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17Calculus Precalculus - Completing The Square

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The idea of completing the square is just to move the terms around so that an expression is in the form we want, without changing the problem. What we want to end up with is a term that is squared (thus the name completing the square). We may also end up with an extra constant. To do this, we need some basic guidelines.

Basic Guidelines For Completing The Square

guideline 1 - - We may add only zero, i.e. if we add a number somewhere, we are required to also subtract the same number.
Now, this can get confusing if you are not paying attention since there are actually two way to do this. First, is to add and subtract the same number on the same side of the equal sign. For example, in \(2x=7 \to 2x+a-a=7\), we added and subtracted a on the left of the equal sign. This is the best way to do it and, in our opinion, the most understandable. When you are working problems and, especially when you are first learning this technique, it is always best to think about it this way.

This second way is equivalent but looks a bit different. Instead of adding and subtracting the same number on one side of the equal sign, some instructors teach to add the same number on both sides, like this \(2x=7 \to 2x+a=7+a\).

The key is to choose one way and stick with it. Don't try to go back and forth or do both. That's where the confusion comes in.

guideline 2 - - We may multiply by only one, i.e. if we multiply by a number somewhere, we must also divide by the same number.
Just like the first guideline, this can get confusing too, for the exact same reason. The first and best way is to multiply by the same number on one side of the equal sign. It looks like this \(2x=7 \to 2x(b/b)=7\). Notice, that we multiplied and divided by the same number b on the left side.

Of course, we could also multiply both sides of the equal sign by the same number, like this \(2x=7 \to 2x(b)=7(b)\), which is equivalent. The point is to choose one or the other and stick with it. Then, when someone shows you the other way, go back to the way you understand and notice the equivalency.

key to guidelines 1 and 2 - - Now, here is the key to getting your head around the previous two guidelines. If you have an equal sign, think of it as a scale with two sides and the equal sign is the pole in between, like this picture. When you are given an equation in a problem statement, the equation is balanced. Your job is to keep it balanced. The only way to do that is to either not change the weight on one side or by changing the weight on both sides by the same amount.
Of course, if you have no equal sign, your only choice is add/multiply the same number to the equation. So the first technique in both of the first two guidelines works with or without an equal sign. This is another great reason to learn it that way first.

guideline 3 - - Completing the square works only on polynomials whose highest power is \(2\). However, you can sometimes use substitution to get your expression in this form or you can isolate the squared the part of the equation to then be able to use this technique. A quick example of the last idea is if you have an expression like \( 6x^3-36x^2 \), you can factor out an x first to get \( x(6x^2-36x) \) and complete the square on the inside expression.

There are additional guidelines which we explain by working through the next example.

Example - - Complete the square on the expression \( 2x^2-12x \).

First, there must always, always, always, without exception, be a positive one coefficient on the \( x^2 \) term (guideline 4). That means we need to factor out a \(2\) to get \( 2(x^2-6x) \).
Now we take the coefficient of the x-term, which is \(-6\) (notice I took the sign too; you'll see why in a minute) and we divide the \( -6 \) by \(2\) and square it, like this \(\displaystyle{ \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9}\). Okay, then we take that \(9\) and add it and subtract it INSIDE the parentheses to get \( 2(x^2-6x+9-9) \).

Now you are saying to yourself, this instructor is crazy. You didn't really do anything. Aha! That's correct (not about the crazy part but about the part that we didn't do anything). We didn't change the problem. That is important. We don't want to change the problem. By adding and subtracting the same number, we are essentially adding zero. That's good.

Notice we emphasized that both operations had to happen inside the parentheses (guideline 5). Because of the \(2\) that we factored out, if we did one inside and one outside, we would be adding on the inside by \(18\) and subtracting on the outside by \(9\). This would change the problem and our answer would be end up being incorrect. So, we added and subtracted by \(9\), both inside the parentheses.

