Applications of Integration for Trig Functions

Question 1 of 6

Find the area of the shaded regions formed by the curve

y = - \cos x

$y=-cosx$ that intercepts the x-axis at

x = \frac{π}{2}

$x=pi/2$

1. $\frac{2}{\sqrt{2}} {units}^{2}$ $2/(sqrt2) \text(units)^2$
2. $- \frac{2}{\sqrt{2}} {units}^{2}$ $-2/(sqrt2) \text(units)^2$
3. $2 + \frac{1}{\sqrt{2}} {units}^{2}$ $2+1/(sqrt2) \text(units)^2$
4. $2 - \frac{1}{\sqrt{2}} {units}^{2}$ $2-1/(sqrt2) \text(units)^2$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

First, identify the limits of the shaded regions

First region

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$\frac{π}{2}$ $pi/2$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$\frac{π}{4}$ $pi/4$

\int_{\frac{π}{2}}^{π} - cos x d x

$\int_{\frac{\pi}{2}}^{\pi}\;-\text{cos}\;x\;dx$

Second region

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$π$ $pi$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$\frac{π}{2}$ $pi/2$

\int_{\frac{π}{4}}^{\frac{π}{2}} - cos x d x

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;-\text{cos}\;x\;dx$

Next, integrate the trigonometric functions

First term

\int_{\frac{π}{4}}^{\frac{π}{2}} - cos x d x

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;-\text{cos}\;x\;dx$

=

$=$

- [sin x]_{\frac{π}{4}}^{\frac{π}{2}}

$-\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

Integrate

cos x

$\text(cos) x$

Second term

\int_{\frac{π}{2}}^{π} - cos x d x

$\int_{\frac{\pi}{2}}^{\pi}\;-\text{cos}\;x\;dx$

=

$=$

- [sin x]_{\frac{π}{2}}^{π}

$-\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{2}}^{\pi}$

Integrate

cos x

$\text(cos)x$

Next, get the difference of the upper and lower limits substituted to the integral as

x

$x$ for each term.

First term

	$\bigg[-\text{sin}\;x\bigg]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

$=$ $=$	$- sin (\frac{π}{2}) - [- sin (\frac{π}{4})]$ $-\text(sin)(pi/2)-[-\text(sin)(pi/4)]$	Substitute the limits

$=$ $=$	$- 1 + \frac{1}{\sqrt{2}}$ $-1+1/(sqrt2)$ $\times (- 1)$ $times(-1)$	Multiply the values by $- 1$ $-1$ to make it positive

$=$ $=$	$1 - \frac{1}{\sqrt{2}}$ $1-1/(sqrt2)$

Second term

	$\bigg[-\text{sin}\;x\bigg]_{\frac{\pi}{2}}^{\pi}$

$=$ $=$	$- sin (π) - [- sin (\frac{π}{2})]$ $-\text(sin)(pi)-[-\text(sin)(pi/2)]$	Substitute the limits

$=$ $=$	$0 + 1$ $0+1$	Evaluate
$=$ $=$	$1$ $1$

Finally, add the values of the first and second term to get the area of the shaded regions

$Area$ $\text(Area)$		$1 - \frac{1}{\sqrt{2}}$ $1-1/(sqrt2)$ $+$ $+$ $1$ $1$

	$=$ $=$	$2 - \frac{1}{\sqrt{2}}$ $2-1/(sqrt2)$

2 - \frac{1}{\sqrt{2}}

$2-1/(sqrt2)$

Question 2 of 6

Find the area of the shaded regions formed by the curves

y = \sin 2 x

$y=sin2x$ and

y = \sin x

$y=sinx$ that intercepts at

x = \frac{π}{3}

$x=pi/3$

Write fractions in the format “a/b”

$A =$ $A=$ (1/4) ${units}^{2}$ $\text(units)^2$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

First, identify the limits of the shaded region

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$\frac{π}{3}$ $pi/3$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$0$ $0$

\int_{0}^{\frac{π}{3}} sin 2 x - sin x d x

$\int_{0}^{\frac{\pi}{3}}\;\text{sin}\;2x-\text{sin}\;x\;dx$

Next, integrate the trigonometric function

$\int_{0}^{\frac{\pi}{3}}\;\text{sin}\;2x-\text{sin}\;x\;dx$	$=$ $=$	$\bigg[-\frac{1}{2}\text{cos}\;2x-(-\text{cos}\;x)\bigg]_{0}^{\frac{\pi}{3}}$	Integrate $sin 2 x - sin x$ $\text(sin) 2x-\text(sin) x$

	$=$ $=$	$\bigg[-\frac{1}{2}\text{cos}\;2x+\text{cos}\;x\bigg]_{0}^{\frac{\pi}{3}}$	Simplify

Finally, get the difference of the upper and lower limits substituted to the integral as

x

$x$ for each term.

