3.5 X 3.5 X 3.14

Learning Objectives

• iii.3.1 Solve integration issues involving the foursquare root of a sum or difference of 2 squares.

In this department, nosotros explore integrals containing expressions of the form $\sqrt{a^{\mathrm{two}}-x^2},$ $\sqrt{a^2+{\mathrm{ten}}^{\mathrm{two}}},$ and $\sqrt{x^2-a^{\mathrm{ii}}},$ where the values of $a$ are positive. We have already encountered and evaluated integrals containing some expressions of this type, simply many notwithstanding remain inaccessible. The technique of trigonometric exchange comes in very handy when evaluating these integrals. This technique uses substitution to rewrite these integrals as trigonometric integrals.

Integrals Involving $\sqrt{a^{\mathrm{ii}}-x^{\mathrm{ii}}}$

Earlier developing a general strategy for integrals containing $\sqrt{a^2-x^{\mathrm{two}}},$ consider the integral ${\displaystyle \int \sqrt{9-x^2}}dx.$ This integral cannot be evaluated using any of the techniques nosotros take discussed so far. Notwithstanding, if we make the substitution $\mathrm{10}=3\text{sin}\theta ,$ we have $d\mathrm{10}=3\text{cos}\theta d\theta .$ After substituting into the integral, we take

$${\displaystyle \int \sqrt{9-x^2}dx}={{\displaystyle \int }}^{\text{}}\sqrt{\mathrm{ix}-{(3\text{sin}\theta )}^{\mathrm{two}}}3\text{cos}\theta d\theta .$$

Afterward simplifying, nosotros take

$${{\displaystyle \int }}^{\text{}}\sqrt{9-x^2}dx={{\displaystyle \int }}^{\text{}}9\sqrt{\mathrm{one}-{\text{sin}}^2\theta }\text{cos}\theta d\theta .$$

Letting $\mathrm{ane}-{\text{sin}}^2\theta ={\text{cos}}^2\theta ,$ nosotros at present have

$${{\displaystyle \int }}^{\text{}}\sqrt{9-x^2}d\mathrm{10}={{\displaystyle \int }}^{\text{}}\mathrm{ix}\sqrt{{\text{cos}}^{\mathrm{ii}}\theta }\text{cos}\theta d\theta .$$

Assuming that $\text{cos}\theta \ge 0,$ we accept

$${{\displaystyle \int }}^{\text{}}\sqrt{\mathrm{nine}-{\mathrm{ten}}^2}dx={{\displaystyle \int }}^{\text{}}9{\text{cos}}^2\theta d\theta .$$

At this point, we can evaluate the integral using the techniques developed for integrating powers and products of trigonometric functions. Before completing this instance, allow'southward take a look at the general theory behind this idea.

To evaluate integrals involving $\sqrt{a^2-x^2},$ nosotros make the substitution $x=a\text{sin}\theta$ and $d\mathrm{10}=a\text{cos}\theta .$ To see that this actually makes sense, consider the following argument: The domain of $\sqrt{a^2-{\mathrm{ten}}^2}$ is $[\text{−}a,a].$ Thus, $\text{−}a\le x\le a.$ Consequently, $\mathrm{-i}\le \frac{\mathrm{10}}{a}\le 1.$ Since the range of $\text{sin}x$ over $\left[\text{−}(\pi \text{/}2),\pi \text{/}\mathrm{ii}\right]$ is $[\mathrm{-1},1],$ in that location is a unique angle $\theta$ satisfying $\text{−}(\pi \text{/}2)\le \theta \le \pi \text{/}2$ then that $\text{sin}\theta =x\text{/}a,$ or equivalently, so that $x=a\text{sin}\theta .$ If we substitute $\mathrm{10}=a\text{sin}\theta$ into $\sqrt{a^2-x^2},$ we become

