx(cos^{-1}x)dx$

Make the following substitution:

Apply the substitution to the integrand: $-3\int sinx(cos^{-1}x)dx=3\int u^{-1}du$

Evaluate the integral: $3\int u^{-1}du=3ln\left | u\right |+C$

Re-substitute the value for u back into the equation: $3ln\left | u\right |+C=3ln\left | cosx\right |+C$

Solution: $\int-3tanxdx=3ln\left | cosx\right |+C$

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Question

Evaluate the integral $\int \frac{2x[\ln(x^2-4)]^5}{x^2-4}dx$

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Answer

Evaluating integrals requires knowledge of the basic integral forms. In this problem, there is a group raised to an exponent, which is the $[\ln(x^2-4)]^5$ .

This arrangement typically follows the basic integral form $\int u^n du = \frac{1}{n+1}u^{n+1}+ C$ , where is a variable expression, is some constant number and is the constant of integration, which stays unknown. In this case, we can use u-substitution to match this form.

The exponent is 5, so the in the basic integral form will be5.

Since the inside of the exponent group is $\ln(x^2-4)$ , set $u = \ln(x^2-4)$ .

Now we differentiate to find . Recall that the derivative of $\ln(v)= \frac{dv}{v}$ .

$du = \frac{2x}{x^2-4}dx$

Notice that the parts of our match up with the integral we are evaluating, and there are no variables that aren't accounted for. This is clearer if we write out and simplify what our substitution says.

$\int u^n du = \int [\ln(x^2-4)]^5\frac{2x}{x^2-4}dx$

$\int \frac{2x[\ln(x^2-4)]^5}{x^2-4}dx$

This is exactly what we are asked to integrate.

This means we can evaluate the integral by using basic integral form directly. Plugging in our and into the right side of $\int u^n du = \frac{1}{n+1}u^{n+1}+C$ , we get

$\frac{1}{5+1}[\ln(x^2-4)]^{5+1}$

$\frac{1}{6}[\ln(x^2-4)]^6+C$

This is equivalent to the answer $\frac{[\ln(x^2-4)]^6}{6}+C$ , which is the correct answer.

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Question

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

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

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Answer

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

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, , 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:

(1)

Now differentiate with respect to to write the differential for ,

(2)

Looking at equation (2), we can solve for , 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 .

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

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

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

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

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

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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Question

Solve the following integral using substitution:

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

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Answer

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 .

Next, we must take the derivative of u. Its derivative is .

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

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

The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{4}$ . 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.

3.

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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Question

Solve the following integral using substitution:

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

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Answer

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 . Next, we must take the derivative of u. Its derivative is . Next, solve this equation for so that we may replace it in the integral. Plug in place of and $\frac{du}{3x^2}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining in the integral, leaving behind a $\frac{1}{3}$ . 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. =

3.

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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Question

Solve the following integral using substitution:

$\int{cos^3xsinxdx}$

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Answer

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 . 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 and $\frac{du}{-sinx}$ in place of into the original integral and simplify. The in the denominator cancels out the remaining 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.

3.

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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Question

Solve the following integral using substitution:

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

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Answer

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 . 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 and $\frac{du}{e^x}$ 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. $\int{e^xcos(e^x)dx}$

2.

3.

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

5. $\int{cosudu}$

6.

7.

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Question

Solve the following integral using substitution:

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

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Answer

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 . 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 and 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.

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

4.

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

6. $\int{sinudu}$

7.

8.

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Question

Solve the following integral using substitution:

$\int{(x^2+e^{3x})(x^3+e^{3x})^{\frac{6}{7}}dx}$

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Answer

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 $x^3+e^{3x}$ . Next, we must take the derivative of . Its derivative is $3x^2+e^{3x}$ . Next, solve this equation for so that we may replace it in the integral. Plug in place of $x^3+e^{3x}$ and $\frac{du}{3x^2+e^{3x}}$ in place of into the original integral and simplify. The $x^2+e^{3x}$ in the denominator cancels out the remaining $x^2+e^{3x}$ 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. $\int{(x^2+e^{3x})(x^3+e^{3x})^{\frac{6}{7}}dx}$

2. $u=x^3+e^{3x}$

3. $du=(3x^2+3e^{3x})dx$

4. $\int(x^2+e^{3x})u^{\frac{6}{7}(\frac{du}{x^2+e^{3x}})}$

5. $\frac{1}{3}\int{u^{\frac{6}{7}}}du$

6. $\frac{1}{3}(\frac{7}{13})u^{\frac{13}{7}}$

7. $\frac{7}{39}u^{\frac{13}{7}}+c$

8. $\frac{7}{39}(x^3+e^{3x})^{\frac{13}{7}}+c$

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Question

Use u-subtitution to fine

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

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Answer

Let

Then

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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Question

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

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Answer

We can use substitution for this integral.

Let ,

then .

Multiplying this last equation by , we get .

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 with , and with . Make sure you also plug the bounds on the integral into 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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Question

$\int e^{4x-2}dx=?$

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Answer

This is a u-substitution integral. We need to substitute the new function, which is modifying our base function (the exponential).

$\int e^xdx=e^x+C$ , but instead of that, our problem is . We can solve this integral by completing the substitution.

$u=4x-2.; du=4dx. ;dx=\frac{du}{4}.$

Now, we can replace everything in our integrand and rewrite in terms of our new variables:

$\int e^{4x-2}dx=\int e^u\frac{du}{4}=\frac{1}{4}\int e^udu.$

$=\frac{1}{4}e^u+C=\frac{1}{4}e^{4x-2}+C$ .

Remember to plug your variable back in and include the integration constant since we have an indefinite integral.

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