Consider Again the Condition of Exercise 14 in Sec. 2.1

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Subsection two.1.i Substitution Rule for Indefinite Integrals

Needless to say, near integration problems we will encounter will not exist so uncomplicated. That is to say nosotros will require more than the bones integration rules we have seen. Here's a slightly more complicated example: Notice

\begin{equation*} \int 2x\cos(ten^ii)\,dx\text{.} \finish{equation*}

This is non a "simple" derivative, but a fiddling thought reveals that it must accept come from an application of the Chain Rule. Multiplied on the "outside" is \(2x\text{,}\) which is the derivative of the "inside" function \(\ds x^two\text{.}\) Checking:

\begin{equation*} {d\over dx}\sin(x^2) = \cos(10^ii){d\over dx}x^2 = 2x\cos(x^2) \end{equation*}

so

\begin{equation*} \int 2x\cos(x^2)\,dx=\sin(x^2)+C\text{.} \stop{equation*}

To summarize: If we suspect that a given role is the derivative of another via the Concatenation Rule, we let \(u\) announce a likely candidate for the inner function, then translate the given office and then that it is written entirely in terms of \(u\text{,}\) with no \(x\) remaining in the expression. If we can integrate this new function of \(u\text{,}\) then the antiderivative of the original function is obtained by replacing \(u\) by the equivalent expression in \(x\text{.}\) This result leads u.s. to the following theorem.

Theorem ii.1. Substitution Rule for Indefinite Integrals.

If \(u=k(x)\) is a differentiable function whose range is an interval \(I\) and \(f\) is continuous on \(I\text{,}\) then

\brainstorm{equation*} \int f(one thousand(10))grand'(x)\,dx=\int f(u)\,du\text{.} \end{equation*}

Nosotros can describe two methods how the Commutation Rule may unfold in an integration process.

Method ane: Even in simple cases y'all may prefer to apply this mechanical process, since information technology often helps to avoid light-headed mistakes. For instance, consider again this uncomplicated problem:

\begin{equation*} \int 2x\cos(x^2)\,dx\text{.} \cease{equation*}

Let \(\ds u=x^2\text{,}\) then \(du/dx = 2x\) or \(du = 2x\,dx\text{.}\) Since we take exactly \(2x\,dx\) in the original integral, nosotros can supplant it by \(du\text{:}\)

\begin{equation*} \int 2x\cos(ten^ii)\,dx=\int \cos u\,du=\sin u +C = \sin(x^two)+C\text{.} \end{equation*}

Method 2: This is not the merely way to do the algebra, and typically at that place are many paths to the correct answer. For case, since \(du/dx = 2x\text{,}\) we have that \(dx=du/2x\text{,}\) and so the integral becomes

\begin{equation*} \int 2x\cos(x^2)\,dx=\int 2x\cos u\,{du\over 2x}=\int \cos u\,du\text{.} \end{equation*}

Guideline for Substitution Dominion.

Given the integral

\begin{equation*} I=\int f(g(ten))g'(ten)\,dx \stop{equation*}

with \(f\) continuous and \(chiliad\) differentiable, the following steps outline the Substitution Rule process for integrating \(I\text{.}\)

1. Let \(u=g(ten)\text{,}\) which is typically the inside office of the function composition in the integrand.

2. Compute \(du=g'(10)dx\text{.}\)

3. Write the integral \(I\) solely in terms of \(u\) using either method:

   1. use the substitution \(u=g(x)\) and \(du=g'(x)dx\text{;}\) or

   2. replace \(dx\) with \(du/g'(ten)\) and cancel \(g'(10)\text{.}\)

4. Integrate with respect to \(u\text{.}\)

5. Replace \(u\) by \(g(ten)\) to write the event in terms of \(x\) only.

Note:

1. Yet the Exchange Rule unfolds in the solution process, the important part is that y'all must eliminate all instances of the original variable, often \(ten\text{,}\) during the process and so write your result dorsum in terms of the original variable \(x\text{.}\)

2. Sometimes, the integrand has to be rearranged to see whether the Substitution Dominion is a possible integration technique.

3. If a get-go substitution did not piece of work out, and then attempt to simplify or rearrange the integrand to see if a different substitution tin can exist used.

4. As a general guideline for the Commutation Dominion, we expect for the within part \(u\) to exist

   • the radicand under a root: east.g.,

     \brainstorm{equation*} \text{ when } \int x^3\sqrt{ten^2-5}\,dx \text{ we choose } u=x^2-5; \cease{equation*}

   • the base of operations in a power with a real exponent: eastward.g.,

     \begin{equation*} \text{ when } \int x\left(10^2-5\correct)^v\,dx \text{ nosotros choose } u=x^2-5; \end{equation*}

   • the exponent in a power with a existent base: e.g.,

     \begin{equation*} \text{ when } \int x5^{x^2-v}\,dx \text{ we cull } u=x^2-5; \end{equation*}

   • the denominator in a fraction: e.g.,

     \begin{equation*} \text{ when } \int \frac{ten}{x^2-5}\,dx \text{ we choose } u=x^2-5\text{.} \finish{equation*}

However, sometimes \(u\) can be something different than is suggested above, then be open up minded about this process.