$$\begin{array}{ccccc}\hfill \sqrt{a^2-{\mathrm{ten}}^2}& =\sqrt{a^{\mathrm{two}}-{(a\text{sin}\theta )}^{\mathrm{two}}}\hfill & & & \text{Allow}x=a\text{sin}\theta \text{where}-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.\text{Simplify.}\hfill \\ & =\sqrt{a^2-a^{\mathrm{ii}}{\text{sin}}^2\theta }\hfill & & & \text{Cistron out}{a}^{\mathrm{ii}}.\hfill \\ & =\sqrt{a^{\mathrm{two}}(\mathrm{ane}-{\text{sin}}^2\theta )}\hfill & & & \text{Substitute}1-{\text{sin}}^{2}x={\text{cos}}^{2}x.\hfill \\ & =\sqrt{a^2{\text{cos}}^2\theta }\hfill & & & \text{Accept the square root.}\hfill \\ & =\left|a\text{cos}\theta \right|\hfill & & & \\ & =a\text{cos}\theta .\hfill & & & \end{array}$$

Since $\text{cos}\theta \ge 0$ on $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ and $a>0,$ $\left|a\text{cos}\theta \right|=a\text{cos}\theta .$ We tin can see, from this word, that by making the substitution $\mathrm{ten}=a\text{sin}\theta ,$ nosotros are able to catechumen an integral involving a radical into an integral involving trigonometric functions. Afterwards we evaluate the integral, nosotros can convert the solution back to an expression involving $\mathrm{10}.$ To see how to do this, let'due south begin by assuming that $0<x<a.$ In this case, $0<\theta <\frac{\pi }{\mathrm{two}}.$ Since $\text{sin}\theta =\frac{x}{a},$ we tin draw the reference triangle in Figure three.4 to assist in expressing the values of $\text{cos}\theta ,$ $\text{tan}\theta ,$ and the remaining trigonometric functions in terms of $x.$ Information technology tin can exist shown that this triangle actually produces the correct values of the trigonometric functions evaluated at $\theta$ for all $\theta$ satisfying $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}.$ It is useful to observe that the expression $\sqrt{a^2-x^2}$ actually appears as the length of one side of the triangle. Last, should $\theta$ appear by itself, we employ $\theta ={\text{sin}}^{\mathrm{-1}}\left(\frac{x}{a}\right).$

Figure iii.four A reference triangle tin aid express the trigonometric functions evaluated at $\theta$ in terms of $\mathrm{10}.$

The essential part of this give-and-take is summarized in the following problem-solving strategy.

Problem-Solving Strategy

Trouble-Solving Strategy: Integrating Expressions Involving $\sqrt{a^{\mathrm{two}}-x^{\mathrm{two}}}$

1. It is a skillful idea to make sure the integral cannot be evaluated easily in another style. For example, although this method tin exist applied to integrals of the grade ${\displaystyle \int \frac{\mathrm{i}}{\sqrt{a^2-x^{\mathrm{two}}}}d\mathrm{ten}},$ ${\displaystyle \int \frac{x}{\sqrt{a^2-x^2}}d\mathrm{10}},$ and ${\displaystyle \int x\sqrt{a^2-x^2}dx,}$ they tin each exist integrated straight either past formula or past a elementary u-substitution.
2. Make the commutation $x=a\text{sin}\theta$ and $d\mathrm{10}=a\text{cos}\theta d\theta .$ Note: This exchange yields $\sqrt{a^2-x^2}=a\text{cos}\theta .$
3. Simplify the expression.
4. Evaluate the integral using techniques from the section on trigonometric integrals.
5. Use the reference triangle from Figure 3.4 to rewrite the event in terms of $x.$ You lot may also need to apply some trigonometric identities and the relationship $\theta ={\text{sin}}^{\mathrm{-one}}\left(\frac{x}{a}\right).$

The following case demonstrates the application of this trouble-solving strategy.

