5.5 Substitution  (Page 4/6)

${\displaystyle \int x\sqrt{4x^2+9}dx=\frac{1}{12}{\left(4x^2+9\right)}^{3\text{/}2}+C};u=4x^2+9$

${\displaystyle \int \frac{x}{\sqrt{4x^2+9}}dx=\frac{1}{4}\sqrt{4x^2+9}+C};u=4x^2+9$

${\displaystyle \int \frac{x}{{(4x^2+9)}^2}dx=-\frac{1}{8(4x^2+9)}};u=4x^2+9$

${\displaystyle \int {\left(x+1\right)}^{4}dx};u=x+1$

${\displaystyle \int {\left(x-1\right)}^{5}dx};u=x-1$

${\displaystyle \int {\left(2x-3\right)}^{\mathrm{-7}}dx};u=2x-3$

${\displaystyle \int {\left(3x-2\right)}^{\mathrm{-11}}dx};u=3x-2$

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

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

${\displaystyle \int \left(x-1\right){\left(x^2-2x\right)}^{3}dx};u=x^2-2x$

${\displaystyle \int \left(x^2-2x\right){\left(x^3-3x^2\right)}^{2}dx};u=x^3=3x^2$

${\displaystyle \int {\text{cos}}^3\theta d\theta };u=\text{sin}\theta$ $\text{(}Hint\text{:}{\text{cos}}^2\theta =1-{\text{sin}}^2\theta \text{)}$

${\displaystyle \int {\text{sin}}^3\theta d\theta };u=\text{cos}\theta$ $\text{(}Hint\text{:}{\text{sin}}^2\theta =1-{\text{cos}}^2\theta \text{)}$

${\displaystyle \int x{\left(1-x\right)}^{99}dx}$

${\displaystyle \int t{\left(1-t^2\right)}^{10}dt}$

${\displaystyle \int {\left(11x-7\right)}^{\mathrm{-3}}dx}$

${\displaystyle \int {\left(7x-11\right)}^{4}dx}$

${\displaystyle \int {\text{cos}}^3\theta \text{sin}\theta d\theta }$

${\displaystyle \int {\text{sin}}^7\theta \text{cos}\theta d\theta }$

${\displaystyle \int {\text{cos}}^2\left(\pi t\right)\text{sin}\left(\pi t\right)dt}$

${\displaystyle \int {\text{sin}}^{2}x{\text{cos}}^{3}xdx}$ $\text{(}Hint\text{:}{\text{sin}}^{2}x+{\text{cos}}^{2}x=1\text{)}$

${\displaystyle \int t\text{sin}\left(t^2\right)\text{cos}\left(t^2\right)dt}$

${\displaystyle \int t^2}{\text{cos}}^2\left(t^3\right)\text{sin}\left(t^3\right)dt$

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

${\displaystyle \int \frac{x^3}{\sqrt{1-x^2}}dx}$

${\displaystyle \int \frac{y^5}{{\left(1-y^3\right)}^{3\text{/}2}}dy}$

${{\displaystyle \int \text{cos}\theta \left(1-\text{cos}\theta \right)}}^{99}\text{sin}\theta d\theta$

${{\displaystyle \int \left(1-{\text{cos}}^3\theta \right)}}^{10}{\text{cos}}^2\theta \text{sin}\theta d\theta$

${\displaystyle \int \left(\text{cos}\theta -1\right){\left({\text{cos}}^2\theta -2\text{cos}\theta \right)}^3\text{sin}\theta d\theta }$

${\displaystyle \int \left({\text{sin}}^2\theta -2\text{sin}\theta \right){\left({\text{sin}}^3\theta -3{\text{sin}}^2\theta \right)}^3\text{cos}\theta d\theta }$

[T] $y=3{\left(1-x\right)}^2$ over $\left[0,2\right]$

[T] $y=x{\left(1-x^2\right)}^3$ over $\left[\mathrm{-1},2\right]$

[T] $y=\text{sin}x{\left(1-\text{cos}x\right)}^2$ over $\left[0,\pi \right]$

[T] $y=\frac{x}{{\left(x^2+1\right)}^2}$ over $\left[\mathrm{-1},1\right]$

${\displaystyle {\int }_0^{1}x\sqrt{1-x^2}}dx$

${\displaystyle {\int }_0^1\frac{x}{\sqrt{1+x^2}}dx}$

${\displaystyle {\int }_0^2\frac{t}{\sqrt{5+t^2}}dt}$

${\displaystyle {\int }_0^1\frac{t}{\sqrt{1+t^3}}dt}$

${\displaystyle {\int }_0^{\pi \text{/}4}{\text{sec}}^2\theta \text{tan}\theta d\theta }$

