Substitution Method for Definite Integrals

A Practical Example

Let’s evaluate the following definite integral:

$$ \int_0^{\sqrt{\pi}} x \cdot \cos x^2 dx $$

We begin by substituting \( t = x^2 \):

$$ t = x^2 $$

Taking the differential:

$$ dt = t' \:dx \\ dt = 2x \: dx $$

We now rewrite the integral using this substitution.

Since \( x^2 = t \), the integral becomes:

$$ \int_{x=0}^{x=\sqrt{\pi}} x \cdot \cos t \: dx $$

We also replace \( x\,dx \) using the differential relation:

$$ dt = 2x \, dx \quad \Rightarrow \quad \frac{dt}{2} = x\, dx $$

Substituting this into the integral:

$$ \int_{x=0}^{x=\sqrt{\pi}} \cos t \cdot \frac{dt}{2} $$

We can factor out the constant:

$$ \frac{1}{2} \int_{x=0}^{x=\sqrt{\pi}} \cos t \, dt $$

At this point, we convert the limits of integration from \( x \) to \( t \):

$$ t = x^2 \quad \Rightarrow \quad t = 0^2 = 0,\quad t = (\sqrt{\pi})^2 = \pi $$

So we now have:

$$ \frac{1}{2} \int_0^\pi \cos t \, dt $$

This integral is straightforward:

$$ \frac{1}{2} [\sin t]_0^\pi = \frac{1}{2} (\sin \pi - \sin 0) = \frac{1}{2} (0 - 0) = 0 $$

So the value of the definite integral is zero.

Example 2

Let’s solve the same integral using a different approach:

$$ \int_0^{\sqrt{\pi}} x \cdot \cos(x^2)\,dx $$

This time, we use the substitution \( t = x^2 \), so \( x = \sqrt{t} \):

$$ x = g(t) = \sqrt{t} $$

Taking the differential:

$$ dx = \frac{1}{2\sqrt{t}}\,dt $$

Substituting \( x = \sqrt{t} \) and \( dx = \frac{1}{2\sqrt{t}}\,dt \) into the integrand:

$$ \int_{x=0}^{x=\sqrt{\pi}} \sqrt{t} \cdot \cos(t)\,\frac{1}{2\sqrt{t}}\,dt $$

Simplifying:

$$ \int_{x=0}^{x=\sqrt{\pi}} \cos(t)\,\frac{1}{2}\,dt = \frac{1}{2} \int_{x=0}^{x=\sqrt{\pi}} \cos(t)\,dt $$

The integral is now written in terms of \( t \), but the limits are still expressed in terms of \( x \). To convert them:

Since \( x = \sqrt{t} \), the inverse function is \( t = x^2 \):

• When \( x = 0 \), then \( t = 0 \)
• When \( x = \sqrt{\pi} \), then \( t = \pi \)

Updating the limits accordingly:

$$ \frac{1}{2} \int_0^\pi \cos(t)\,dt $$

This is again a basic integral:

$$ \frac{1}{2} \left[ \sin(t) \right]_0^\pi = \frac{1}{2} (0 - 0) = 0 $$

So we reach the same conclusion with this alternative method:

$$ \int_0^{\sqrt{\pi}} x \cdot \cos(x^2)\,dx = 0 $$

Note. This second method is more formal - it uses the substitution \( x = g(t) \), which requires the explicit inverse \( x = \sqrt{t} \). It's helpful for understanding the general substitution formula involving \( g^{-1}(a) \) and \( g^{-1}(b) \). However, in practical computations, the approach using \( t = x^2 \) is often more intuitive and direct. Ultimately, the choice depends on which method simplifies the process more effectively: some problems lend themselves better to one strategy than the other.

And so on.