Case iii.21

Integrating an Expression Involving $\sqrt{a^2-x^{\mathrm{ii}}}$

Evaluate ${{\displaystyle \int }}^{\text{}}\sqrt{\mathrm{nine}-x^2}d\mathrm{10}.$

Example iii.22

Integrating an Expression Involving $\sqrt{a^2-x^{\mathrm{two}}}$

Evaluate ${\displaystyle \int \frac{\sqrt{\mathrm{iv}-{\mathrm{ten}}^{\mathrm{ii}}}}{x}dx.}$

In the next example, we run across that we sometimes take a pick of methods.

Example 3.23

Integrating an Expression Involving $\sqrt{a^2-x^2}$ 2 Ways

Evaluate ${{\displaystyle \int }}^{\text{}}{x}^3\sqrt{\mathrm{i}-{\mathrm{ten}}^2}d\mathrm{10}$ two ways: first by using the substitution $u=\mathrm{ane}-{\mathrm{10}}^{\mathrm{ii}}$ then by using a trigonometric commutation.

Checkpoint three.14

Rewrite the integral ${\displaystyle \int \frac{x^3}{\sqrt{25-{\mathrm{10}}^2}}dx}$ using the appropriate trigonometric substitution (do not evaluate the integral).

Integrating Expressions Involving $\sqrt{a^2+{\mathrm{10}}^2}$

For integrals containing $\sqrt{a^{\mathrm{ii}}+x^{\mathrm{two}},}$ permit's start consider the domain of this expression. Since $\sqrt{a^2+{\mathrm{ten}}^{\mathrm{two}}}$ is defined for all existent values of $x,$ we restrict our choice to those trigonometric functions that have a range of all real numbers. Thus, our choice is restricted to selecting either $x=a\text{tan}\theta$ or $\mathrm{10}=a\text{cot}\theta .$ Either of these substitutions would actually work, simply the standard substitution is $x=a\text{tan}\theta$ or, equivalently, $\text{tan}\theta =\mathrm{10}\text{/}a.$ With this substitution, nosotros make the assumption that $\text{−}(\pi \text{/}2)<\theta <\pi \text{/}\mathrm{two},$ so that we besides have $\theta ={\text{tan}}^{\mathrm{-ane}}\left(x\text{/}a\right).$ The procedure for using this commutation is outlined in the post-obit problem-solving strategy.

Problem-Solving Strategy

Problem-Solving Strategy: Integrating Expressions Involving $\sqrt{a^2+x^2}$

1. Check to see whether the integral can be evaluated easily by using another method. In some cases, information technology is more convenient to use an alternative method.
2. Substitute $x=a\text{tan}\theta$ and $dx=a{\text{sec}}^{\mathrm{ii}}\theta d\theta .$ This substitution yields
   $\sqrt{a^2+x^{\mathrm{two}}}=\sqrt{a^2+{(a\text{tan}\theta )}^2}=\sqrt{a^2(\mathrm{one}+{\text{tan}}^2\theta )}=\sqrt{a^{\mathrm{two}}{\text{sec}}^2\theta }=\left|a\text{sec}\theta \right|=a\text{sec}\theta .$ (Since $-\frac{\pi }{2}<\theta <\frac{\pi }{\mathrm{ii}}$ and $\text{sec}\theta >0$ over this interval, $\left|a\text{sec}\theta \right|=a\text{sec}\theta .)$
3. Simplify the expression.
4. Evaluate the integral using techniques from the section on trigonometric integrals.
5. Utilise the reference triangle from Figure 3.7 to rewrite the result in terms of $x.$ Yous may as well need to utilise some trigonometric identities and the relationship $\theta ={\text{tan}}^{\mathrm{-one}}\left(\frac{x}{a}\right).$ (Note: The reference triangle is based on the assumption that $x>0;$ all the same, the trigonometric ratios produced from the reference triangle are the aforementioned as the ratios for which $\mathrm{ten}\le 0.)$

Figure 3.vii A reference triangle tin can be constructed to express the trigonometric functions evaluated at $\theta$ in terms of $x.$

Instance 3.24

Integrating an Expression Involving $\sqrt{a^{\mathrm{ii}}+x^2}$

Evaluate ${\displaystyle \int \frac{dx}{\sqrt{1+x^{\mathrm{two}}}}}$ and check the solution past differentiating.

