17Calculus Integrals - Volumes
This page gets you started on calculating volumes of the first type, volumes with known cross-section. For other ways to calculate volume, see the links in the related topics panel.
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Exam Preparation Suggestions |
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1. Volume integral problems are notoriously long and most will take more time than you are probably used to to work all the way through. So, once you are down to a few hours before your exam, just take the time to set up the integrals but not evaluate them completely. Most of your points will probably come from setting up the integral. By now, you know how to evaluate most integrals that you will see in class and evaluating them is the most time-consuming part. So focus on where you can get the most points and make sure you are confident setting up the integrals. |
Volume With A Known Cross-Section
built with GeoGebra |
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To get a feel for how to use integrals to find volumes, we will start with finding volumes when we know the equation of a cross-sectional area. Let's start with some examples that you are familiar with.
You know that the volume of a rectangular box is length times width times height or, in equation terms, if l is the length, w is the width and h is the height, the volume is \(V=lwh\). Let's look at this from an integral calculus point of view. Let's take a cross-section parallel to the top of the box. The cross-section is the same no matter where we take it. In equation form, the area of the cross-section is \(A=lw\).
The volume of a solid with cross-section \(A(x)\) is \(V=\int_a^b{A(x)~dx}\). In our example, our height goes from \(0\) to \(h\). So the volume is \(V=\int_0^h{lw~dx} = \) \( [lwx]_0^h = \) \(lwh\), i.e. \(V=lwh\) as we would expect.
In our example, the area, \(A=lw\), was constant. However, usually, the cross-section will be different depending on where we take the cross-section. That's why we write the area as \(A(x)\). Of course, if we take the cross-section perpendicular to the y-axis, the volume integral would be \(A(y)=\int_c^d{A(y)~dy}\).
Use the volume integral to calculate the volume of a cylinder with radius r and height h.
\(V=\pi r^2 h\)
Use the volume integral to calculate the volume of a cylinder with radius r and height h.
Solution |
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Problem Statement - Use the volume integral to calculate the volume of a cylinder with radius r and height h.
Solution - If you take the cross-section parallel to the base, the area is a circle, which is \(A=\pi r^2\). We integrate from \(0\) to \(h\) to get \(V = \int_0^h{\pi r^2~dx} = \) \( \pi r^2 x|_0^h = \) \(\pi r^2 h\).
For a more complete explanation including a 3dim graph, see this Pauls Online Notes page.
Final Answer |
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\(V=\pi r^2 h\) |
Okay, so far the cross-section has been constant making the equation of the area quite simple. Let's look at an example where the cross-section is not constant.
Calculate the volume of the solid of a pyramid with a square base of side L and height h.
\(V=L^2h/3\)
Calculate the volume of the solid of a pyramid with a square base of side L and height h.
Solution |
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Problem Statement - Calculate the volume of the solid of a pyramid with a square base of side L and height h.
Solution - We will outline the solution here but if you want a complete, well-written solution with 3dim plots, see this Pauls Online Notes page.
For this problem, it helps to think of the pyramid as upside-down with it's point sitting at the origin and the base up in the air. Since the cross-section is a square, we know the area of the cross-section is \(A=s^2\), where \(s\) is the length of side. However, \(s\) is different depending on where the cross-section is taken. Using similar triangles, we can get \(s=Ly/h\) where \(y\) is the height above the origin.
The volume integral is then \(V=\int_0^h{L^2y^2/h^2~dy}\) which solves to \(L^2h/3\).
Final Answer |
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\(V=L^2h/3\) |
That example is just one way to work the problem. Here are several videos showing other ways to solve it.
Michel vanBiezen - Calculus - Integration: Volume of a Pyramid - Square Base Pyramid [8mins-12secs]
video by Michel vanBiezen |
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MIP4U - Volume of a Solid with Known Cross Section Using Integration - Pyramid [6mins-42secs]
video by MIP4U |
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David Lippman - Volumes by integrating cross sections - square pyramid [8mins-16secs]
video by David Lippman |
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Integrals and Volumes: Volume of a Pyramid [7mins-35secs]
There are several more examples on this Pauls Online Notes page that we recommend. Before we jump into the practice problems, take a few minutes to watch this video. It has a good explanation and several examples.
ProfRobBob - Volumes of Solids with Known Cross Sections 3 Examples
video by ProfRobBob |
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Practice
The base of a solid is bounded by the curves \(f(x)=2-2x\) and \(g(x)=x^2-1\) and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.
Problem Statement |
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The base of a solid is bounded by the curves \(f(x)=2-2x\) and \(g(x)=x^2-1\) and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.