Example 2.2. Substitution Rule.

Evaluate \(\ds\int(ax+b)^n\,dx\text{,}\) assuming \(a,b\) are constants, \(a\not=0\text{,}\) and \(n\) is a positive integer.

Solution

We let \(u=ax+b\) so \(du=a\,dx\) or \(dx=du/a\text{.}\) Then

\begin{equation*} \begin{split} \int(ax+b)^n\,dx\amp=\int {i\over a} u^n\,du\\ \amp={one\over a(n+1)}u^{n+i}+C\\ \amp = {1\over a(n+one)}(ax+b)^{n+1}+C. \end{split} \terminate{equation*}

Instance 2.3. Exchange Rule.

Evaluate \(\ds\int \sin(ax+b)\,dx\text{,}\) assuming that \(a\) and \(b\) are constants and \(a\non=0\text{.}\)

Solution

Once again we permit \(u=ax+b\) so \(du=a\,dx\) or \(dx=du/a\text{.}\) Then

\begin{equation*} \brainstorm{divide} \int\sin(ax+b)\,dx\amp=\int {1\over a} \sin u\,du\\ \amp={1\over a}(-\cos u)+C\\ \amp= -{ane\over a}\cos(ax+b)+C.\end{split} \finish{equation*}

Example 2.4. Substitution in Denominator.

Evaluate the following integral: \(\ds\int \frac{2y}{\sqrt{1-4y^two}}\,dy\text{.}\)

Solution

We try the exchange:

\begin{equation*} u=one-4y^2\text{.} \end{equation*}

And then,

\begin{equation*} du=-8y~dy \cease{equation*}

In the numerator nosotros take \(2y~dy\text{,}\) and so rewriting the differential gives:

\brainstorm{equation*} -\frac{i}{iv}du=2y~dy\text{.} \finish{equation*}

Then the integral is:

\brainstorm{equation*} \begin{split} \int \frac{2y}{\sqrt{1-4y^2}}\,dy\amp = \int \left(i-4y^2\right)^{-i/2}(2y~dy)\\ \amp = \int u^{-ane/two}\left(-\frac{1}{four}du\right)\\ \amp =\left(\frac{-1}{4}\right)\frac{u^{1/two}}{1/two}+C\\ \amp =-\frac{\sqrt{1-4y^2}}{2}+C\finish{separate} \end{equation*}

Example 2.v. Substitution in Base.

Evaluate the following integral: \(\ds\int \cos ten(\sin x)^5\,dx\text{.}\)

Solution

In this question we will let \(u=\sin 10\text{.}\) Then,

\begin{equation*} du=\cos x~dx\text{.} \end{equation*}

Thus, the integral becomes:

\brainstorm{equation*} \begin{split up} \int \cos x(\sin x)^5\,dx\amp =\int u^5\,du\\ =\amp \frac{u^half-dozen}{6}+C\\ =\amp \frac{(\sin 10)^6}{six}+C\end{split} \end{equation*}

Example 2.6. Substitution.

Evaluate the following integral: \(\ds\int \frac{\cos(\sqrt ten)}{\sqrt x}\,dx\text{.}\)

Solution

We use the substitution:

\begin{equation*} u=x^{ane/2}\text{.} \end{equation*}

Then,

\begin{equation*} du=\frac{1}{2}10^{-one/2}\,dx\text{.} \end{equation*}

Rewriting the differential we become:

\brainstorm{equation*} ii~du=\frac{1}{\sqrt x}~dx\text{.} \end{equation*}

The integral becomes:

\brainstorm{equation*} \begin{carve up} \int \frac{\cos(\sqrt x)}{\sqrt x}\,dx\amp = 2\int \cos u\,du\\ =\amp two\sin u+C\\ =\amp ii\sin(\sqrt x)+C\terminate{split} \end{equation*}

Example 2.7. Exchange.

Evaluate the following integral: \(\ds\int 2x^3\sqrt{x^2+1}\,dx\text{.}\)

Solution

This problem is a lilliputian bit different than the previous ones. It makes sense to let:

\brainstorm{equation*} u=x^2+one\text{,} \end{equation*}

and then

\begin{equation*} du=2x~dx\text{.} \end{equation*}

Making this exchange gives:

\brainstorm{equation*} \begin{split} \int 2x^3\sqrt{x^2+one}\,dx\amp = \int 10^two\sqrt{x^2+ane}(2x)\,dx\\ =\amp \int ten^2u^{1/2}\,du\end{split} \end{equation*}

This is a problem because our integrals tin't have a mixture of two variables in them. Usually this ways we chose our \(u\) incorrectly. Still, in this case we can eliminate the remaining \(x\)'s from our integral past using:

\brainstorm{equation*} u=x^2+one \implies x^ii=u-one\text{.} \stop{equation*}

Nosotros become:

\begin{equation*} \int x^2u^{1/2}\,du = \int (u-1)u^{1/2}\,du\text{.} \end{equation*}

Now we proceed by simplifying the integrand and noticing we are left with ability functions, which are readily integrated.

\begin{equation*} \int (u-1)u^{1/ii}\,du=\int u^{iii/ii}-u^{one/ii}\,du=\frac{two}{5}u^{5/two}-\frac{ii}{three}u^{3/2}+C\text{.} \end{equation*}

Therefore, our original integral becomes

\begin{equation*} \int 2x^3\sqrt{ten^2+1}\,dx=\frac{2}{5}(ten^two+1)^{v/2}-\frac{2}{iii}(10^two+one)^{3/ii}+C \terminate{equation*}

Instance two.8. Two Choices for Substitution.