Example 3.25

Evaluating ${\displaystyle \int \frac{dx}{\sqrt{1+x^2}}}$ Using a Different Substitution

Employ the substitution $x=\text{sinh}\theta$ to evaluate ${\displaystyle \int \frac{dx}{\sqrt{\mathrm{one}+x^{\mathrm{ii}}}}.}$

Analysis

This reply looks quite different from the answer obtained using the substitution $\mathrm{10}=\text{tan}\theta .$ To see that the solutions are the same, set $y={\text{sinh}}^{\mathrm{-1}}\mathrm{10}.$ Thus, $\text{sinh}y=x.$ From this equation nosotros obtain:

$$\frac{{\mathrm{east}}^y-e^{\text{−}y}}{\mathrm{two}}=\mathrm{ten}.$$

After multiplying both sides past $2{\mathrm{due\; east}}^y$ and rewriting, this equation becomes:

$$e^{\mathrm{ii}y}-\mathrm{ii}\mathrm{10}{e}^y-1=0.$$

Use the quadratic equation to solve for $e^y\text{:}$

$${\mathrm{east}}^y=\frac{2x\pm \sqrt{4{\mathrm{10}}^2+\mathrm{four}}}{2}.$$

Simplifying, we take:

$$e^y=\mathrm{ten}\pm \sqrt{x^2+\mathrm{i}}.$$

Since $x-\sqrt{{\mathrm{10}}^2+1}<0,$ information technology must exist the case that ${\mathrm{east}}^y=x+\sqrt{x^2+1}.$ Thus,

$$y=\text{ln}\left(\mathrm{ten}+\sqrt{x^{\mathrm{two}}+1}\right).$$

Concluding, we obtain

$${\text{sinh}}^{\mathrm{-1}}\mathrm{10}=\text{ln}\left(\mathrm{10}+\sqrt{{\mathrm{10}}^2+1}\right).$$

Afterwards we make the terminal observation that, since $\mathrm{10}+\sqrt{x^2+\mathrm{ane}}>0,$

$$\text{ln}\left(x+\sqrt{x^{\mathrm{two}}+1}\right)=\text{ln}\left|\sqrt{1+{\mathrm{10}}^2}+x\right|,$$

we come across that the two different methods produced equivalent solutions.

Case 3.26

Finding an Arc Length

Discover the length of the bend $y={\mathrm{ten}}^2$ over the interval $[0,\frac{\mathrm{one}}{2}].$

Checkpoint 3.15

Rewrite ${{\displaystyle \int }}^{\text{}}{x}^3\sqrt{{\mathrm{10}}^2+\mathrm{iv}}dx$ by using a exchange involving $\text{tan}\theta .$

Integrating Expressions Involving $\sqrt{x^2-a^2}$

The domain of the expression $\sqrt{x^2-a^2}$ is $\left(\text{−}\infty ,\text{−}a\right]\cup \left[a,\text{+}\infty \right).$ Thus, either $\mathrm{ten}\le \text{−}a$ or $x\ge a.$ Hence, $\frac{x}{a}\le -1$ or $\frac{x}{a}\ge 1.$ Since these intervals correspond to the range of $\text{sec}\theta$ on the set up $\left[0,\frac{\pi }{2}\right)\cup \left(\frac{\pi }{2},\pi \right],$ it makes sense to use the commutation $\text{sec}\theta =\frac{\mathrm{10}}{a}$ or, equivalently, $x=a\text{sec}\theta ,$ where $0\le \theta <\frac{\pi }{\mathrm{two}}$ or $\frac{\pi }{2}<\theta \le \pi .$ The respective substitution for $d\mathrm{10}$ is $dx=a\text{sec}\theta \text{tan}\theta d\theta .$ The procedure for using this substitution is outlined in the following problem-solving strategy.

