Trig Applications of Integration

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

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

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

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

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

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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}

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

Trig Applications of Integration

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Quiz summary

Information

1. Question

Hint

Great Work!

Integrals of Trigonometric Functions

2. Question

Hint

Correct!

Integrals of Trigonometric Functions

3. Question

Hint

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

Hint

Nice Job!

Integrals of Trigonometric Functions

Volume of a Solid of Revolution

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