Evaluate \(\ds \int \frac{2x}{\sqrt[3]{ten^ii-5}}\,dx\text{.}\)

Solution

Method 1: Let \(u=10^ii-5\text{,}\) \(du=2xdx\text{.}\)

\begin{equation*} \begin{split} \int \frac{2x}{\sqrt[3]{10^2-v}}\,dx \amp = \int \frac{du}{u^{1/3}} \\[1ex] \amp = \frac{u^{two/three}}{ii/three} + C \\[1ex] \amp = \frac{iii}{2} u^{ii/three} + C \\[1ex] \amp = \frac{iii}{2} \left(x^2-5\right)^{two/3} + C \end{split} \end{equation*}

Method two: Let \(u= \sqrt[iii]{x^2-five}\text{,}\) \(u^iii = x^2-5\text{,}\) \(3u^2du=2xdx\text{.}\)

\begin{equation*} \begin{split} \int \frac{2x}{\sqrt[3]{10^two-5}}\,dx \amp = \int \frac{3u^2du}{u} \\[1ex] \amp = three\int u du\\[1ex] \amp = 3\frac{u^2}{2} + C \\[1ex] \amp = \frac{3}{two} \left(x^ii-5\right)^{2/3} + C \stop{dissever} \end{equation*}

Example ii.9. Application in Sales Projection.

An airline wants to predict the number of tickets on a particular route that will be sold. They find that over the adjacent year, ticket sales growth after \(t\) weeks tin be modelled by

\brainstorm{equation*} g-500e^{-0.025t} \end{equation*}

tickets per week for \(t \in [0,52]\text{.}\)

1. Determine the predicted total number of tickets sold later the first \(t\) weeks.

2. Determine the predicted total number of tickets sold in the next year.

Solution

1. Let \(Northward(t)\) exist the total number of tickets predicted to be sold afterwards the first \(t\) months. We are given

   \begin{equation*} N'(t)= one thousand-500e^{-0.025t}\text{.} \terminate{equation*}

   Thus,

   \brainstorm{equation*} N(t) = \int \left(grand-500e^{-0.025t}\right)\,dt\text{.} \end{equation*}

   We integrate by substitution. Let \(u=-0.025t\text{,}\) \(du=-0.025 dt\text{:}\)

   \begin{equation*} \begin{split} Due north(t) \amp = \int \left(1000-500e^{-0.025t}\correct)\,dt \\[1ex] \amp = \int g dt + \frac{500}{0.025} \int e^{u}\,du \\[1ex] \amp = 1000t + xx,000e^{-0.025t} + C \end{split} \end{equation*}

   Information technology remains to make up one's mind the value of \(C\text{.}\) We crave

   \brainstorm{equation*} N(0) = 0 \implies C=-20,000\text{.} \end{equation*}

   Therefore, the total number of tickets sold afterwards the first \(t\) months is predicted to be

   \begin{equation*} Northward(t)=1000t+20,000e^{-0.025t}-twenty,000\text{.} \stop{equation*}

2. Since

   \begin{equation*} N(52)=1000(52)+20,000e^{-0.025(52)}-20,000 = 37450.64\text{,} \stop{equation*}

   we have that the model predicts that 37,451 tickets will be sold on this route in the side by side yr.

Subsection 2.i.two Substitution Dominion for Definite Integrals

The next example shows how to apply the Substitution Rule when dealing with definite integrals.

Example ii.x. Substitution Rule with Two Methods.

Evaluate \(\ds\int_2^4 x\sin(x^2)\,dx\text{.}\)

Solution

Method 1: First Compute the Antiderivative, then Evaluate the Definite Integral.

Let \(\ds u=ten^2\text{,}\) and then \(du=2x\,dx\) or \(x\,dx=du/2\text{.}\) Then

\begin{equation*} \int x\sin(x^2)\,dx=\int {i\over 2} \sin u\,du={ane\over two}(-\cos u)+C= -{1\over 2}\cos(x^2)+C\text{.} \finish{equation*}

At present

\brainstorm{equation*} \begin{split} \int_2^4 x\sin(10^2)\,dx\amp=\left.-{1\over two}\cos(10^2)\correct|_2^4\\ \amp =-{1\over 2}\cos(16)+{1\over ii}\cos(four)\end{split}. \finish{equation*}

Method two: Changing the Limits of Integration with the Substitution.