Trouble-Solving Strategy

Problem-Solving Strategy: Integrals Involving $\sqrt{{\mathrm{ten}}^{\mathrm{two}}-a^2}$

1. Check to see whether the integral cannot exist evaluated using another method. If so, we may wish to consider applying an alternative technique.
2. Substitute $x=a\text{sec}\theta$ and $dx=a\text{sec}\theta \text{tan}\theta d\theta .$ This substitution yields
   

   $$\sqrt{{\mathrm{10}}^{\mathrm{two}}-a^2}=\sqrt{{\left(a\text{sec}\theta \right)}^{\mathrm{ii}}-a^2}=\sqrt{a^2({\text{sec}}^2\theta –1)}=\sqrt{a^2{\text{tan}}^2\theta }=\left|a\text{tan}\theta \right|.$$

   
   For $\mathrm{ten}\ge a,$ $\left|a\text{tan}\theta \right|=a\text{tan}\theta$ and for $x\le -a,$ $\left|a\text{tan}\theta \right|=\text{−}a\text{tan}\theta .$
3. Simplify the expression.
4. Evaluate the integral using techniques from the department on trigonometric integrals.
5. Utilise the reference triangles from Figure iii.9 to rewrite the upshot in terms of $x.$ You may also demand to utilize some trigonometric identities and the relationship $\theta ={\text{sec}}^{\mathrm{-one}}\left(\frac{\mathrm{10}}{a}\right).$ (Note: We need both reference triangles, since the values of some of the trigonometric ratios are different depending on whether $x\ge a$ or $\mathrm{10}\le \text{−}a.)$

Figure iii.9 Use the appropriate reference triangle to express the trigonometric functions evaluated at $\theta$ in terms of $x.$

Example iii.27

Finding the Area of a Region

Find the expanse of the region between the graph of $f\left(x\right)=\sqrt{x^{\mathrm{ii}}-\mathrm{ix}}$ and the 10-axis over the interval $[3,\mathrm{v}].$

Checkpoint three.16

Evaluate ${\displaystyle \int \frac{dx}{\sqrt{x^2-\mathrm{iv}}}}.$ Presume that $x>2.$

Section 3.iii Exercises

Simplify the following expressions past writing each 1 using a single trigonometric function.

126 .

$4-\mathrm{four}{\text{sin}}^2\theta$

127.

$9{\text{sec}}^2\theta -9$

128 .

$a^{\mathrm{two}}+a^{\mathrm{two}}{\text{tan}}^2\theta$

129.

$a^2+a^{\mathrm{ii}}{\text{sinh}}^{\mathrm{ii}}\theta$

130 .

$16{\text{cosh}}^{\mathrm{two}}\theta -\mathrm{xvi}$

Employ the technique of completing the square to express each trinomial every bit the square of a binomial or the foursquare of a binomial plus a constant.

131.

$4x^{\mathrm{ii}}-4x+\mathrm{one}$

132 .

$2x^2-8\mathrm{ten}+\mathrm{iii}$

133.

$\text{−}{\mathrm{10}}^2-\mathrm{ii}x+4$

Integrate using the method of trigonometric substitution. Express the last answer in terms of the variable.

134 .

${\displaystyle \int \frac{dx}{\sqrt{\mathrm{four}-x^2}}}$

135.

${\displaystyle \int \frac{d\mathrm{10}}{\sqrt{x^2-a^2}}}$

136 .

${\displaystyle \int \sqrt{4-x^{\mathrm{two}}}dx}$

137.

${\displaystyle \int \frac{d\mathrm{10}}{\sqrt{\mathrm{i}+9x^2}}}$

138 .