Okay, now we are going to take the \(9\) we added and associate it with the terms with x in them. Look closely. \( x^2-6x+9=(x-3)^2 \) Because of the way we got the number \(9\) (by dividing by two and squaring) this creates a squared term, thus 'completing' the square. Do you see that?

Now let's see what we have. \( 2(x^2-6x+9-9)=2[(x-3)^2-9] \) We can distribute the \(2\) on the right and end up with \( 2x^2-12x = 2(x-3)^2-18 \). And that's it. (If you need to, take a minute to multiply out the expression on the right side of the equal sign to convince yourself that it is the same as what we started out with.)

Final Answer

\( 2x^2-12x = 2(x-3)^2-18 \)

One question we often get is, how do you know if the sign inside squared term is positive or negative? In the above example, you might ask whether the squared term is \((x-3)\) or \((x+3)\). Well, the sign in the squared term should be the same sign as the coefficient of the x-term in the original problem (after we factor out what we need to get a positive one in front of the squared term). In this case, the coefficient of the x-term is negative (\(-6\)) and so the squared term is \((x-3)\). If you multiply out \(2(x-3)^2-18\) to get \(2x^2-12x\) in the final answer above, you may be able to see this more clearly.

Here is an interesting video that may help you visualize what is going on with this technique.

cylurian - Completing The Square [5min-52secs]

video by cylurian

As a side note, you can use completing the square to derive the quadratic formula. This is important to know since knowing where an equation comes from means you really understand when and how to use an equation. This next video shows how to do this. Before watching this video, try it on your own.

cylurian - completing the square → quadratic formula [9min-53secs]

video by cylurian

Okay, time for some practice problems.

Practice

Instructions - - Unless otherwise instructed, solve these problems by completing the square, giving your answers in exact form.

Express \( x^2 + 8x - 1 \) in the form \( (x+a)^2 + b \) by completing the square.

Problem Statement

Express \( x^2 + 8x - 1 \) in the form \( (x+a)^2 + b \) by completing the square.

Final Answer

\( (x+4)^2 - 17 \)

Problem Statement

Express \( x^2 + 8x - 1 \) in the form \( (x+a)^2 + b \) by completing the square.

Solution

2803 video

Final Answer

\( (x+4)^2 - 17 \)

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Express \( x^2 - 10x + 3 \) in the form \( (x+a)^2 + b \) by completing the square.

Problem Statement

Express \( x^2 - 10x + 3 \) in the form \( (x+a)^2 + b \) by completing the square.

Final Answer

\( (x-5)^2 - 22 \)

Problem Statement

Express \( x^2 - 10x + 3 \) in the form \( (x+a)^2 + b \) by completing the square.

Solution

2804 video

Final Answer

\( (x-5)^2 - 22 \)

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Express \( x^2 +3x - 2 \) in the form \( (x+a)^2 + b \) by completing the square.

Problem Statement

Express \( x^2 +3x - 2 \) in the form \( (x+a)^2 + b \) by completing the square.

Final Answer

\( (x + 3/2)^2 - 17/4 \)

Problem Statement

Express \( x^2 +3x - 2 \) in the form \( (x+a)^2 + b \) by completing the square.

Solution

2805 video

Final Answer

\( (x + 3/2)^2 - 17/4 \)

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Express \( x^2 + 12x - 5 \) in the form \( (x+a)^2 + b \) by completing the square.

Problem Statement

Express \( x^2 + 12x - 5 \) in the form \( (x+a)^2 + b \) by completing the square.

Final Answer

\( (x + 6)^2 - 41 \)

Problem Statement

Express \( x^2 + 12x - 5 \) in the form \( (x+a)^2 + b \) by completing the square.

Solution

2806 video

Final Answer

\( (x + 6)^2 - 41 \)

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Express \( x^2 - 5x + 1 \) in the form \( (x+a)^2 + b \) by completing the square.

Problem Statement

Express \( x^2 - 5x + 1 \) in the form \( (x+a)^2 + b \) by completing the square.

