Integrating X Sin X: Step-By-Step Solution

The integration of $x \sin x$ represents a classic problem in calculus, and it exemplifies the application of integration by parts, which is an important technique for handling integrals, where integrand is a product of functions. Integration by parts involves selecting parts of the integrand to differentiate and integrate, typically using the formula $\int u \, dv = uv – \int v \, du$. In this case, the goal is to simplify the original integral into a more manageable form that can be evaluated using standard integral formulas. After the successful evaluation, the result of integrating $x \sin x$ can be expressed in closed form using trigonometric functions and algebraic terms, making it an instructive example for students learning integral calculus.

Ever stumbled upon a problem that looks intimidating at first glance? Well, integrating (x \sin(x)) might just be one of those moments! But don’t worry, it’s not as scary as it looks.

Think of it this way: Imagine you’re designing a bridge, calculating the movement of a pendulum, or even analyzing sound waves. Believe it or not, integrals like (x \sin(x)) pop up in these scenarios. Integrating (x \sin(x)) isn’t just a classroom exercise; it’s a valuable skill with real-world implications.

Why Bother with This Integral?

So, why should you care about integrating (x \sin(x))? Because it’s a fantastic way to sharpen your calculus skills!

Mastering integration techniques like the one we’ll use here (integration by parts) is crucial for anyone diving deeper into calculus, physics, engineering, or any field that relies on mathematical modeling. It’s like learning a secret code that unlocks solutions to complex problems.

The Relevance and Applications

This type of integral is more than just abstract math. It shows up in:

• Physics: Calculating centers of mass and moments of inertia.
• Engineering: Analyzing vibrations, signal processing, and electrical circuits.
• Applied Mathematics: Solving differential equations and modeling various phenomena.

By the end of this post, you’ll not only know how to integrate (x \sin(x)) but also understand why it matters. Let’s get started!

Foundation: Essential Calculus Concepts

Alright, before we jump headfirst into the thrilling world of integrating (x \sin(x)), let’s make sure we’ve got our calculus toolkit ready. Think of it like prepping ingredients before you bake a cake – you wouldn’t want to start without flour, right?

Antiderivatives: The Building Blocks

So, what’s an antiderivative, anyway? It’s basically the reverse process of taking a derivative. Imagine you have a function, and you magically need to find another function whose derivative is the one you started with. That’s the antiderivative! It’s like saying, “Hey, what function, when I differentiate it, gives me this?”. Finding an antiderivative is the whole goal when solving indefinite integrals.

Indefinite Integrals: A Family of Functions

Now, let’s talk indefinite integrals. When we find an antiderivative, we’re not just finding one function, but a whole family of them. Why? Because the derivative of a constant is always zero! This is why we always, and I mean always, add that little “+ C” at the end – the constant of integration. It’s like the secret ingredient that makes the solution complete! Don’t forget that C!

Product Rule of Differentiation: The Reverse Connection

Remember the product rule from differentiation? It states that ( (uv)’ = u’v + uv’ ). It might seem unrelated, but trust me, it’s the backbone of integration by parts. Integration by parts is essentially the reverse of the product rule. We’re using the product rule to cleverly undo a derivative and find an integral. It’s like watching a magic trick in reverse!

Trigonometric and Algebraic Functions: The Players

Time to meet our stars: the sine function, ( \sin(x) ), and the algebraic function ( x ). We need to know their derivatives and integrals inside and out. The derivative of ( \sin(x) ) is ( \cos(x) ), and the integral of ( \sin(x) ) is ( -\cos(x) ). And for ( x ), its derivative is 1, and its integral is ( \frac{x^2}{2} ). These are the players in our integral game, and knowing their moves is crucial.

Variable of Integration: Understanding (dx)

Finally, let’s talk about ( dx ). It might look like a random tag-along, but it’s super important. It tells us that x is our variable of integration. The ( dx ) basically says, “Hey, we’re integrating with respect to x!” It’s part of the notation and understanding what it means will help you keep everything straight as we move through the steps. Think of it as a guide.

With these fundamental concepts under our belts, we’re all set to tackle the integration by parts technique and conquer that ( \int x \sin(x) \, dx ) integral! Let’s do this!