	$\bigg[-\frac{1}{2}\text{cos}\;2x+\text{cos}\;x\bigg]_{0}^{\frac{\pi}{3}}$

$=$ $=$	$[- \frac{1}{2} cos 2 (\frac{π}{3}) + cos \frac{π}{3}] - [- \frac{1}{2} cos 2 (0) + cos 0]$ $[-1/2\text(cos) 2(pi/3)+\text(cos) pi/3]-[-1/2\text(cos) 2(0)+\text(cos) 0]$	Substitute the limits

$=$ $=$	$[(- \frac{1}{2} \cdot - \frac{1}{2}) + \frac{1}{2}] - [- \frac{1}{2} + 1]$ $[(-1/2*-1/2)+1/2]-[-1/2+1]$	Substitute known trigonometric values

$=$ $=$	$\frac{1}{4} + \frac{1}{2} + \frac{1}{2} - 1$ $1/4+1/2+1/2-1$	Evaluate

$=$ $=$	$\frac{1}{4}$ $1/4$

A = \frac{1}{4} {units}^{2}

$A=1/4 \text(units)^2$

Question 3 of 6

Find the area of the shaded regions formed by the curve

y = \cos x

$y=cosx$ that intercepts the straight line

y = \frac{1}{2}

$y=1/2$ and the x-axis at

x = \frac{π}{2}

$x=pi/2$

1. $A = \frac{π}{6} + 1 + (\sqrt{3}) {units}^{2}$ $A=pi/6+1+(sqrt3) \text(units)^2$
2. $A = \frac{π}{6} + 1 - \frac{\sqrt{3}}{2} {units}^{2}$ $A=pi/6+1-(sqrt3)/2 \text(units)^2$
3. $A = π + 1 - 2 {units}^{2}$ $A=pi+1-2 \text(units)^2$
4. $A = \frac{π}{6} - \frac{\sqrt{3}}{2} {units}^{2}$ $A=pi/6-(sqrt3)/2 \text(units)^2$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

First, find the value of the x-intercept

$y$ $y$	$=$ $=$	$cos x$ $\text(cos) x$

$y$ $y$	$=$ $=$	$\frac{1}{2}$ $1/2$

$cos x$ $\text(cos) x$	$=$ $=$	$\frac{1}{2}$ $1/2$

$x$ $x$	$=$ $=$	$\frac{π}{3}$ $pi/3$	Compute using calculator

Next, solve the red-shaded region by using the formula for area of a rectangle

$length$ $\text(length)$	$=$ $=$	$x$ $x$	$=$ $=$	$\frac{π}{3}$ $pi/3$

$height$ $\text(height)$	$=$ $=$	$y$ $y$	$=$ $=$	$\frac{1}{2}$ $1/2$

$Area$ $\text(Area)$	$=$ $=$	$length \times height$ $\text(length)xx\text(height)$

	$=$ $=$	$\frac{π}{3} \times \frac{1}{2}$ $pi/3xx1/2$	Substitute known values

	$=$ $=$	$\frac{π}{6}$ $pi/6$

Next, identify the limits of the yellow-shaded region

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$\frac{π}{2}$ $pi/2$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$\frac{π}{}$ $pi/$

\int_{\frac{π}{3}}^{\frac{π}{2}} cos x d x

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$

Then, integrate the trigonometric function

\int_{\frac{π}{3}}^{\frac{π}{2}} cos x d x

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$

=

$=$

[sin x]_{\frac{π}{3}}^{\frac{π}{2}}

$\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{3}}^{\frac{\pi}{2}}$

Integrate

cos x

$\text(cos) x$

Next, get the difference of the upper and lower limits substituted to the integral as

x

$x$ for each term.

	$-\bigg[\text{sin}\;x\bigg]_{\frac{\pi}{2}}^{\pi}$

$=$ $=$	$sin (\frac{π}{2}) - [sin (\frac{π}{3})]$ $\text(sin)(pi/2)-[\text(sin)(pi/3)]$	Substitute the limits

$=$ $=$	$1 - \frac{\sqrt{3}}{2}$ $1-(sqrt3)/2$	Substitute known trigonometric values

Finally, add the areas of the rectangle and the yellow-shaded region term to get the total area

