Solving The Integral Of X² Arcsin(x) / √(1 - X²)

October 12, 2025 by TextBrain Team 49 views

Hey guys! Today, we're diving deep into a fascinating integral problem: evaluating the integral of x² * arcsin(x) divided by the square root of (1 - x²). This looks like a beast, I know, but don't worry! We'll break it down step-by-step and conquer it together. Integrals involving inverse trigonometric functions often require a bit of cleverness, and this one is no exception. So, buckle up and let's get started!

Understanding the Challenge

Before we jump into the solution, let's quickly understand why this integral isn't straightforward. The combination of the x² term, the arcsin(x) function (which is the inverse sine), and the square root in the denominator makes it a tricky customer. We can't directly apply simple integration rules here. Our main keyword here is integral, and we will try to use it as much as possible. The integral is a fundamental concept in calculus that represents the area under a curve. It's the reverse process of differentiation, and mastering integration techniques is crucial for various applications in physics, engineering, and other fields.

The presence of arcsin(x) suggests that integration by parts might be a viable strategy, but we need to carefully choose our 'u' and 'dv' to simplify the problem. The square root term hints at a possible trigonometric substitution, which could help us get rid of the square root and express the integral in terms of trigonometric functions. So, we have two potential avenues to explore, and the key is to make the right choices to make our lives easier. We aim to transform the integral into a form that we can readily solve using standard techniques. Remember, the goal isn't just to get the answer but also to understand why the method works. This will help you tackle similar problems in the future with confidence. So, let’s delve deeper into the techniques we can employ to solve this integral.

Strategy: Integration by Parts and Trigonometric Substitution

Our plan of attack involves a two-pronged approach:

Integration by Parts: We'll use this technique to deal with the arcsin(x) function. Remember the formula: ∫u dv = uv - ∫v du. The trick is to select 'u' and 'dv' strategically. In this case, a good choice for 'u' will be arcsin(x) because its derivative simplifies things. For 'dv', we'll take the remaining part of the integral. This method will help us to reduce the complexity of the integral by shifting the difficulty from the original integral to another, hopefully simpler, integral.
Trigonometric Substitution: The square root in the denominator (√(1 - x²)) is a classic indicator for trigonometric substitution. We'll substitute x with sin(θ). This substitution will allow us to use trigonometric identities to simplify the expression under the square root. The Pythagorean identity, sin²(θ) + cos²(θ) = 1, will be our best friend here. This substitution essentially changes the variable of integration, making the integral more manageable. The combination of these two techniques often proves to be a powerful strategy for solving complex integrals. By breaking down the problem into smaller, more manageable parts, we can tackle even the most daunting integrals. Now, let's put these strategies into action and see how they work in practice.

Step-by-Step Solution

Let's execute our plan. First, we'll tackle the integration by parts. Let:

u = arcsin(x)
dv = (x² / √(1 - x²)) dx

Now we need to find du and v:

du = (1 / √(1 - x²)) dx

To find 'v', we need to integrate dv, which is ∫(x² / √(1 - x²)) dx. This is where our trigonometric substitution comes into play. Let's substitute:

x = sin(θ)
dx = cos(θ) dθ

Substituting these into the integral for 'v', we get:

v = ∫(sin²(θ) / √(1 - sin²(θ))) cos(θ) dθ

Using the Pythagorean identity (1 - sin²(θ) = cos²(θ)), we simplify the square root:

v = ∫(sin²(θ) / cos(θ)) cos(θ) dθ

The cos(θ) terms cancel out:

v = ∫sin²(θ) dθ

To integrate sin²(θ), we use the double-angle identity: sin²(θ) = (1 - cos(2θ)) / 2

v = ∫(1 - cos(2θ)) / 2 dθ

v = (1/2) ∫(1 - cos(2θ)) dθ

v = (1/2) [θ - (1/2)sin(2θ)] + C

Now, we need to convert back to x. Since x = sin(θ), θ = arcsin(x). Also, sin(2θ) = 2sin(θ)cos(θ) = 2x√(1 - x²). So,

v = (1/2) [arcsin(x) - x√(1 - x²)]

We can ignore the constant of integration for now, as we'll add it at the very end.

Now that we have 'u', 'dv', 'du', and 'v', we can apply the integration by parts formula: ∫u dv = uv - ∫v du

∫(x² arcsin(x) / √(1 - x²)) dx = arcsin(x) * (1/2) [arcsin(x) - x√(1 - x²)] - ∫(1/2) [arcsin(x) - x√(1 - x²)] * (1 / √(1 - x²)) dx

This looks intimidating, but we've made significant progress! Let's simplify this expression step by step.

Simplifying the Resulting Integral

Let's rewrite the equation from the previous step:

∫(x² arcsin(x) / √(1 - x²)) dx = (1/2) arcsin²(x) - (1/2) x arcsin(x) √(1 - x²) - (1/2) ∫[arcsin(x) / √(1 - x²) - x] dx

Now, let's distribute the integral:

∫(x² arcsin(x) / √(1 - x²)) dx = (1/2) arcsin²(x) - (1/2) x arcsin(x) √(1 - x²) - (1/2) ∫arcsin(x) / √(1 - x²) dx + (1/2) ∫x dx

We now have two simpler integrals to deal with. The integral ∫x dx is straightforward:

∫x dx = (1/2)x² + C

For the integral ∫arcsin(x) / √(1 - x²) dx, we can use a simple u-substitution:

Let u = arcsin(x)
du = (1 / √(1 - x²)) dx

So, the integral becomes:

∫u du = (1/2)u² + C = (1/2)arcsin²(x) + C

Now we have all the pieces of the puzzle! Let's substitute these results back into our main equation.

Putting It All Together

Substituting the results of the simpler integrals back into our equation, we get:

∫(x² arcsin(x) / √(1 - x²)) dx = (1/2) arcsin²(x) - (1/2) x arcsin(x) √(1 - x²) - (1/2) [(1/2)arcsin²(x)] + (1/2) [(1/2)x²] + C

Now, let's simplify and combine like terms:

∫(x² arcsin(x) / √(1 - x²)) dx = (1/2) arcsin²(x) - (1/2) x arcsin(x) √(1 - x²) - (1/4) arcsin²(x) + (1/4) x² + C

Combining the arcsin²(x) terms:

∫(x² arcsin(x) / √(1 - x²)) dx = (1/4) arcsin²(x) - (1/2) x arcsin(x) √(1 - x²) + (1/4) x² + C

And there we have it! The final result of our integral is:

(1/4) arcsin²(x) - (1/2) x arcsin(x) √(1 - x²) + (1/4) x² + C

Conclusion

Wow, that was quite a journey! We successfully evaluated the integral of x² arcsin(x) / √(1 - x²) dx by using a combination of integration by parts and trigonometric substitution. We saw how breaking down a complex problem into smaller, more manageable steps can make even the trickiest integrals solvable. The key takeaways are the strategic use of integration techniques and the power of trigonometric substitutions in simplifying expressions involving square roots. Remember, practice makes perfect! So, keep practicing these techniques, and you'll become a master of integration in no time. Hope you guys found this helpful and happy integrating!