3.3 Trigonometric substitution  (Page 4/5)

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

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

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

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

$16{\text{cosh}}^2\theta -16$

$4x^2-4x+1$

$2x^2-8x+3$

$\text{−}{x}^2-2x+4$

${\displaystyle \int \frac{dx}{\sqrt{4-x^2}}}$

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

${\displaystyle \int \sqrt{4-x^2}dx}$

${\displaystyle \int \frac{dx}{\sqrt{1+9x^2}}}$

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

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

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

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

${\displaystyle \int \frac{\sqrt{x^2-25}}{x}dx}$

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

${\displaystyle \int \frac{dx}{\sqrt{x^6-x^2}}}$

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

${\displaystyle \int \frac{dx}{{\left(1+x^2\right)}^{3\text{/}2}}}$

${\displaystyle \int \frac{dx}{{\left(x^2-9\right)}^{3\text{/}2}}}$

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

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

${\displaystyle \int \frac{x^{2}dx}{x^2+4}}$

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

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

${\displaystyle {\int }_{\mathrm{-1}}^1{(1-x^2)}^{3\text{/}2}dx}$

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

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

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

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

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

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

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

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

${\displaystyle \int \frac{1}{\sqrt{\text{−}{x}^2+2x+8}}dx}$

${\displaystyle \int \frac{1}{\sqrt{\text{−}{x}^2+10x}}dx}$

${\displaystyle \int \frac{1}{\sqrt{x^2+4x-12}}}dx$

Evaluate the integral without using calculus: ${\displaystyle {\int }_{\mathrm{-3}}^3\sqrt{9-x^2}dx.}$

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

Evaluate the integral ${\displaystyle \int \frac{dx}{\sqrt{1-x^2}}}$ using two different substitutions. First, let $x=\text{cos}\theta$ and evaluate using trigonometric substitution. Second, let $x=\text{sin}\theta$ and use trigonometric substitution. Are the answers the same?

Evaluate the integral ${\displaystyle \int \frac{dx}{x\sqrt{x^2-1}}}$ using the substitution $x=\text{sec}\theta .$ Next, evaluate the same integral using the substitution $x=\text{csc}\theta .$ Show that the results are equivalent.

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

State the method of integration you would use to evaluate the integral ${\displaystyle \int x\sqrt{x^2+1}dx}.$ Why did you choose this method?

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

Evaluate ${\displaystyle {\int }_{\mathrm{-1}}^1\frac{xdx}{x^2+1}}$

Find the length of the arc of the curve over the specified interval: $y=\text{ln}x,[1,5].$ Round the answer to three decimal places.

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

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

$\left(x^2+36\right)\frac{dy}{dx}=1,y(6)=0$

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

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

An oil storage tank can be described as the volume generated by revolving the area bounded by $y=\frac{16}{\sqrt{64+x^2}},x=0,y=0,x=2$ about the x -axis. Find the volume of the tank (in cubic meters).

During each cycle, 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 $s=0$ when $t=0.$

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