A somewhat neater culling to this method is to change the original limits to friction match the variable \(u\text{.}\) Since \(\ds u=ten^two\text{,}\) when \(x=2\text{,}\) \(u=4\text{,}\) and when \(x=four\text{,}\) \(u=xvi\text{.}\) So we tin can do this:

\brainstorm{equation*} \brainstorm{split} \int_2^4 ten\sin(x^2)\,dx \amp= \int_4^{16} {ane\over 2} \sin u\,du\\ \amp =\left.-{one\over two}(\cos u)\right|_4^{16} \\ \amp =-{1\over two}\cos(xvi)+{1\over 2}\cos(4).\end{split up} \end{equation*}

An wrong, and dangerous, culling is something like this:

\brainstorm{equation*} \brainstorm{split up} \int_2^four x\sin(x^two)\,dx\amp =\int_2^4 {1\over ii} \sin u\,du\\ \amp= \left.-{1\over 2}\cos (u)\right|_2^four\\ \amp= \left.-{i\over 2}\cos(x^2)\correct|_2^4\\ \amp =-{1\over ii}\cos(16)+{1\over two}\cos(4).\end{carve up} \finish{equation*}

This is wrong because \(\ds\int_2^iv {1\over ii} \sin u\,du\) ways that \(u\) takes on values between 2 and 4, which is wrong. Information technology is unsafe, because it is very piece of cake to get to the point \(\ds\left.-{1\over two}\cos (u)\right|_2^4\) and forget to substitute \(\ds ten^2\) dorsum in for \(u\text{,}\) thus getting the incorrect respond \(\ds -{1\over 2}\cos(four)+{1\over 2}\cos(2)\text{.}\) An acceptable culling is to conspicuously indicate that the limit exchange is to be done with respect to the \(x\) variable, using something like:

\brainstorm{equation*} \brainstorm{split} \int_2^4 x\sin(x^2)\,dx\amp=\int_{x=2}^{x=4} {1\over 2} \sin u\,du\\ \amp = \left.-{1\over 2}\cos (u)\correct|_{ten=2}^{ten=4}\\ \amp= \left.-{1\over two}\cos(10^2)\right|_2^4\\ \amp=-{\cos(sixteen)\over 2}+{\cos(4)\over2}.\end{separate} \cease{equation*}

To summarize, we have the following.

Theorem 2.11. Substitution Rule for Definite Integrals.

If \(g'\) is continuous on \([a,b]\) and \(f\) is continuous on the range of \(u=chiliad(x)\text{,}\) then

\begin{equation*} \int_a^b f(chiliad(x))g'(x)\,dx=\int_{g(a)}^{one thousand(b)}f(u)\,du\text{.} \cease{equation*}

Notation: When using the Substitution Rule for integrating definite integrals, it is important to modify the limits of integration from those of the original part to those of the substituted part. Otherwise, the definite integral volition evaluate to an wrong result.

Instance 2.12. Commutation Rule for Definite Integrals.

Evaluate \(\ds\int_{1/4}^{1/2}{\cos(\pi t)\over\sin^2(\pi t)}\,dt\text{.}\)

Solution

Allow \(u=\sin(\pi t)\) so \(du=\pi\cos(\pi t)\,dt\) or \(du/\pi=\cos(\pi t)\,dt\text{.}\) We change the limits to \(\ds u(1/4)=\sin(\pi/four)=\sqrt2/two\) and \(u(1/2)=\sin(\pi/two)=1\text{.}\) And then

\begin{equation*} \begin{split} \int_{1/4}^{1/2}{\cos(\pi t)\over\sin^two(\pi t)}\,dt \amp= \int_{\sqrt2/ii}^{1}{1\over \pi}{1\over u^2}\,du\\ \amp= \int_{\sqrt2/two}^{1} {ane\over \pi}u^{-2}\,du\\ \amp = \left.{ane\over \pi}{u^{-i}\over -one}\right|_{\sqrt2/ii}^{1}\\ \amp= -{one\over\pi}+{\sqrt2\over\pi}.\end{carve up} \cease{equation*}

Case two.13. Exchange Dominion for Definite Integrals.

Evaluate \(\ds\int_{-1}^{i} \left(y+ane\right)\left(y^2+2y\right)^8dy\text{.}\)

Solution

Although we could expand the integrand, since this would yield powers of \(y\) which we tin certainly integrate without using the Substitution Rule at all, the exponent 8 would make this a rather messy process that is surely prone to errors. Instead we proceed with the obvious choice of substitution and let \(u=y^two+2y\text{,}\) and so

\begin{equation*} dy=(2y+2)dy = ii(y+1)dy \implies \frac{du}{two}=(y+1)dy \end{equation*}

with limits of integration

\begin{equation*} u(-1)=(-1)^two+2(-i)=-ane \text{ and } u(1)=(i)^two+2(1)=three\text{.} \end{equation*}

Take notation that sometimes the value of a limit of integration does non change. The point is that one nonetheless needs to substitute the values of the original variable of integration, in this example \(y\text{,}\) to piece of work with the limits of integration for the substituted variable, in this case \(u\text{.}\)

Then the integral evaluates as follows

\begin{equation*} \begin{split} \int_{-ane}^{ane} \left(y+1\right)\left(y^2+2y\right)^8dy \amp= \int_{-ane}^{3}\frac{u^8}{2}du\\ \amp = \frac{u^ix}{18}\bigg\vert_{-1}^3\\ \amp= \frac{3^9}{18}-\frac{(-1)^nine}{18}\\ \amp = \frac{xix,684}{18}.\terminate{split} \end{equation*}

Exercises for Section ii.ane.