${\displaystyle \int \frac{x^{2}dx}{\sqrt{1-x^2}}}$

139.

${\displaystyle \int \frac{dx}{x^{\mathrm{ii}}\sqrt{1-x^{\mathrm{two}}}}}$

140 .

${\displaystyle \int \frac{d\mathrm{10}}{{(1+x^{\mathrm{ii}})}^2}}$

141.

${\displaystyle \int \sqrt{{\mathrm{10}}^{\mathrm{ii}}+\mathrm{ix}}}dx$

142 .

${\displaystyle \int \frac{\sqrt{{\mathrm{10}}^{\mathrm{two}}-25}}{x}dx}$

143.

${\displaystyle \int \frac{{\theta }^{3}d\theta }{\sqrt{9-{\theta }^2}}}$

144 .

${\displaystyle \int \frac{d\mathrm{ten}}{\sqrt{{\mathrm{10}}^6-x^2}}}$

145.

${\displaystyle \int \sqrt{x^6-x^8}}dx$

146 .

${\displaystyle \int \frac{d\mathrm{10}}{{\left(\mathrm{one}+x^2\right)}^{3\text{/}\mathrm{two}}}}$

147.

${\displaystyle \int \frac{d\mathrm{10}}{{\left(x^2-9\right)}^{\mathrm{iii}\text{/}2}}}$

148 .

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

149.

${\displaystyle \int \frac{x^{2}dx}{\sqrt{x^2-1}}}$

150 .

${\displaystyle \int \frac{x^{\mathrm{ii}}dx}{{\mathrm{10}}^{\mathrm{ii}}+\mathrm{four}}}$

151.

${\displaystyle \int \frac{dx}{{\mathrm{ten}}^{\mathrm{two}}\sqrt{x^2+\mathrm{i}}}}$

152 .

${\displaystyle \int \frac{x^{2}dx}{\sqrt{\mathrm{one}+x^{\mathrm{two}}}}}$

153.

${\displaystyle {\int }_{\mathrm{-1}}^{\mathrm{i}}{(1-x^2)}^{\mathrm{iii}\text{/}2}dx}$

In the following exercises, utilise the substitutions $\mathrm{10}=\text{sinh}\theta ,\text{cosh}\theta ,$ or $\text{tanh}\theta .$ Express the terminal answers in terms of the variable x.

154 .

${\displaystyle \int \frac{dx}{\sqrt{x^2-1}}}$

155.

${\displaystyle \int \frac{dx}{x\sqrt{\mathrm{one}-{\mathrm{10}}^2}}}$

156 .

${\displaystyle \int \sqrt{{\mathrm{ten}}^2-1}}dx$

157.

${\displaystyle \int \frac{\sqrt{x^2-\mathrm{one}}}{x^{\mathrm{two}}}dx}$

158 .

${\displaystyle \int \frac{dx}{1-x^2}}$

159.

${\displaystyle \int \frac{\sqrt{\mathrm{one}+{\mathrm{ten}}^{\mathrm{ii}}}}{x^{\mathrm{ii}}}d\mathrm{ten}}$

Use the technique of completing the square to evaluate the post-obit integrals.

160 .

${\displaystyle \int \frac{1}{x^2-6\mathrm{10}}dx}$

161.

${\displaystyle \int \frac{\mathrm{one}}{x^2+2x+1}dx}$

162 .

${\displaystyle \int \frac{1}{\sqrt{\text{−}{\mathrm{10}}^2+\mathrm{two}x+\mathrm{viii}}}d\mathrm{ten}}$

163.

${\displaystyle \int \frac{\mathrm{ane}}{\sqrt{\text{−}{x}^2+10x}}dx}$

164 .

${\displaystyle \int \frac{1}{\sqrt{{\mathrm{ten}}^{\mathrm{ii}}+4x-12}}}dx$

165.

Evaluate the integral without using calculus: ${\displaystyle {\int }_{\mathrm{-three}}^{\mathrm{three}}\sqrt{9-{\mathrm{ten}}^2}dx.}$

166 .

