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17Calculus Integrals - Integral Tricks

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Here are some integral tricks that are usually not taught by instructors.

Why Don't Teachers Teach These Techniques?

In a perfect world, everything would be easy and there would be tricks to calculus. However, this is not a perfect world. So, teachers do not always teach the easiest ways to do things. There are several reasons for this.

1. There just isn't enough time.

2. Some tricks, like the ones on this page, teach you to get answers without learning the technique that the instructor is required to teach you and test you on.

3. Most tricks are useful only for very specific types of problems. You will often find that learning tricks takes more time and energy to learn than they are worth. Therefore the time and energy to learn them is better spent on learning the actual technique required to pass the course.

As an instructor, I believe that each of these are valid reasons not to teach these tricks. In my opinion, tricks keep you from learning the basics and can make future use of calculus difficult when you run into a problem that does not fit a specific trick.

That said, if you want to take the time to learn tricks, you can use them to check your answers. This is a very good use of tricks and, as an instructor, I encourage you to do this. However, do not expect to get full credit on your assignments and exams if you use these techniques as your answers, since your instructor probably requires you to use another specific technique. Check with your instructor to see what they require.

Technique 1 - Integrating an Inverse Function

This video clip introduces the equation \[ \int{ f^{-1}(x) ~dx } = x \cdot f^{-1}(x) - (F \circ f^{-1})(x) + C \]

In this equation, we have
\( F(x) = \int{f(x) ~dx} \)
\( f^{-1}(x) \) is the inverse function of \(f(x)\)
\( (F \circ f^{-1})(x) \) is the composition of \(F(x)\) and \(f^{-1}(x)\)

So how do we use this equation? Here he evaluates the integral \(\int{ \arccos x ~ dx }\) using this formula.

BriTheMathGuy - Integration Tricks (That Teachers Won't Tell You) for Integral Calculus

video by BriTheMathGuy

Okay, work this practice problem to check that the answer is the same as he got in the video.

Evaluate \(\int{ \arccos x ~ dx }\) using integration by parts.

Problem Statement

Evaluate \(\int{ \arccos x ~ dx }\) using integration by parts.

Final Answer

\( x\arccos x - \sqrt{1-x^2} + C \)

Problem Statement

Evaluate \(\int{ \arccos x ~ dx }\) using integration by parts.

Solution

The integration by parts formula is \( \int{u~dv} = uv - \int{v~du} \).

\(\displaystyle{ \int{ \arccos x ~ dx } }\)

\(\displaystyle{ u = \arccos x \to du = \frac{-1}{\sqrt{1-x^2}} ~dx }\)

\(\displaystyle{ dv = dx \to v = x }\)

\(\displaystyle{ x\arccos x - \int{ \frac{-x}{\sqrt{1-x^2}} ~dx } }\)

Use integration by substitution. \(\displaystyle{ u=1-x^2 \to du = -2x~dx \to du/2 = -x~dx }\)
This \(u\) is different than the \(u\) used in integration by parts above.

\(\displaystyle{ x\arccos x - \int{ \frac{1}{u^{1/2}} \frac{du}{2} } }\)

\(\displaystyle{ x \arccos x - \frac{1}{2}\int{ u^{-1/2} ~ du } }\)

\(\displaystyle{ x \arccos x - \frac{1}{2}\frac{u^{1/2}}{1/2} + C }\)

\(\displaystyle{ x \arccos x - \sqrt{1-x^2} + C }\)

Final Answer

\( x\arccos x - \sqrt{1-x^2} + C \)

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Technique 2 - Adjusting the Integrand by a Constant

This trick does not seem to be very useful but it does work. As an instructor, this is one of the few 'tricks' that I believe may be useful to teach. We actually use it in the solution for the practice problem for technique 4. \[ \int_a^b{ f(x)~dx } = \int_a^b{ f(a+b-x) ~ dx } \]

In this next video clip, he uses this trick to evaluate \(\displaystyle{ \int_1^3{ (4-x)^3 ~ dx } }\)

BriTheMathGuy - Integration Tricks (That Teachers Won't Tell You) for Integral Calculus

video by BriTheMathGuy

Okay, work this practice problem to check that the answer is the same as he got in the video.

Evaluate \(\displaystyle{ \int_1^3{ (4-x)^3 ~dx } }\) using integration by substitution.