${\displaystyle {\int }_0^{\pi \text{/}4}\frac{\text{sin}\theta }{{\text{cos}}^4\theta }d\theta }$

[T] ${\displaystyle \int \left(2x+1\right)e^{x^2+x-6}dx}$ over $\left[\mathrm{-3},2\right]$

[T] ${\displaystyle \int \frac{\text{cos}\left(\text{ln}\left(2x\right)\right)}{x}dx}$ on $\left[0,2\right]$

[T] ${\displaystyle \int \frac{3x^2+2x+1}{\sqrt{x^3+x^2+x+4}}dx}$ over $\left[\mathrm{-1},2\right]$

[T] ${\displaystyle \int \frac{\text{sin}x}{{\text{cos}}^{3}x}dx}$ over $\left[-\frac{\pi }{3},\frac{\pi }{3}\right]$

[T] ${\displaystyle \int \left(x+2\right)e^{\text{−}{x}^2-4x+3}dx}$ over $\left[\mathrm{-5},1\right]$

[T] ${\displaystyle \int 3x^2\sqrt{2x^3+1}}dx$ over $\left[0,1\right]$

If $h\left(a\right)=h\left(b\right)$ in ${\displaystyle {\int }_a^{b}g\text{'}\left(h\left(x\right)\right)h\left(x\right)dx},$ what can you say about the value of the integral?

Is the substitution $u=1-x^2$ in the definite integral ${\displaystyle {\int }_0^2\frac{x}{1-x^2}dx}$ okay? If not, why not?

${\displaystyle {\int }_0^{\pi }{\text{cos}}^2\left(2\theta \right)\text{sin}\left(2\theta \right)d\theta }$

${\displaystyle {\int }_0^{\sqrt{\pi }}t\text{cos}\left(t^2\right)\text{sin}\left(t^2\right)dt}$

${\displaystyle {\int }_0^1\left(1-2t\right)dt}$

${\displaystyle {\int }_0^1\frac{1-2t}{\left(1+{\left(t-\frac{1}{2}\right)}^2\right)}dt}$

${\displaystyle {\int }_0^{\pi }\text{sin}\left({\left(t-\frac{\pi }{2}\right)}^3\right)\text{cos}\left(t-\frac{\pi }{2}\right)dt}$

${\displaystyle {\int }_0^2\left(1-t\right)\text{cos}\left(\pi t\right)dt}$

${\displaystyle {\int }_{\pi \text{/}4}^{3\pi \text{/}4}{\text{sin}}^{2}t\text{cos}tdt}$

Show that the average value of $f\left(x\right)$ over an interval $\left[a,b\right]$ is the same as the average value of $f\left(cx\right)$ over the interval $\left[\frac{a}{c},\frac{b}{c}\right]$ for $c>0.$

Find the area under the graph of $f\left(t\right)=\frac{t}{{\left(1+t^2\right)}^a}$ between $t=0$ and $t=x$ where $a>0$ and $a\ne 1$ is fixed, and evaluate the limit as $x\to \infty .$

Find the area under the graph of $g\left(t\right)=\frac{t}{{\left(1-t^2\right)}^a}$ between $t=0$ and $t=x,$ where $0<x<1$ and $a>0$ is fixed. Evaluate the limit as $x\to 1.$

The area of a semicircle of radius 1 can be expressed as ${\displaystyle {\int }_{\mathrm{-1}}^1\sqrt{1-x^2}dx}.$ Use the substitution $x=\text{cos}t$ to express the area of a semicircle as the integral of a trigonometric function. You do not need to compute the integral.

The area of the top half of an ellipse with a major axis that is the x -axis from $x=\mathrm{-1}$ to a and with a minor axis that is the y -axis from $y=\text{−}b$ to b can be written as ${\displaystyle {\int }_{\text{−}a}^{a}b\sqrt{1-\frac{x^2}{a^2}}dx}.$ Use the substitution $x=a\text{cos}t$ to express this area in terms of an integral of a trigonometric function. You do not need to compute the integral.

[T] The following graph is of a function of the form $f(t)=a\text{sin}(nt)+b\text{sin}(mt).$ Estimate the coefficients a and b , and the frequency parameters n and m . Use these estimates to approximate ${\displaystyle {\int }_0^{\pi }f\left(t\right)dt}.$

[T] The following graph is of a function of the form $f\left(x\right)=a\text{cos}\left(nt\right)+b\text{cos}\left(mt\right).$ Estimate the coefficients a and b and the frequency parameters n and m . Use these estimates to approximate ${\displaystyle {\int }_0^{\pi }f\left(t\right)dt}.$