Trig Applications of Integration

Question 1 of 6

Find the area of the shaded regions formed by the curve

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

$x=pi/2$

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

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Integrals of Trigonometric Functions

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

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

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

First, identify the limits of the shaded regions

First region

$\text(Upper Limit)$	$=$	$pi/2$

$\text(Lower Limit)$	$=$	$pi/4$

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

Second region

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

$\text(Lower Limit)$	$=$	$pi/2$

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

Next, integrate the trigonometric functions

First term

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

$=$

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

Integrate

$\text(cos) x$

Second term

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

$=$

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

Integrate

$\text(cos)x$

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

$x$ for each term.

First term

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

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

$=$	$-1+1/(sqrt2)$ $times(-1)$	Multiply the values by $-1$ to make it positive

$=$	$1-1/(sqrt2)$

Second term

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

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

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

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

$\text(Area)$		$1-1/(sqrt2)$ $+$ $1$

	$=$	$2-1/(sqrt2)$

$2-1/(sqrt2)$

Question 2 of 6

Find the area of the shaded regions formed by the curves

$y=sin2x$ and

$y=sinx$ that intercepts at

$x=pi/3$

Write fractions in the format “a/b”

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

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Integrals of Trigonometric Functions

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

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

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

First, identify the limits of the shaded region

$\text(Upper Limit)$	$=$	$pi/3$

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

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

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 $\text(sin) 2x-\text(sin) x$

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

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

$x$ for each term.

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

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

$=$	$[(-1/2*-1/2)+1/2]-[-1/2+1]$	Substitute known trigonometric values

$=$	$1/4+1/2+1/2-1$	Evaluate

$=$	$1/4$

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

Question 3 of 6

Find the area of the shaded regions formed by the curve

$y=cosx$ that intercepts the straight line

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

$x=pi/2$

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

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Integrals of Trigonometric Functions

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

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

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

First, find the value of the x-intercept

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

$y$	$=$	$1/2$

$\text(cos) x$	$=$	$1/2$

$x$	$=$	$pi/3$	Compute using calculator

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

$\text(length)$	$=$	$x$	$=$	$pi/3$

$\text(height)$	$=$	$y$	$=$	$1/2$

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

	$=$	$pi/3xx1/2$	Substitute known values

	$=$	$pi/6$

Next, identify the limits of the yellow-shaded region

$\text(Upper Limit)$	$=$	$pi/2$

$\text(Lower Limit)$	$=$	$pi/$

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

Then, integrate the trigonometric function

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

$=$

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

Integrate

$\text(cos) x$

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

$x$ for each term.

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

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

$=$	$1-(sqrt3)/2$	Substitute known trigonometric values

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

$\text(Area)$

$pi/6$

$+$ $1-(sqrt3)/2$

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

Question 4 of 6

Find the area of the shaded regions formed by the curve

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

$x=pi/2$ and the straight line

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

$x=-1/2$

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

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Integrals of Trigonometric Functions

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

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

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

First, identify the limits of the shaded regions

Yellow region

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

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

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

Green region

$\text(Upper Limit)$	$=$	$pi/2$

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

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

Next, integrate the trigonometric functions

Yellow region

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

$=$

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

Integrate

$2x+1$

Green region

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

$=$

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

Integrate

$\text(cos)x$

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

$x$ for each term.

Yellow region

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

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

$=$	$-(1/4-1/2)$	Evaluate

$=$	$-(-1/4)$

$=$	$1/4$

Green region

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

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

$=$	$1-0$	$\text(sin) pi/2=1$
$=$	$1$

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

$\text(Area)$		$1/4$ $+$ $1$

	$=$	$1 1/4$

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

Question 5 of 6

A region formed by the curve

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

$x=0$ to

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

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

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Integrals of Trigonometric Functions

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

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

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

Volume of a Solid of Revolution

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

First, identify the upper and lower limits

$\text(Upper Limit)$	$=$	$pi/6$

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

Next, substitute the known values to the volume formula

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

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

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

	$=$	$2\pi\cdot\sqrt3$	$\text(tan) pi/3=sqrt3$

	$=$	$2\pi\sqrt3$

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

Question 6 of 6

A region formed by the curve

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

$x=0$ to

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

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

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Integrals of Trigonometric Functions

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

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

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

Volume of a Solid of Revolution

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

First, remove the surd from the function

$y$	$=$	$sqrt(sinx)$
$y^2$	$=$	$(sqrt(sinx))^2$	Raise both sides by $2$
$y^2$	$=$	$sinx$

Next, identify the upper and lower limits

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

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

Next, substitute the known values to the volume formula

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

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

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

	$=$	$2\pi$

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

Trig Applications of Integration

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Quiz summary

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1. Question

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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

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Fantastic!

Integrals of Trigonometric Functions

5. Question

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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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