Do two.1.ane.

Evaluate the post-obit indefinite integrals.

1. \(\ds\int (1-t)^9\,dt\)

   Reply Solution

   Let \(u=ane-t\text{.}\) Then \(du = -dt\) and

   \begin{equation*} \int (1-t)^nine\,dt = -\int u^9 \,du = -\frac{i}{x} u^{10} + C = -\frac{(1-t)^{10}}{10} + C\text{.} \end{equation*}

2. \(\ds\int (x^two+1)^2\,dx\)

   Reply Solution

   First, try \(u=ten^2+1\text{.}\) Then \(du = 2x\,dx\) and

   \brainstorm{equation*} \int (x^two+1)^2\,dx = \int u^2 \frac{du}{2x}\text{,} \stop{equation*}

   and and so this does choice of commutation will not piece of work. Instead, we solve directly:

   \begin{equation*} \int (x^ii+1)^two\,dx = \int (ten^4+2x^two+i)\,dx = \frac{x^5}{five}+\frac{2x^three}{3}+10+C\text{.} \end{equation*}

3. \(\ds\int x(x^2+ane)^{100}\,dx\)

   Respond

   \(\ds (ten^2+1)^{101}/202+C\)

   Solution

   Allow \(u=x^2+1\text{.}\) Then \(du = 2x\,dx\) and

   \brainstorm{equation*} \int 10(x^2+1)^{100}\,dx = \int xu^{100} \, \frac{du}{2x} = \frac{i}{ii} \int u^{100}\,du\text{.} \finish{equation*}

   This is now an integral we tin solve:

   \begin{equation*} \int u^{100}\,du = \frac{1}{101} u^{101} + C\text{,} \terminate{equation*}

   and then nosotros have

   \begin{equation*} \int 10(x^2+one)^{100}\,dx = \frac{1}{202} (x^2+1)^{101} + C\text{.} \finish{equation*}

4. \(\ds\int {1\over\root 3 \of {1-5t}}\,dt\)

   Answer

   \(\ds -3(1-5t)^{2/3}/x+C\)

   Solution

   Let \(u=1-5t\text{.}\) Then \(du= -five\,dt\) and

   \begin{equation*} \int \frac{1}{\sqrt[iii]{1-5t}}\,dt = \int \frac{1}{-5\sqrt[3]{u}}\,du\text{.} \cease{equation*}

   We now carry out the integration with respect to \(u\text{,}\) and and then rewrite our respond in terms of \(t\text{:}\)

   \begin{equation*} \int \frac{1}{-v\sqrt[3]{u}}\,du = \frac{-3}{10}u^{two/three}+C = \frac{-3}{ten}(i-5t)^{2/iii}+C\text{.} \end{equation*}

5. \(\ds\int \sin^3x\cos x\,dx\)

   Respond Solution

   Let \(u=\sin x\text{.}\) Then \(du = \cos x\,dx\) and

   \begin{equation*} \begin{split} \int \sin^3 10\cos x\,dx \amp = \int u^iii \,du\\ \amp = \frac{one}{4} u^iv + C\\ \amp = \frac{1}{4} \sin^4x + C. \stop{split up} \end{equation*}

6. \(\ds\int x\sqrt{100-ten^ii}\,dx\)

   Answer

   \(\ds -(100-x^two)^{three/ii}/3+C\)

   Solution

   Permit \(u=100-x^2\text{.}\) Then \(du = -2x\,dx\) and

   \brainstorm{equation*} \begin{split} \int x \sqrt{100-x^2}\,dx \amp = -\frac{1}{2}\int \sqrt{u}\,du \\ \amp = -\frac{one}{iii} u^{three/ii} + C\\ \amp = -\frac{1}{3} \left(100-ten^two\right)^{3/two} + C. \cease{split} \end{equation*}

7. \(\ds\int {x^ii\over\sqrt{1-x^3}}\,dx\)

   Answer

   \(\ds \ds -2\sqrt{1-10^3}/3+C\)

   Solution

   Allow \(u=ane-x^3\text{,}\) with \(du=-3x^2\text{:}\)

   \brainstorm{equation*} \begin{split up} \int \frac{x^ii}{\sqrt{one-10^iii}}\,dx \amp = -\frac{1}{3}\int \frac{1}{\sqrt{u}}\,du\\ \amp = -\frac{2}{3} \sqrt{u} + C\\ \amp= -\frac{2}{3}\sqrt{ane-ten^three} + C.\cease{dissever} \cease{equation*}