Area

$\text(Area)$

$\frac{π}{6}$ $pi/6$

+

$+$ $1 - \frac{\sqrt{3}}{2}$ $1-(sqrt3)/2$

A = \frac{π}{6} + 1 - \frac{\sqrt{3}}{2} {units}^{2}

$A=pi/6+1-(sqrt3)/2 \text(units)^2$

Question 4 of 6

Find the area of the shaded regions formed by the curve

y = \cos x

$y=cosx$ that intercepts the x-axis at

x = \frac{π}{2}

$x=pi/2$ and the straight line

y = 2 x + 1

$y=2x+1$ that intercepts the x-axis at

x = - \frac{1}{2}

$x=-1/2$

1. $A = 1 \frac{1}{4} {units}^{2}$ $A=1 1/4 \text(units)^2$
2. $A = - \frac{1}{4} {units}^{2}$ $A=-1/4 \text(units)^2$
3. $A = \frac{1}{4} {units}^{2}$ $A=1/4 \text(units)^2$
4. $A = 1 {units}^{2}$ $A=1 \text(units)^2$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

First, identify the limits of the shaded regions

Yellow region

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$0$ $0$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$- \frac{1}{2}$ $-1/2$

\int_{- \frac{1}{2}}^{0} (2 x + 1) d x

$\int_{-\frac{1}{2}}^{0}\;(2x+1)\;dx$

Green region

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$\frac{π}{2}$ $pi/2$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	`0

\int_{0}^{\frac{π}{2}} cos x d x

$\int_{0}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$

Next, integrate the trigonometric functions

Yellow region

\int_{- \frac{1}{2}}^{0} (2 x + 1) d x

$\int_{-\frac{1}{2}}^{0}\;(2x+1)\;dx$

=

$=$

- [x^{2} + x]_{- \frac{1}{2}}^{0}

$-\bigg[x^2+x\bigg]_{-\frac{1}{2}}^{0}$

Integrate

2 x + 1

$2x+1$

Green region

\int_{0}^{\frac{π}{2}} cos x d x

$\int_{0}^{\frac{\pi}{2}}\;\text{cos}\;x\;dx$

=

$=$

[sin x]_{0}^{\frac{π}{2}}

$\bigg[\text{sin}\;x\bigg]_{0}^{\frac{\pi}{2}}$

Integrate

cos x

$\text(cos)x$

Next, get the difference of the upper and lower limits substituted to the integral as

x

$x$ for each term.

Yellow region

	$-\bigg[x^2+x\bigg]_{-\frac{1}{2}}^{0}$

$=$ $=$	$[{(0)}^{2} + 0] - [{(- \frac{1}{2})}^{2} + \frac{1}{2}]$ $[(0)^2+0]-[(-1/2)^2+1/2]$	Substitute the limits

$=$ $=$	$- (\frac{1}{4} - \frac{1}{2})$ $-(1/4-1/2)$	Evaluate

$=$ $=$	$- (- \frac{1}{4})$ $-(-1/4)$

$=$ $=$	$\frac{1}{4}$ $1/4$

Green region

	$\bigg[\text{sin}\;x\bigg]_{0}^{\frac{\pi}{2}}$

$=$ $=$	$[sin \frac{π}{2}] - [sin (0)]$ $[\text(sin) (pi)/2]-[\text(sin)(0)]$	Substitute the limits

$=$ $=$	$1 - 0$ $1-0$	$sin \frac{π}{2} = 1$ $\text(sin) pi/2=1$
$=$ $=$	$1$ $1$

Finally, add the values of the shaded regions to get the total area

$Area$ $\text(Area)$		$\frac{1}{4}$ $1/4$ $+$ $+$ $1$ $1$

	$=$ $=$	$1 \frac{1}{4}$ $1 1/4$

A = 1 \frac{1}{4} {units}^{2}

$A=1 1/4 \text(units)^2$

Question 5 of 6

A region formed by the curve

y = 2 \sec 2 x

$y=2sec2x$ and x-axis is rotated about the x-axis from

x = 0

$x=0$ to

x = \frac{π}{6}

$x=pi/6$ . Find the volume of this solid of revolution.