Final Answer

\( (x - 5/2)^2 - 21/4 \)

Problem Statement

Express \( x^2 - 5x + 1 \) in the form \( (x+a)^2 + b \) by completing the square.

Solution

2807 video

Final Answer

\( (x - 5/2)^2 - 21/4 \)

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Express \( 3x^2 - 12x + 5 \) in the form \( a(x+b)^2 + c \) by completing the square.

Problem Statement

Express \( 3x^2 - 12x + 5 \) in the form \( a(x+b)^2 + c \) by completing the square.

Final Answer

\( 3(x - 2)^2 - 7 \)

Problem Statement

Express \( 3x^2 - 12x + 5 \) in the form \( a(x+b)^2 + c \) by completing the square.

Solution

2808 video

Final Answer

\( 3(x - 2)^2 - 7 \)

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Express \( 2x^2 - x - 1 \) in the form \( a(x+b)^2 + c \) by completing the square.

Problem Statement

Express \( 2x^2 - x - 1 \) in the form \( a(x+b)^2 + c \) by completing the square.

Final Answer

\( 2(x - 1/4)^2 - 9/8 \)

Problem Statement

Express \( 2x^2 - x - 1 \) in the form \( a(x+b)^2 + c \) by completing the square.

Solution

2809 video

Final Answer

\( 2(x - 1/4)^2 - 9/8 \)

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Express \( 2x^2 + 12x - 3 \) in the form \( a(x+b)^2 + c \) by completing the square.

Problem Statement

Express \( 2x^2 + 12x - 3 \) in the form \( a(x+b)^2 + c \) by completing the square.

Final Answer

\( 2(x+3)^2 - 21 \)

Problem Statement

Express \( 2x^2 + 12x - 3 \) in the form \( a(x+b)^2 + c \) by completing the square.

Solution

2810 video

Final Answer

\( 2(x+3)^2 - 21 \)

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Express \( 5x^2 + 3x - 2 \) in the form \( a(x+b)^2 + c \) by completing the square.

Problem Statement

Express \( 5x^2 + 3x - 2 \) in the form \( a(x+b)^2 + c \) by completing the square.

Final Answer

\( 5(x + 3/10)^2 - 49/20 \)

Problem Statement

Express \( 5x^2 + 3x - 2 \) in the form \( a(x+b)^2 + c \) by completing the square.

Solution

2811 video

Final Answer

\( 5(x + 3/10)^2 - 49/20 \)

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Express \( 5 - 8x - x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Problem Statement

Express \( 5 - 8x - x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Final Answer

\( 21 - (x + 4)^2 \)

Problem Statement

Express \( 5 - 8x - x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Solution

2812 video

Final Answer

\( 21 - (x + 4)^2 \)

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Express \( 5 + 12x - 2x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Problem Statement

Express \( 5 + 12x - 2x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Final Answer

\( 23 - 2(x-3)^2 \)

Problem Statement

Express \( 5 + 12x - 2x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Solution

2813 video

Final Answer

\( 23 - 2(x-3)^2 \)

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Express \( 4 + 10x - x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Problem Statement

Express \( 4 + 10x - x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Final Answer

\( 29 - (x - 5)^2 \)

Problem Statement

Express \( 4 + 10x - x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Solution

2814 video

Final Answer

\( 29 - (x - 5)^2 \)

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Express \( 1 - 3x - 5x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Problem Statement

Express \( 1 - 3x - 5x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Final Answer

\( 29/20 - 5(x + 3/10)^2 \)

Problem Statement

Express \( 1 - 3x - 5x^2 \) in the form \( a - b(x+c)^2 \) by completing the square.

Solution

2815 video

Final Answer

\( 29/20 - 5(x + 3/10)^2 \)

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Solve \( 2x^2 + 7x + 5= 0 \) by completing the square.

Problem Statement

Solve \( 2x^2 + 7x + 5= 0 \) by completing the square.

Final Answer

\( x = -1 \) or \( x = -5/2 \)

Problem Statement

Solve \( 2x^2 + 7x + 5= 0 \) by completing the square.