8. \(\ds\int \cos(\pi t)\cos\bigl(\sin(\pi t)\bigr)\,dt\)

   Answer

   \(\ds \sin(\sin\pi t)/\pi+C\)

   Solution

   Permit \(u=\sin(\pi t)\text{.}\) And then \(du = \pi\cos(\pi t)\) and

   \begin{equation*} \brainstorm{split up} \int \cos(\pi t)\cos(\sin(\pi t))\,dt \amp = \frac{1}{\pi} \int \cos u \,du \\ \amp = \frac{ane}{\pi} \sin u + C\\ \amp = \frac{1}{\pi} \sin(\sin(\pi t)) + C. \end{split up} \stop{equation*}

9. \(\ds\int {\sin x\over\cos^3 10}\,dx\)

   Answer

   \(\ds \ds one/(two\cos^two ten)=(one/2)\sec^2x+C\)

   Solution

   Let \(u=\cos ten\text{.}\) And then \(du = -\sin 10\,dx\) and

   \begin{equation*} \begin{separate} \int \frac{\sin 10}{\cos^3 10}\,dx \amp = -\int \frac{1}{u^three}\,du \\ \amp = \frac{1}{2 u^two} + C\\ \amp = \frac{i}{2 \cos^two x} + C. \terminate{divide} \end{equation*}

10. \(\ds\int\tan x\,dx\)

    Answer Solution

    Permit \(u=\cos 10\text{.}\) And so \(du=-\sin x\,dx\) and

    \brainstorm{equation*} \begin{split} \int \frac{\sin x}{\cos x}\,dx \amp = -\int \frac{1}{u}\,du \\ \amp = -\ln|u|+C\\ \amp = -\ln|\cos x| + C. \end{split} \end{equation*}

11. \(\ds\int\sec^2x\tan 10\,dx\)

    Answer Solution

    Let \(u=\sec x\,dx\text{.}\) And so \(du = \tan x \sec x\,dx\) and

    \brainstorm{equation*} \begin{split} \int \sec^2 10\tan x\,dx \amp = \int u\,du\\ \amp =\frac{1}{two} u^2 + C\\ \amp =\frac{i}{2} \sec^2x + C. \end{split} \end{equation*}

12. \(\ds\int {\sin(\tan 10)\over\cos^2x}\,dx\)

    Answer Solution

    Let \(u=\tan x\text{.}\) Then \(du = \sec^2 x\,dx\) and

    \brainstorm{equation*} \brainstorm{split up} \int \frac{\sin(\tan x)}{\cos^2 x}\,dx \amp = \int \sin u\,du\\ \amp = -\cos u + C\\ \amp = -\cos(\tan x) + C. \cease{split} \end{equation*}

13. \(\ds\int {6x\over(x^two - 7)^{i/9}}\,dx\)

    Answer

    \(\ds (27/8)(10^2-7)^{8/9}\)

    Solution

    Allow \(u = x^2-7\text{.}\) Then \(du = 2x\,dx\) and

    \begin{equation*} \brainstorm{split} \int \frac{6x}{(x^2-seven)^{1/9}} \,dx \amp = \int \frac{3}{u^{one/9}}\,du\\ \amp = \frac{27}{viii} u^{8/9} + C\\ \amp = \frac{27}{8} (x^2-7)^{8/9}. \stop{carve up} \terminate{equation*}

14. \(\ds\int f(10) f'(x)\,dx\)

    Respond Solution

    Let \(u = f(x)\text{.}\) Then \(du = f'(x)\,dx\) and

    \begin{equation*} \begin{separate} \int f(x) f'(x)\,dx \amp = \int u\,du\\ \amp = \frac{1}{2} u^2 + C\\ \amp = \frac{one}{two} \left[f(x)\right]^two + C. \finish{split} \finish{equation*}

Exercise 2.ane.2.

Evaluate the post-obit definite integrals.

1. \(\ds\int_0^\pi\sin^v(3x)\cos(3x)\,dx\)

   Answer Solution

   Allow \(u = \sin(3x)\text{.}\) And so \(du = 3\cos(3x)\,dx\text{,}\) and the integration bounds are from

   \begin{equation*} \sin(0) = 0 \text{ to } \sin(3\pi) = 0\text{.} \end{equation*}

   Therefore, nosotros have

   \brainstorm{equation*} \int_0^\pi \sin^5(3x)\cos(3x)\,dx = \int_0^0 u^5 \left(\frac{du}{iii}\correct) = 0\text{.} \end{equation*}

2. \(\ds\int_0^{\sqrt{\pi}/2} x\sec^2(x^2)\tan(x^2)\,dx\)

   Answer Solution

   We know that

   \brainstorm{equation*} \diff{}{x} \sec(ten) = \tan(x)\sec(10)\text{,} \end{equation*}

   so accept the substitution

   \begin{equation*} u=\sec(x^2) \implies du = 2x\sec(x^2)\tan(ten^2)\,dx\text{.} \cease{equation*}

