Integral Calculation - Exercise 31

We’re asked to evaluate the indefinite integral:

$$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \ dx $$

There are several possible approaches.

Method 1

We begin with the integral:

$$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \ dx $$

Using the substitution method, we introduce a new variable: \( t = \sqrt{x} \).

$$ t = \sqrt{x} $$

We compute the differential:

$$ dt = d( \sqrt{x} ) = \frac{1}{2 \sqrt{x}} \ dx $$

Solving for \( dx \):

$$ dx = 2 \sqrt{x} \ dt $$

Now substitute into the original integral:

$$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \cdot (2 \sqrt{x} \ dt) $$

The \( \sqrt{x} \) terms cancel out:

$$ \int 2 \cos \sqrt{x} \ dt $$

Since \( t = \sqrt{x} \), we rewrite the integrand in terms of \( t \):

$$ 2 \int \cos t \ dt $$

The integral of \( \cos t \) is \( \sin t + c \), so we get:

$$ 2 \sin(t) + c $$

Finally, substituting back \( t = \sqrt{x} \):

$$ 2 \sin(\sqrt{x}) + c $$

This is the solution.

Method 2

Once again, we aim to solve:

$$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \ dx $$

We make the substitution \( t = \sqrt{x} \):

$$ t = \sqrt{x} $$

Then compute the differential:

$$ dt = \frac{1}{2 \sqrt{x}} \ dx $$

From this we deduce:

$$ \frac{1}{\sqrt{x}} \ dx = 2 dt $$

Substituting into the integral:

$$ \int \cos \sqrt{x} \cdot \frac{1}{\sqrt{x}} \ dx = \int \cos \sqrt{x} \cdot (2 \ dt) $$

Which simplifies to:

$$ 2 \int \cos \sqrt{x} \ dt $$

Replacing \( \sqrt{x} \) with \( t \):

$$ 2 \int \cos t \ dt = 2 \sin t + c $$

Substitute back \( t = \sqrt{x} \):

$$ 2 \sin(\sqrt{x}) + c $$

This confirms the same result.

Method 3

We again consider the integral:

$$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \ dx $$

Observe that the derivative of \( \sin(\sqrt{x}) \) is:

$$ \frac{d}{dx} [\sin(\sqrt{x})] = \cos(\sqrt{x}) \cdot \frac{1}{2 \sqrt{x}} $$

Multiplying both sides by 2:

$$ 2 \cdot d(\sin(\sqrt{x})) = \frac{\cos(\sqrt{x})}{\sqrt{x}} \ dx $$

So we can rewrite the integral as:

$$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \ dx = \int 2 \cdot d(\sin(\sqrt{x})) $$

Which becomes:

$$ 2 \int d(\sin(\sqrt{x})) $$

Letting \( t = \sin(\sqrt{x}) \):

$$ 2 \int dt = 2t + c $$

Substituting back:

$$ 2 \sin(\sqrt{x}) + c $$

Once again, we arrive at the same result.

And so on...