Hint |
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This solid is difficult to visualize. However, the cross sections are squares. So, just find an equation for the length in the y-direction (perpendicular to the x-axis) and square it to get an equation for the area.
Problem Statement |
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The base of a solid is bounded by the curves \(f(x)=2-2x\) and \(g(x)=x^2-1\) and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.
Final Answer |
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\(512/15\)
Problem Statement |
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The base of a solid is bounded by the curves \(f(x)=2-2x\) and \(g(x)=x^2-1\) and the cross sections perpendicular to the x-axis are perfect squares. Find the volume of the solid.
Hint |
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This solid is difficult to visualize. However, the cross sections are squares. So, just find an equation for the length in the y-direction (perpendicular to the x-axis) and square it to get an equation for the area.
Solution |
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2262 video
video by PatrickJMT |
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Final Answer |
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\(512/15\) |
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Calculate the volume of the solid whose cross-section is a rectangle with a height of \(y=\cos(x)\) and width of 2 over the interval \(0\leq x \leq \pi/2\).
Problem Statement |
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Calculate the volume of the solid whose cross-section is a rectangle with a height of \(y=\cos(x)\) and width of 2 over the interval \(0\leq x \leq \pi/2\).
Final Answer |
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\(V=2\)
Problem Statement |
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Calculate the volume of the solid whose cross-section is a rectangle with a height of \(y=\cos(x)\) and width of 2 over the interval \(0\leq x \leq \pi/2\).
Solution |
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2263 video
video by MIP4U |
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Final Answer |
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\(V=2\) |
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Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of \(y=x^2\) inches on the interval \([0,2]\).
Problem Statement |
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Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of \(y=x^2\) inches on the interval \([0,2]\).
Hint |
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Plot the equation \(y=x^2\) from \(x=0\) to \(x=2\). The height of the triangle is the distance from the x-axis to the curve. Remember the equation of the area of a triangle is \(A=(1/2)bh\) where b is the base and h is the height.
Don’t forget to give units in your answer.
Problem Statement |
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Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of \(y=x^2\) inches on the interval \([0,2]\).
Final Answer |
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\(16/3\) in3
Problem Statement |
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Calculate the volume of the solid where the cross-section is a triangle with a base of 4 inches and height of \(y=x^2\) inches on the interval \([0,2]\).
Hint |
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Plot the equation \(y=x^2\) from \(x=0\) to \(x=2\). The height of the triangle is the distance from the x-axis to the curve. Remember the equation of the area of a triangle is \(A=(1/2)bh\) where b is the base and h is the height.
Don’t forget to give units in your answer.
Solution |
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2264 video
video by MIP4U |
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Final Answer |
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\(16/3\) in3 |
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Calculate the volume of the solid where the cross-section perpendicular to the x-axis is a circle with diameter \(y=\sqrt{x}\) on the interval \([0,4]\).
Problem Statement |
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Calculate the volume of the solid where the cross-section perpendicular to the x-axis is a circle with diameter \(y=\sqrt{x}\) on the interval \([0,4]\).
Final Answer |
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\(V=2\pi\)
Problem Statement |
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Calculate the volume of the solid where the cross-section perpendicular to the x-axis is a circle with diameter \(y=\sqrt{x}\) on the interval \([0,4]\).
Solution |
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2265 video
video by MIP4U |
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Final Answer |
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\(V=2\pi\) |
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Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8-x^2\) and cross-sections parallel to the y-axis that are semi-circles.
Problem Statement |
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Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8-x^2\) and cross-sections parallel to the y-axis that are semi-circles.
Hint |
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The diameter of the semi-circle is the distance between the curves.
Problem Statement |
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Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8-x^2\) and cross-sections parallel to the y-axis that are semi-circles.
Final Answer |
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\(\displaystyle{ V = \frac{256\pi}{15} }\)
Problem Statement |
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Calculate the volume of a solid with base bounded by the curves \(y=x^2\) and \(y=8-x^2\) and cross-sections parallel to the y-axis that are semi-circles.
Hint |
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The diameter of the semi-circle is the distance between the curves.
Solution |
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This video sets up the integral but does not evaluate it. Here is the remainder of the solution.