1. $V = \sqrt{3} {units}^{3}$ $V=sqrt3 \text(units)^3$
2. $V = π \sqrt{3} {units}^{3}$ $V=pisqrt3 \text(units)^3$
3. $V = 2 \sqrt{3} {units}^{3}$ $V=2sqrt3 \text(units)^3$
4. $V = 2 π \sqrt{3} {units}^{3}$ $V=2pisqrt3 \text(units)^3$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

Volume of a Solid of Revolution

V = \int_{a}^{b} π y^{2} d x

$V=\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$

First, identify the upper and lower limits

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$\frac{π}{6}$ $pi/6$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$0$ $0$

Next, substitute the known values to the volume formula

$V$ $V$	$=$ $=$	$\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$

	$=$ $=$	$\int_{\color{#D800AD}{0}}^{\color{#9a00c7}{\frac{\pi}{6}}}\;\pi\cdot(\color{#004ec4}{2\text{sec}})^2\;2x\;dx$	Substitute known values

	$=$ $=$	$\int_{0}^{\frac{\pi}{6}}\;\pi\cdot 4\text{sec}^2\;2x\;dx$	Evaluate

	$=$ $=$	$4\pi\int_{0}^{\frac{\pi}{6}}\;\text{sec}^2\;2x\;dx$

	$=$ $=$	$4\pi\bigg[\frac{1}{2}\;\text{tan}\;2x\bigg]_{0}^{\frac{\pi}{6}}$	Integrate

	$=$ $=$	$2\pi\bigg[\text{tan}\;2x\bigg]_{0}^{\frac{\pi}{6}}$	Simplify

	$=$ $=$	$2\pi\bigg[\text{tan}\;2\left(\color{#9a00c7}{\frac{\pi}{6}}\right)-\text{tan}\;2(\color{#D800AD}{0})\bigg]$	Substitute the limits

	$=$ $=$	$2\pi\bigg[\text{tan}\left(\frac{\pi}{3}\right)-0\bigg]$

	$=$ $=$	$2\pi\cdot\sqrt3$	$tan \frac{π}{3} = \sqrt{3}$ $\text(tan) pi/3=sqrt3$

	$=$ $=$	$2\pi\sqrt3$

V = 2 π \sqrt{3} {units}^{3}

$V=2pisqrt3 \text(units)^3$

Question 6 of 6

A region formed by the curve

y = \sqrt{\sin x}

$y=sqrt(sinx)$ and x-axis is rotated about the x-axis from

x = 0

$x=0$ to

x = π

$x=pi$ . Find the volume of this solid of revolution.

1. $V = 2 π {units}^{3}$ $V=2pi \text(units)^3$
2. $V = π {units}^{3}$ $V=pi \text(units)^3$
3. $V = 2 {units}^{3}$ $V=2 \text(units)^3$
4. $V = \frac{2}{π} {units}^{3}$ $V=2/pi \text(units)^3$

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Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

Volume of a Solid of Revolution

V = \int_{a}^{b} π y^{2} d x

$V=\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$

First, remove the surd from the function

$y$ $y$	$=$ $=$	$\sqrt{\sin x}$ $sqrt(sinx)$
$y^{2}$ $y^2$	$=$ $=$	${(\sqrt{\sin x})}^{2}$ $(sqrt(sinx))^2$	Raise both sides by $2$ $2$
$y^{2}$ $y^2$	$=$ $=$	$\sin x$ $sinx$

Next, identify the upper and lower limits

$Upper Limit$ $\text(Upper Limit)$	$=$ $=$	$π$ $pi$

$Lower Limit$ $\text(Lower Limit)$	$=$ $=$	$0$ $0$

Next, substitute the known values to the volume formula

$V$ $V$	$=$ $=$	$\int_{\color{#D800AD}{a}}^{\color{#9a00c7}{b}}\;\pi\color{#004ec4}{y}^2\;dx$

	$=$ $=$	$\int_{\color{#D800AD}{0}}^{\color{#9a00c7}{\pi}}\;\pi\cdot(\color{#004ec4}{\text{sin}})\;x\;dx$	Substitute known values


	$=$ $=$	$\pi\bigg[-\text{cos}\;x\bigg]_{0}^{\pi}$	Integrate

	$=$ $=$	$\pi[- \text{cos} (\color{#9a00c7}{\pi}) ]-[-\text{cos}(\color{#D800AD}{0})]$	Substitute the limits

	$=$ $=$	$\pi[-(-1)-(-1)]$	Substitute known trigonometric values

	$=$ $=$	$\pi(1+1)$

	$=$ $=$	$2\pi$

V = 2 π {units}^{3}

$V=2pi \text(units)^3$

Applications of Integration for Trig Functions

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Quiz summary

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1. Question

Hint

Great Work!

Integrals of Trigonometric Functions

2. Question

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Correct!

Integrals of Trigonometric Functions

3. Question

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Keep Going!

Integrals of Trigonometric Functions

4. Question

Hint

Fantastic!

Integrals of Trigonometric Functions

5. Question

Hint

Excellent!

Integrals of Trigonometric Functions

Volume of a Solid of Revolution

6. Question

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Nice Job!

Integrals of Trigonometric Functions

Volume of a Solid of Revolution

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