Find the area enclosed by the ellipse $\frac{x^{\mathrm{ii}}}{4}+\frac{y^2}{9}=1.$

167.

Evaluate the integral ${\displaystyle \int \frac{dx}{\sqrt{\mathrm{one}-{\mathrm{ten}}^2}}}$ using two different substitutions. Showtime, let $\mathrm{10}=\text{cos}\theta$ and evaluate using trigonometric substitution. Second, permit $x=\text{sin}\theta$ and utilize trigonometric substitution. Are the answers the same?

168 .

Evaluate the integral ${\displaystyle \int \frac{dx}{x\sqrt{{\mathrm{10}}^2-\mathrm{one}}}}$ using the substitution $x=\text{sec}\theta .$ Next, evaluate the same integral using the substitution $\mathrm{ten}=\text{csc}\theta .$ Show that the results are equivalent.

169.

Evaluate the integral ${\displaystyle \int \frac{x}{x^2+1}}dx$ using the form ${\displaystyle \int \frac{\mathrm{ane}}{u}du}.$ Next, evaluate the aforementioned integral using $\mathrm{10}=\text{tan}\theta .$ Are the results the same?

170 .

State the method of integration yous would use to evaluate the integral ${\displaystyle \int x\sqrt{x^{\mathrm{ii}}+1}dx}.$ Why did you choose this method?

171.

State the method of integration yous would use to evaluate the integral ${\displaystyle \int {\mathrm{10}}^2\sqrt{x^2-1}d\mathrm{10}}.$ Why did yous choose this method?

172 .

Evaluate ${\displaystyle {\int }_{\mathrm{-1}}^1\frac{\mathrm{10}d\mathrm{10}}{x^{\mathrm{ii}}+1}}$

173.

Observe the length of the arc of the bend over the specified interval: $y=\text{ln}x,[1,5].$ Round the respond to iii decimal places.

174 .

Find the surface area of the solid generated past revolving the region bounded by the graphs of $y={\mathrm{ten}}^{\mathrm{ii}},y=0,\mathrm{10}=0,\text{and}\mathrm{ten}=\sqrt{2}$ about the x-axis. (Round the answer to iii decimal places).

175.

The region bounded by the graph of $f(x)=\frac{\mathrm{i}}{1+x^2}$ and the x-centrality between $x=0$ and $x=\mathrm{i}$ is revolved about the x-centrality. Find the volume of the solid that is generated.

Solve the initial-value problem for y as a function of 10.

176 .

$\left({\mathrm{ten}}^{\mathrm{two}}+36\right)\frac{dy}{dx}=\mathrm{ane},y(6)=0$

177.

$\left(64-x^2\right)\frac{dy}{dx}=1,y(0)=3$

178 .

Discover the area bounded by $y=\frac{2}{\sqrt{64-4x^2}},x=0,y=0,\text{and}\mathrm{ten}=2.$

179.

An oil storage tank can be described equally the book generated past revolving the area bounded by $y=\frac{\mathrm{sixteen}}{\sqrt{64+{\mathrm{ten}}^{\mathrm{ii}}}},\mathrm{10}=0,y=0,x=\mathrm{two}$ about the x-axis. Observe the book of the tank (in cubic meters).

180 .

During each bike, the velocity v (in feet per second) of a robotic welding device is given by $v=2t-\frac{14}{4+t^2},$ where t is time in seconds. Find the expression for the displacement s (in feet) as a function of t if $\mathrm{south}=0$ when $t=0.$

181.

Find the length of the curve $y=\sqrt{16-{\mathrm{10}}^2}$ between $x=0$ and $x=2.$

3.5 X 3.5 X 3.14,

Source: https://openstax.org/books/calculus-volume-2/pages/3-3-trigonometric-substitution

Posted by: freemanhilows.blogspot.com

Share this post