Problem Statement

Evaluate \(\displaystyle{ \int_1^3{ (4-x)^3 ~dx } }\) using integration by substitution.

Final Answer

\(\displaystyle{ \frac{3^4}{4} - \frac{1}{4} = \frac{3^4-1}{4} = 20 }\)

Problem Statement

Evaluate \(\displaystyle{ \int_1^3{ (4-x)^3 ~dx } }\) using integration by substitution.

Solution

\(\displaystyle{ \int_1^3{ (4-x)^3 ~dx } }\)

\(\displaystyle{ u = 4-x \to du = -dx }\)

upper limit: \( x = 3 \to u = 4-3 = 1\)

lower limit: \( x = 1 \to u = 4-1 = 3 \)

\(\displaystyle{ \int_3^1{ u^3 (-du)} }\)

Switch the limits of integration to get rid of the negative sign.

\(\displaystyle{ \int_1^3{u^3~du} }\)

\(\displaystyle{ \left. \frac{u^4}{4} \right|_1^3 }\)

\(\displaystyle{ \frac{3^4}{4} - \frac{1^4}{4} }\)

Final Answer

\(\displaystyle{ \frac{3^4}{4} - \frac{1}{4} = \frac{3^4-1}{4} = 20 }\)

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Technique 3 - Leibniz Integral Rule

This technique requires the use of a calculus 3 technique, partial integration. It also involves a derivative. It is tricky to use and, as you will see in the video clip below, it is not always clear how to use it. Here is the formula. \[ \int_a^b{f(x,y)~dx} \to F'(y) = \int_a^b{ f_y(x,y)~dx } \]

So how does this work? Here is a quick video clip showing the explanation behind this rule. He also works an example of finding \(F'(x)\) where \(\displaystyle{ F(x) = \int_1^2{ \frac{\cos(tx)}{t} ~dt } }\)

Dr Chris Tisdell - Differentiate under integral signs: Leibniz rule

video by Dr Chris Tisdell

So how do we use this to evaluate a more basic integral? In this video clip, he uses this to evaluate \(\displaystyle{ \int_0^1{ \frac{x^5-1}{\ln x} ~dx } }\). It's strange how he changes the problem to fit the formula but it seems to work.

BriTheMathGuy - Integration Tricks (That Teachers Won't Tell You) for Integral Calculus

video by BriTheMathGuy

built with GeoGebra

Unfortunately, this integral is not easy to evaluate using another method. Notice this is an improper integral at both end points. To check his answer, I plotted the function using GeoGebra and then told GeoGebra to integrate from zero to one. Here is the result.

Technique 4 - Simplifying \(x\) Times a Function of Sine

In this video clip, he proves the formula \[ \int_0^{\pi}{ x f(\sin x) ~dx } = \frac{\pi}{2} \int_0^{\pi}{ f(\sin x) ~dx } \]

Michael Penn - A nice integration trick! [clip 1]

video by Michael Penn

He uses this formula to solve the integral \(\displaystyle{ \int_0^{\pi}{ (1+2x) \frac{\sin^3 x}{1+\cos^2x} ~dx } }\)

Michael Penn - A nice integration trick! [clip 2]

video by Michael Penn

It is possible to evaluate this integral using substitution and integration by parts.

Evaluate \(\displaystyle{ \int_0^{\pi}{ (1+2x)\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\) using integration by substitution and integration by parts to show that the above trick works on this problem.

Problem Statement

Evaluate \(\displaystyle{ \int_0^{\pi}{ (1+2x)\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\) using integration by substitution and integration by parts to show that the above trick works on this problem.

Hint

Use technique 2 discussed above to evaluate the integral involving \(\arctan(\cos x)\).

Problem Statement

Evaluate \(\displaystyle{ \int_0^{\pi}{ (1+2x)\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\) using integration by substitution and integration by parts to show that the above trick works on this problem.

Final Answer

\( (\pi + 1)(\pi - 2) \)

Problem Statement

Evaluate \(\displaystyle{ \int_0^{\pi}{ (1+2x)\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\) using integration by substitution and integration by parts to show that the above trick works on this problem.

Hint

Use technique 2 discussed above to evaluate the integral involving \(\arctan(\cos x)\).