The Star: Integration by Parts Technique

Alright, folks, buckle up because we’re about to dive into one of the coolest tools in the calculus toolbox: Integration by Parts. Think of it as the MacGyver of integration – when you’re staring down an integral that looks impossible, this is often the trick that saves the day. It’s especially handy when you’ve got a product of two different types of functions hanging out together under the integral sign, like our star of the show, ( x \sin(x) ).

The Integration by Parts Formula: Unveiled

So, what’s the secret formula? Here it is, in all its glory:

( \int u \, dv = uv – \int v \, du )

Now, I know what you’re thinking: “Whoa, that looks complicated!” But trust me, it’s not as scary as it seems. Let’s break it down:

• u: This is a part of your original integral that you choose strategically (more on that in a bit).
• dv: This is the remaining part of your integral. Think of it as dv is derivative.
• du: This is the derivative of u. So, if you pick u, you need to find its derivative to get du.
• v: This is the integral of dv. So, if you pick dv, you need to integrate it to get v.

The formula essentially says: “The integral of ( u \, dv ) is equal to u times v, minus the integral of ( v \, du ).” The goal is to pick u and dv in such a way that the new integral, ( \int v \, du ), is easier to solve than the original one! Think of it like trading a difficult task for an easier one.

Applying Integration by Parts to ( \int x \sin(x) \, dx )

Okay, let’s put this into action with our integral: ( \int x \sin(x) \, dx ).

Choosing ( u ) and ( dv ): A Strategic Decision

This is where the magic happens! We need to decide what part of ( x \sin(x) \, dx ) will be our u and what will be our dv. Here’s the golden rule: **Choose *u so that its derivative is simpler than u itself***.

In this case, let’s pick:

• ( u = x )
• ( dv = \sin(x) \, dx )

Why? Because when we take the derivative of ( x ), we get ( 1 ), which is much simpler!

There’s also a handy mnemonic device called ILATE or LIATE that can help you make this decision. It stands for:

• I: Inverse trigonometric functions (like arcsin(x), arctan(x))
• L: Logarithmic functions (like ln(x))
• A: Algebraic functions (like x, ( x^2 ), polynomials)
• T: Trigonometric functions (like sin(x), cos(x))
• E: Exponential functions (like ( e^x ))

The rule says: choose u in the order of these types of functions. Since algebraic functions (( x )) come before trigonometric functions (( \sin(x) )), we choose ( x ) as u.

Finding ( du ) and ( v ): The Transformation

Now that we’ve made our choices, let’s find du and v:

• If ( u = x ), then ( du = dx )
• If ( dv = \sin(x) \, dx ), then ( v = \int \sin(x) \, dx = -\cos(x) )

Remember, when finding v, we technically get ( -\cos(x) + C ), but we can ignore the “+ C” for now because it will be taken care of later.

Applying the Formula: Step-by-Step

Now we substitute everything into our integration by parts formula:

( \int u \, dv = uv – \int v \, du )

( \int x \sin(x) \, dx = (x)(-\cos(x)) – \int (-\cos(x)) \, dx )

Simplifying, we get:

( \int x \sin(x) \, dx = -x \cos(x) + \int \cos(x) \, dx )

Simplifying and Solving: The Final Stretch

Look at that! Our integral is now much simpler. We just need to integrate ( \cos(x) ):

( \int \cos(x) \, dx = \sin(x) + C )

The Solution: Putting It All Together

Finally, let’s put it all together. Our solution is:

( \int x \sin(x) \, dx = -x \cos(x) + \sin(x) + C )

And there you have it! We’ve successfully integrated ( x \sin(x) ) using integration by parts. Wasn’t that fun? (Okay, maybe not fun, but definitely rewarding!).

Definite Integral: Taking Our Integral for a Walk (Over an Interval!)

Alright, so we’ve mastered the art of indefinite integrals – finding that general antiderivative. But what if we want something more specific? What if we want to know the exact area under the curve of (x \sin(x)) between two points? That’s where definite integrals strut onto the stage! They allow us to put boundaries on our integration, giving us a numerical answer instead of a whole family of functions. Think of it as putting fences around our wild, unbounded integral!

Imagine you’re an architect, and you need to calculate the area of a curved wall to figure out how much paint to buy. You can’t just use rectangles and triangles! Definite integrals come to the rescue! Evaluating an integral over an interval ([a, b]) basically means finding the net area trapped between the curve of our function and the x-axis, from point a to point b.

