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Trig Applications of IntegrationTrig Applications of Integration
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Question 1 of 6
1. Question
Find the area of the shaded regions formed by the curve `y=-cosx` that intercepts the x-axis at `x=pi/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 regionsFirst 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 functionsFirst 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
2. Question
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
3. Question
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`
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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
4. Question
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`
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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 regionsYellow 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 functionsYellow 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
5. Question
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.
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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
6. Question
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.
Hint
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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`
Quizzes
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