Hey there! Let's walk through solving this integral step by step—you mentioned getting stuck on the cosine term, so we'll break down exactly how to tackle this using substitution, integration by parts, and even touch on special functions since this integral doesn't have a straightforward elementary solution.

Step 1: Simplify with Substitution #

First off, let's use substitution to simplify that tricky cosine argument. Let:
u = 1/x²
Now compute the derivative of u with respect to x:
du/dx = -2/x³ ⇒ du = -2/x³ dx

We need to rewrite x dx (the other part of our integrand) in terms of u. From u = 1/x², we can rearrange to get x² = 1/u, so x = u^(-1/2). Plug this into our du equation to solve for dx:
du = -2/(u^(-3/2)) dx ⇒ dx = -u^(3/2)/2 du

Multiply by x to get x dx:
x dx = u^(-1/2) * (-u^(3/2)/2) du = -1/(2u²) du

Now substitute everything back into the original integral:
∫x cos(1/x²) dx = ∫cos(u) * (-1/(2u²)) du = -1/2 ∫cos(u)/u² du

Step 2: Use Integration by Parts #

The integral ∫cos(u)/u² du isn't something we can solve directly, so integration by parts is the way to go. Remember the integration by parts formula:
∫f dg = fg - ∫g df

Let's define our terms:

• f = cos(u) (so df = -sin(u) du)
• dg = 1/u² du (so g = -1/u—since integrating 1/u² gives -1/u)

Applying the formula:

∫cos(u)/u² du = cos(u)*(-1/u) - ∫(-1/u)*(-sin(u)) du
= -cos(u)/u - ∫sin(u)/u du

Step 3: Introduce the Sine Integral Special Function #

Here's the thing: ∫sin(u)/u du can't be expressed using basic elementary functions (like polynomials, trig functions, or exponentials). Instead, we use a special function called the Sine Integral, denoted Si(u). By definition:
Si(u) = ∫₀^u sin(t)/t dt

Putting this back into our expression:
∫cos(u)/u² du = -cos(u)/u - Si(u) + C

Now plug this back into the integral we had after substitution:

-1/2 ∫cos(u)/u² du = -1/2*(-cos(u)/u - Si(u)) + C
= cos(u)/(2u) + (1/2)Si(u) + C

Step 4: Substitute Back to the Original Variable #

Finally, swap u back out for 1/x² to get our solution in terms of x:

cos(1/x²)/(2*(1/x²)) + (1/2)Si(1/x²) + C
= (x²/2)cos(1/x²) + (1/2)Si(1/x²) + C

Bonus: Series Expansion (If Special Functions Aren't Allowed) #

If you're in a context where you can't use special functions, you can use the Taylor series expansion of cos(1/x²) instead. Recall that:
cos(z) = 1 - z²/2! + z⁴/4! - z⁶/6! + ...
Let z = 1/x², so:
cos(1/x²) = 1 - 1/(2x⁴) + 1/(24x⁸) - 1/(720x¹²) + ...

Multiply by x and integrate term-by-term:

∫x cos(1/x²) dx = ∫x(1 - 1/(2x⁴) + 1/(24x⁸) - ...) dx
= ∫(x - 1/(2x³) + 1/(24x⁷) - ...) dx
= x²/2 + 1/(4x²) - 1/(144x⁶) + ... + C

This gives you an infinite series approximation of the integral, which can be useful for numerical calculations.

内容的提问来源于stack exchange，提问作者Nehal Samee