And who’s the superhero behind this magic? None other than the Fundamental Theorem of Calculus! This theorem is the secret sauce that connects derivatives and integrals, allowing us to use the antiderivative we found earlier to calculate these areas.

Limits of Integration: Setting Boundaries

These aren’t just suggestions; they’re rules! The limits of integration, denoted as a and b in our interval ([a, b]), tell us exactly where to start and stop calculating the area. Think of them as the start and end lines of a race. Without them, we’d be running forever!

For example, let’s say we want to integrate from (0) to ( \pi/2 ). That means we’re only interested in the area under the curve of (x \sin(x)) between (x = 0) and (x = \pi/2). This specification is super important – changing the limits changes the area (and the final answer!).

Evaluating ( \int_{0}^{\pi/2} x \sin(x) \, dx )

Time to get our hands dirty! Remember that beautiful antiderivative we found using integration by parts? It was (-x \cos(x) + \sin(x) ). Now, we’re going to use it to calculate our definite integral: ( \int_{0}^{\pi/2} x \sin(x) \, dx ).

Here’s the magic trick:

1. Plug in the Upper Limit: Substitute (x = \pi/2) into our antiderivative: ([-\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})])
2. Plug in the Lower Limit: Substitute (x = 0) into our antiderivative: ([-(0) \cos(0) + \sin(0)])
3. Subtract: Subtract the result from step 2 from the result of step 1.

So, we get: ([-\frac{\pi}{2} \cos(\frac{\pi}{2}) + \sin(\frac{\pi}{2})] – [-(0) \cos(0) + \sin(0)] ). Let’s simplify this bad boy!

Remember that ( \cos(\frac{\pi}{2}) = 0 ), ( \sin(\frac{\pi}{2}) = 1 ), ( \cos(0) = 1 ), and ( \sin(0) = 0 ). Substituting these values:

( [-\frac{\pi}{2} (0) + 1] – [-(0)(1) + 0] = [0 + 1] – [0 + 0] = 1 )

The Result: A Numerical Value

Drumroll, please! The definite integral of (x \sin(x)) from (0) to (\pi/2) is… (1)!

( \int_{0}^{\pi/2} x \sin(x) \, dx = 1 )

That means the area under the curve of (x \sin(x)) between (x = 0) and (x = \pi/2) is exactly 1 square unit. How cool is that? We’ve taken our integral, given it boundaries, and arrived at a precise answer. This opens doors to countless real-world applications, which we’ll peek at next!

Real-World Applications and Further Exploration: Where Does This Stuff Actually Matter?

Okay, so you’ve conquered the integral of xsin(x) – high five! But now you’re probably thinking, “Great, I can do this one specific integral… but when am I ever going to use this?” Fear not, my friend! This isn’t just some abstract math problem cooked up to torture students. Integrals like this pop up in all sorts of real-world scenarios, often in places you wouldn’t expect.

For instance, in the wild world of physics, this type of integral is super helpful when calculating things like the moment of inertia – basically, how hard it is to spin something. Think about designing a figure skater’s spin or ensuring a satellite stays oriented correctly in space; it’s integrals all the way down! And it’s not just physics; engineers use these concepts in signal processing, which is essential for everything from your phone working to medical imaging!

Want More Math Adventures? Your Treasure Map to Further Learning

So, you’ve got the integration bug now, huh? Excellent! The good news is that the internet is overflowing with resources to help you become an integration ninja. Websites like Khan Academy, MIT OpenCourseWare, and Paul’s Online Math Notes are goldmines of information, offering tutorials, practice problems, and even video lectures. Don’t be shy – dive in and explore!

And hey, the best way to truly master integration is to, well, integrate! Grab some similar integrals (maybe try xcos(x) next!), roll up your sleeves, and get practicing. Trust me; the more you do, the more comfortable and confident you’ll become. Who knows, you might even start enjoying it… maybe. No promises! But you’ll definitely be a calculus rockstar, and that’s a pretty awesome title to hold.

So, next time you stumble upon ∫x sinx dx in your calculus adventures, don’t panic! Break it down with integration by parts, and you’ll be enjoying that -x cosx + sinx + C in no time. Happy integrating!