   When \(x=0\text{,}\) we have that \(u=\sec(0) = i\text{,}\) and when \(10=\sqrt{\pi}/two\text{,}\) \(u=\sec(\pi/4)= \sqrt{2}\text{.}\) So in terms of \(u\text{,}\) the integral becomes

   \begin{equation*} \int_1^{\sqrt{two}} u \left(\frac{du}{2}\right) = \frac{1}{two} \left[\frac{i}{2}u^2\right]_1^{\sqrt{2}} = \frac{1}{ii}-\frac{1}{four} = \frac{1}{4}\text{.} \finish{equation*}

3. \(\ds\int_3^4 {1\over(3x-7)^ii}\,dx\)

   Answer Solution

   Accept \(u=3x-7\) with \(du=3\,dx\text{.}\) The integration bounds are thus

   \begin{equation*} 10=3 \implies u = 9-7 = 2, \text{ and } x=four \implies u = 12-seven = 5\text{.} \end{equation*}

   We now compute

   \begin{equation*} \displaystyle \int_3^4 \frac{1}{(3x-seven)^2}\,dx = \int_2^5 \frac{1}{u^2} \left(\frac{du}{iii}\right) = \frac{one}{three}\left[-\frac{1}{u}\right]_2^v = \frac{1}{10}\text{.} \end{equation*}

4. \(\ds\int_0^{\pi/6}(\cos^2x - \sin^2x)\,dx\)

   Answer Solution

   We first use a trigonometric identity to simplify the integrand:

   \begin{equation*} \int_0^{\pi/six} \left(\cos^2 x- \sin^2x\right)\,dx = \int_0^{\pi/6} \cos (2x)\,dx\text{.} \stop{equation*}

   Now let \(u=2x\) and \(du = 2\,dx\text{.}\) So the integral becomes

   \begin{equation*} \int_0^{\pi/iii} \frac{\cos u}{2}\,du = \frac{\sin u}{2} \bigg\vert_0^{\pi/3} = \frac{\sqrt{3}}{4}\text{.} \end{equation*}

5. \(\ds\int_{-one}^1 (2x^3-one)(x^4-2x)^6\,dx\)

   Answer Solution

   Permit \(u=x^4-2x\text{.}\) And then \(du = 4x^3-2 = 2(2x^three-one)\text{.}\) Therefore, when \(10=-1\text{,}\) \(u=1+two=three\text{,}\) and when \(x=1\text{,}\) \(u=ane-ii=-1\text{.}\) The integral then becomes

   \brainstorm{equation*} \begin{split} \int_{-ane}^1 (2x^three-1)(x^four-2x)^6\,dx \amp = -\frac{i}{two}\int_{-1}^{3} u^vi\,du \\ \amp =\frac{-1}{fourteen} u^7 \bigg\vert_{-one}^3\\ \amp = \frac{-i}{fourteen} \left(three^7 +1\correct)\\ \amp = -\frac{1094}{7}. \finish{divide} \end{equation*}

6. \(\ds\int_{-1}^i \sin^7 x\,dx\)

   Answer Solution

   Let \(u=\cos 10\) and \(du = -\sin x\,dx\text{.}\) And then:

   \begin{equation*} \int_{-1}^1 \sin^7 x\,dx = -\int_{\cos(-1)}^{\cos(one)} \left(1-u^2\correct)^three\,du = 0\text{.} \terminate{equation*}

Practice 2.1.3.

A toy truck manufacturer estimates that the number of sales after Christmas declines at a charge per unit of \(-3e^{-0.2t}\) toys per day (\(t \in [0,60]\)). If the manufacturer sells \(10,000\) units on Christmas day (\(t=0\)), determine the number of expected sales after \(t\) days.

Answer

\(15e^{-0.2t}+9985\) trucks.

Solution

Allow \(S(t)\) announce the number of sales after \(t\) days after Christmas. And so we know that \(S'(t) = -3 e^{-0.2t}\) and \(S(0) = ten,000\text{.}\) Therefore,

\begin{equation*} S(t) = -3\int e^{-0.2t}, dt\text{.} \end{equation*}

Let \(u=-0.2t\text{,}\) and \(du=-0.ii\,dt\text{.}\) Then,

\begin{equation*} \int due east^{-0.2t}\,dt = -\frac{ten}{2}\int eastward^{u}\,du = -\frac{10}{ii} e^{u}+C\text{.} \end{equation*}

So we must have

\brainstorm{equation*} S(t) = 15 e^{-0.2t} +C\text{.} \cease{equation*}

Now using the initial status, we find

\begin{equation*} South(0) = 15+C = ten,000 \implies C = 9985\text{.} \end{equation*}

All together, we find that

\begin{equation*} Southward(t) = 15e^{-0.2t}+9985\text{.} \end{equation*}

Practice two.1.4.