Solution

2816 video

Final Answer

\( x = -1 \) or \( x = -5/2 \)

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Solve \( x^2 - 5x = -4 \) by completing the square.

Problem Statement

Solve \( x^2 - 5x = -4 \) by completing the square.

Final Answer

\( x = 1 \) or \( x = 4 \)

Problem Statement

Solve \( x^2 - 5x = -4 \) by completing the square.

Solution

2817 video

Final Answer

\( x = 1 \) or \( x = 4 \)

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Solve \( 2x^2 - 8x + 3 = 0 \) by completing the square.

Problem Statement

Solve \( 2x^2 - 8x + 3 = 0 \) by completing the square.

Final Answer

\(\displaystyle{ x = 2 \pm \frac{\sqrt{10}}{2} }\)

Problem Statement

Solve \( 2x^2 - 8x + 3 = 0 \) by completing the square.

Solution

2818 video

Final Answer

\(\displaystyle{ x = 2 \pm \frac{\sqrt{10}}{2} }\)

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Solve \( x^2 + 4x = 5 \) by completing the square.

Problem Statement

Solve \( x^2 + 4x = 5 \) by completing the square.

Final Answer

\( x = 1 \) or \( x = -5 \)

Problem Statement

Solve \( x^2 + 4x = 5 \) by completing the square.

Solution

2819 video

Final Answer

\( x = 1 \) or \( x = -5 \)

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Solve \( 2x^2 - 12x - 7 =0 \) by completing the square.

Problem Statement

Solve \( 2x^2 - 12x - 7 =0 \) by completing the square.

Final Answer

\(\displaystyle{ x = 3 \pm \frac{5\sqrt{2}}{2} }\)

Problem Statement

Solve \( 2x^2 - 12x - 7 =0 \) by completing the square.

Solution

2820 video

Final Answer

\(\displaystyle{ x = 3 \pm \frac{5\sqrt{2}}{2} }\)

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Solve \( 3x^2 - 5x = 10 \) by completing the square.

Problem Statement

Solve \( 3x^2 - 5x = 10 \) by completing the square.

Final Answer

\(\displaystyle{ x = \frac{5 \pm \sqrt{145}}{6} }\)

Problem Statement

Solve \( 3x^2 - 5x = 10 \) by completing the square.

Solution

2821 video

Final Answer

\(\displaystyle{ x = \frac{5 \pm \sqrt{145}}{6} }\)

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Solve \( x^2 + 6x - 7 = 0 \) by completing the square.

Problem Statement

Solve \( x^2 + 6x - 7 = 0 \) by completing the square.

Final Answer

\( x = 1 \) or \( x = -7 \)

Problem Statement

Solve \( x^2 + 6x - 7 = 0 \) by completing the square.

Solution

2822 video

Final Answer

\( x = 1 \) or \( x = -7 \)

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Solve \( 2x^2 - 10x - 3 = 0 \) by completing the square.

Problem Statement

Solve \( 2x^2 - 10x - 3 = 0 \) by completing the square.

Final Answer

\(\displaystyle{ x = \frac{5 \pm \sqrt{31}}{2} }\)

Problem Statement

Solve \( 2x^2 - 10x - 3 = 0 \) by completing the square.

Solution

2823 video

Final Answer

\(\displaystyle{ x = \frac{5 \pm \sqrt{31}}{2} }\)

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Solve \( -x^2 - 6x + 7 = 0 \) by completing the square.

Problem Statement

Solve \( -x^2 - 6x + 7 = 0 \) by completing the square.

Final Answer

\( x = 9 \) or \( x = -1 \)

Problem Statement

Solve \( -x^2 - 6x + 7 = 0 \) by completing the square.

Solution

2824 video

Final Answer

\( x = 9 \) or \( x = -1 \)

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Solve \( x^2+6x-9=0 \) by completing the square.

Problem Statement

Solve \( x^2+6x-9=0 \) by completing the square.