\(\displaystyle{ V = \int_{-2}^{2}{ \frac{\pi}{2} (4-x^2)^2 ~dx } }\) |
\(\displaystyle{ V = \frac{\pi}{2} \int_{-2}^2{ 16+x^4-8x^2 ~ dx } }\) |
\(\displaystyle{ V = \frac{\pi}{2} \left[ 16x + \frac{x^5}{5} - \frac{8x^3}{3} \right]_{-2}^2 }\) |
\(\displaystyle{ V = \frac{\pi}{2} \left[ 16(2)+ \frac{2^5}{5} - \frac{8}{3}(2^3) \right] - }\) \(\displaystyle{ \frac{\pi}{2} \left[ 16(-2) + \frac{(-2)^5}{5} -\frac{8}{3}(-2)^3 \right] }\) |
\(\displaystyle{ V = \frac{\pi}{2} \left[ 32+\frac{32}{5}-\frac{64}{3} \right] - }\) \(\displaystyle{ \frac{\pi}{2} \left[ -32 - \frac{32}{5} + \frac{64}{3} \right] }\) |
\(\displaystyle{ V = \frac{\pi}{2} \left[ 64+\frac{64}{5}- \frac{128}{3} \right] }\) |
\(\displaystyle{ V = \frac{\pi}{2} \left[ \frac{960}{15} + \frac{192}{15} - \frac{640}{15} \right] }\) |
\(\displaystyle{ V = \frac{\pi}{2}\left[ \frac{512}{15} \right] }\) |
\(\displaystyle{ V = \frac{256\pi}{15} }\) |
2266 video
video by David Lippman |
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Final Answer |
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\(\displaystyle{ V = \frac{256\pi}{15} }\) |
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Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by \(y=\pm (2-x^2/8)\) and \(x=2\). Units are in feet.
Problem Statement |
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Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by \(y=\pm (2-x^2/8)\) and \(x=2\). Units are in feet.
Hint |
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The equation of the area of equilateral triangle with side \(a\) is \(A=\sqrt{3}(a^2/4)\).
Problem Statement |
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Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by \(y=\pm (2-x^2/8)\) and \(x=2\). Units are in feet.
Final Answer |
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\(\displaystyle{\frac{153\sqrt{3}}{10}}\) ft3
Problem Statement |
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Calculate the volume of a solid with an equilateral triangular cross-section parallel to the y-axis bounded by \(y=\pm (2-x^2/8)\) and \(x=2\). Units are in feet.
Hint |
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The equation of the area of equilateral triangle with side \(a\) is \(A=\sqrt{3}(a^2/4)\).
Solution |
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He writes his answer as \(15\frac{3}{10}\sqrt{3}\) which is equal to \((15+3/10)\sqrt{3}\). Do not write it this way. This is only done in basic algebra. After you reach algebra 2, trig and especially calculus, this is interpreted as \(15(3/10)\sqrt{3}\) which is not what he means.
2267 video
Final Answer |
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\(\displaystyle{\frac{153\sqrt{3}}{10}}\) ft3 |
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For \(1 \leq x \leq 4\), determine the volume of the solid where each face is a triangle with base \((x+1)\) ft and height \(\sqrt{x}\) ft.
Problem Statement |
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For \(1 \leq x \leq 4\), determine the volume of the solid where each face is a triangle with base \((x+1)\) ft and height \(\sqrt{x}\) ft.
Final Answer |
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\(128/15\) ft3
Problem Statement |
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For \(1 \leq x \leq 4\), determine the volume of the solid where each face is a triangle with base \((x+1)\) ft and height \(\sqrt{x}\) ft.
Solution |
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2268 video
video by MIP4U |
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Final Answer |
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\(128/15\) ft3 |
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Calculate the volume of the solid where each face at x is a square with a height of \(\sqrt{x}\) on \([0,4]\).
Problem Statement |
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Calculate the volume of the solid where each face at x is a square with a height of \(\sqrt{x}\) on \([0,4]\).
Final Answer |
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\(V=8\)
Problem Statement |
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Calculate the volume of the solid where each face at x is a square with a height of \(\sqrt{x}\) on \([0,4]\).
Solution |
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2269 video
video by MIP4U |
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Final Answer |
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\(V=8\) |
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Calculate the volume of the solid whose base is bounded by \(y=6-2x^2/3\) and \(y=0\) with cross-section parallel to the x-axis is a triangle whose height and base are equal.
Problem Statement |
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Calculate the volume of the solid whose base is bounded by \(y=6-2x^2/3\) and \(y=0\) with cross-section parallel to the x-axis is a triangle whose height and base are equal.
Final Answer |
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\(V=54\) cubic units
Problem Statement |
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Calculate the volume of the solid whose base is bounded by \(y=6-2x^2/3\) and \(y=0\) with cross-section parallel to the x-axis is a triangle whose height and base are equal.
Solution |
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2270 video
video by MIP4U |
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Final Answer |
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\(V=54\) cubic units |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
related topics on other pages |
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external links you may find helpful |
Larson Calculus (Google books) Volume - Disk, Washer Methods and Shell, Cylinder Methods pgs 458-477 |
Interactive Mathematics - Volume of Solid of Revolution by Integration (Disk method) |
Pauls Online Notes - Volumes of Solids of Revolution / Method of Rings |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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