Let \(q\) exist the quantity (in thousands) of a product in the market place when the unit of measurement cost is fix at \(p\) dollars per unit. There are currently \(2000\) units available at a toll of $2/unit. Determine the corresponding supply equation if the cost increases at a charge per unit of

\brainstorm{equation*} p'(q)=\frac{100q}{(1-q)^2}\text{.} \end{equation*}

Answer

\(p(q)= 100(1/(1-q)+\log(q-1))+102\)

Solution

We first solve the indefinite integral:

\begin{equation*} \int p'(q)\,dq = \int \frac{100q}{(one-q)^ii}\,dq\text{.} \finish{equation*}

Let \(u=1-q\) and \(du = -dq\text{.}\) Then \(q=one-u\) and

\brainstorm{equation*} \begin{carve up} 100 \int \frac{q}{(ane-q)^2}\,dq \amp = -100 \int \frac{1-u}{u^2}\,du\\ \amp = -100\left(\int \frac{1}{u^2}\,du - \int \frac{ane}{u}\,du\correct)\\ \amp = -100\left(\frac{-ane}{u} -\ln |u| \right) + C\\ \amp = \frac{100}{i-q} +100\ln|1-q|+C \end{split} \end{equation*}

Therefore, for some abiding \(C\text{,}\) nosotros have

\begin{equation*} p(q) = \frac{100}{1-q} +100\ln|ane-q|+C\text{.} \end{equation*}

Since 2000 units (\(q=2\)) are available at $2 per unit, nosotros know that

\brainstorm{equation*} 2 = p(2) = \frac{100}{-1} + 100 \ln (i)+ C = C-100 \implies C = 102\text{.} \end{equation*}

All together, nosotros find that the supply equation is

\begin{equation*} p(q) = \frac{100}{1-q} +100\ln|1-q|+102\text{.} \end{equation*}

Practice 2.ane.5.

Allow \(q\) stand for the daily quantity demanded (in thousands) of a certain product. When the unit price \(p\) is set at $\(iii\text{,}\) it is observed that \(1000\) units are demanded per day. Make up one's mind the corresponding demand equation if the cost decreases at a rate of

\begin{equation*} p'(q)=\frac{-200q}{(iii+q^ii)^{3/2}}\text{.} \finish{equation*}

Answer

\(p(q)=\frac{200}{\sqrt{q^ii+3}}-97\)

Solution

We are given that a sure demand equation decreases at a rate of

\begin{equation*} p'(q) = \frac{-200q}{(3+q^2)^{iii/2}} \end{equation*}

dollars per day, where \(q\) is the number of units demanded per day, measured in thousands. To find the corresponding demand function, we kickoff set the indefinite integral,

\begin{equation*} p(q) = \int p'(q)\,dq = \int \frac{-200q}{(3+q^2)^{3/2}} \,dq\text{.} \end{equation*}

We solve by substitution: Let

\brainstorm{equation*} u=3+q^2, \ du = 2q\,dq\text{.} \end{equation*}

Thus,

\begin{equation*} p(u) = \int \frac{-200}{u^{iii/ii}} \frac{du}{2} = \frac{200}{\sqrt{u}} + C\text{.} \stop{equation*}

So in terms of \(q\text{,}\)

\brainstorm{equation*} p(q) = \frac{200}{\sqrt{iii+q^2}} + \chapeau{C}\text{.} \end{equation*}

We are farther given that \(p(one) = 3\text{,}\) then we must have that

\begin{equation*} \frac{200}{\sqrt{iv}} + \hat{C} = 3 \implies \lid{C}=-97\text{.} \terminate{equation*}

The demand office of the production is thus

\brainstorm{equation*} p(q) = \frac{200}{\sqrt{3+q^2}} - 97\text{.} \finish{equation*}

Exercise 2.1.6.

In the first month (\(t=0\)) post-obit a successful marketing entrada, a company sells two,000 units of their product. Management predicts that the sales volition reject at a charge per unit of

\brainstorm{equation*} N'(t)=-200(due east^{-2t}+i) \end{equation*}

units per calendar month for the next three months. Decide the full number of expected sales after iii months.

Answer

Approximately 1300 units.

Solution

We kickoff integrate

\begin{equation*} \int -200 \left(due east^{-2t}+ane\correct)\,dt \end{equation*}

Let \(u=-2t\) and \(du = -2\,dt\text{.}\) Then:

\brainstorm{equation*} \begin{separate} \int -200 \left(due east^{-2t}+one\right)\,dt \amp = 100\int east^u\,du + 100\int\,du \\ \amp = 100 eastward^u + 100 u + C\\ \amp = 100 eastward^{-2t}-200 t + C \cease{split} \finish{equation*}

Therefore, if 2,000 units were sold in the get-go month, we must have

\begin{equation*} 2000 = North(0) = 100+C \implies C =1900\text{.} \end{equation*}

Hence, the number of units sold can be described by the equation

\begin{equation*} N(t) = -200 e^{-2t} - 200t + 1900\text{.} \finish{equation*}

Subsequently three months, the number of units sold will and then be

\begin{equation*} N(three) = -200 e^{-6} - 600 + 1900 = 1299.50\text{,} \stop{equation*}

or about 1300 units.

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Source: https://www.sfu.ca/math-coursenotes/Math%20158%20Course%20Notes/sec_SubRule.html

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