Final Answer

\( x = -3 \pm 3\sqrt{2} \)

Problem Statement

Solve \( x^2+6x-9=0 \) by completing the square.

Solution

2825 video

Final Answer

\( x = -3 \pm 3\sqrt{2} \)

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Solve \( 3x^2+ 2x - 9 = 0 \) by completing the square.

Problem Statement

Solve \( 3x^2+ 2x - 9 = 0 \) by completing the square.

Final Answer

\(\displaystyle{ x = \frac{-1 \pm 2\sqrt{7}}{3} }\)

Problem Statement

Solve \( 3x^2+ 2x - 9 = 0 \) by completing the square.

Solution

2826 video

Final Answer

\(\displaystyle{ x = \frac{-1 \pm 2\sqrt{7}}{3} }\)

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Solve \( 3x^2 + 7x - 2 = 0 \) by completing the square.

Problem Statement

Solve \( 3x^2 + 7x - 2 = 0 \) by completing the square.

Final Answer

\(\displaystyle{ x = \frac{7 \pm \sqrt{73}}{6} }\)

Problem Statement

Solve \( 3x^2 + 7x - 2 = 0 \) by completing the square.

Solution

2827 video

Final Answer

\(\displaystyle{ x = \frac{7 \pm \sqrt{73}}{6} }\)

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Solve \( x^2 - 4x - 32 = 0 \) by completing the square.

Problem Statement

Solve \( x^2 - 4x - 32 = 0 \) by completing the square.

Final Answer

\( x= 8 \) or \( x = -4 \)

Problem Statement

Solve \( x^2 - 4x - 32 = 0 \) by completing the square.

Solution

2828 video

Final Answer

\( x= 8 \) or \( x = -4 \)

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Solve \( x^2 + 5x + 2 = 0 \) by completing the square.

Problem Statement

Solve \( x^2 + 5x + 2 = 0 \) by completing the square.

Final Answer

\(\displaystyle{ x = \frac{-5 \pm \sqrt{17}}{2} }\)

Problem Statement

Solve \( x^2 + 5x + 2 = 0 \) by completing the square.

Solution

2829 video

Final Answer

\(\displaystyle{ x = \frac{-5 \pm \sqrt{17}}{2} }\)

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Complete the square on \( x^2 - 2x - 5 = 0 \)

Problem Statement

Complete the square on \( x^2 - 2x - 5 = 0 \)

Solution

2538 video

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Solve \( x^2 + 6x = -8 \) by completing the square.

Problem Statement

Solve \( x^2 + 6x = -8 \) by completing the square.

Solution

2539 video

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Complete the square on \(y=3x^2-2x+1\)

Problem Statement

Complete the square on \(y=3x^2-2x+1\)

Solution

1283 video

video by Krista King Math

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Solve \(x^2-6x+8=0\) by completing the square.

Problem Statement

Solve \(x^2-6x+8=0\) by completing the square.

Solution

1285 video

video by PatrickJMT

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Solve \(x^2+4x-20=12\) by completing the square.

Problem Statement

Solve \(x^2+4x-20=12\) by completing the square.

Solution

1286 video

video by PatrickJMT

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Complete the square on both x and y for \( 0 = x^2 + y^2 - 12x - 22y + 139 \).

Problem Statement

Complete the square on both x and y for \( 0 = x^2 + y^2 - 12x - 22y + 139 \).

Solution

This is an interesting problem from analytic geometry that uses the technique of completing the square to solve the problem.

1284 video

video by Krista King Math

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Find the value of c that makes \( x^2 - 44x + c \) a perfect trinomial and write in factored form.

Problem Statement

Find the value of c that makes \( x^2 - 44x + c \) a perfect trinomial and write in factored form.

Solution

1304 video

video by Khan Academy

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completing the square 17calculus youtube playlist

Here is a playlist of the videos on this page.

Really UNDERSTAND Precalculus

Topics You Need To Understand For This Page

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

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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