Solution

We are going to separate the integral into the two integrals \(\displaystyle{ \int_0^{\pi}{ \frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\) and \(\displaystyle{ \int_0^{\pi}{ 2x\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

\(\displaystyle{ \int_0^{\pi}{ \frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

Drop the limits of integration and integrate the indefinite integral \(\displaystyle{ \int{ \frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

\(\displaystyle{ \int{ \frac{\sin x(1-\cos^2 x)}{1 + \cos^2 x} ~dx } }\)

\(\displaystyle{ u = \cos x \to -du = \sin x ~dx }\)

\(\displaystyle{ -\int{ \frac{1-u^2}{1+u^2} ~du } }\)

\(\displaystyle{ \int{ \frac{u^2-1}{u^2+1} ~du } }\)

Do polynomial long division to get \(\displaystyle{ \int{ 1 - \frac{2}{u^2+1} ~du } }\)

\(\displaystyle{ u - 2\arctan u }\)

\(\displaystyle{ \cos x - 2\arctan(\cos x) }\)

Evaluate this result at the endpoints \(\displaystyle{ \left[ \cos x - 2\arctan(\cos x) \right]_0^{\pi} }\)

\(\displaystyle{ [\cos \pi - 2\arctan(\cos \pi)] - [ \cos 0 - 2 \arctan(\cos 0) ] }\)

\(\displaystyle{ -1 - 2\arctan(-1) - [1 - 2\arctan(1)] }\)

\(\displaystyle{ -1 - 2(-\pi/4) -1 + 2(\pi/4)}\)

\(\displaystyle{ -2+\pi }\)


Now, let's evaluate \(\displaystyle{ \int_0^{\pi}{ 2x\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\). Like we did above, we will drop the limits of integration and evaluate the result at the endpoints once we finish the integration.

\(\displaystyle{ \int{ 2x\frac{\sin^3 x}{1 + \cos^2 x} ~dx } }\)

Use integration by parts \(\displaystyle{ u = 2x \to du = 2dx }\)

\(\displaystyle{ dv = \frac{\sin^3 x}{1+\cos^2 x} ~dx }\)

\(\displaystyle{ v = \int{ \frac{\sin^3 x}{1+\cos^2 x} ~dx } }\)

\(v\) is the integral we did in the first part of the problem, so we have \( v = \cos x - 2\arctan(\cos x) \)

\(\displaystyle{ 2x[ \cos x - 2\arctan(\cos x) ] - 2 \int{ \cos x - 2\arctan(\cos x) ~dx }}\)

Now let's break the last integral into two.

\(\displaystyle{ 2\int{\cos x ~ dx } = 2\sin x }\)

When we evaluate this on the limits of integration, we get \( 2\sin \pi - 2\sin 0 = 0\)

For the second integral we have \(\displaystyle{ \int{ \arctan(\cos x) ~ dx } }\)

We are going to put the limits of integration back in since we will need them for the next step. So we have \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } }\)

We will use technique 2 above to shift this integrand. \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } = \int_0^{\pi}{ \arctan(\cos(\pi-x)) ~dx } }\)

\(\displaystyle{ \int_0^{\pi}{ \arctan(-\cos(x)) ~dx } }\)

\(\displaystyle{ \int_0^{\pi}{ -\arctan(\cos(x)) ~dx } }\)

So now we have \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } = -\int_0^{\pi}{ \arctan(\cos x) ~ dx } }\)

If we add \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } }\) to both sides, we have \(\displaystyle{ 2 \int_0^{\pi}{ \arctan(\cos x) ~ dx } = 0 }\)

So \(\displaystyle{ \int_0^{\pi}{ \arctan(\cos x) ~ dx } = 0 }\)

We are left with \(\displaystyle{ 2x[ \cos x - 2\arctan(\cos x) ] }\). This is the \(uv\) part of the integration by parts.

Evaluating this on the limits of integration gives us \(\displaystyle{ 2x[ \cos x - 2\arctan(\cos x) ]_0^{\pi} }\)

\(\displaystyle{ 2\pi[\cos \pi - 2\arctan(-1)] - 0 }\)

\(\displaystyle{ 2\pi[ -1 + \pi/2] }\)

\(\displaystyle{ \pi(\pi-2) }\)


Combining the two results gives us \( (\pi-2) + \pi(\pi-2) = (\pi+1)(\pi-2) \)

Final Answer

\( (\pi + 1)(\pi - 2) \)

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Really UNDERSTAND Calculus

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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