Integrate the following function: $ \frac{\cos \sqrt{x}}{\sqrt{x}} $

> [!TLDR] > The integral $\int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx$ is solved by substituting $u = \sqrt{x}$, which simplifies the integrand to $2 \cos u$. The final answer is $2 \sin \sqrt{x} + C$. The key to this problem is noticing that the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, and we have a $\frac{1}{\sqrt{x}}$ factor sitting right next to $\cos \sqrt{x}$. That’s a dead giveaway for **U Substitution** — we let the inner function be $u$, and the rest of the integrand becomes its derivative (up to a constant factor). Let’s walk through it. 1. **Set up the substitution.** Let $u = \sqrt{x}$. Then $x = u^2$, so $dx = 2u \, du$. Why this choice? Because $\sqrt{x}$ appears inside the cosine, and its derivative $\frac{1}{2\sqrt{x}}$ is almost exactly the $\frac{1}{\sqrt{x}}$ we have. The substitution will collapse the whole expression into something clean. 2. **Rewrite the integral in terms of $u$.** The original integral is $$ \int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx. $$ Replace $\sqrt{x}$ with $u$, and $dx$ with $2u \, du$: $$ \int \frac{\cos u}{\sqrt{x}} \cdot 2u \, du. $$ But $\sqrt{x} = u$, so $\frac{1}{\sqrt{x}} = \frac{1}{u}$. That gives: $$ \int \frac{\cos u}{u} \cdot 2u \, du = \int 2 \cos u \, du. $$ The $u$ cancels beautifully — that’s the whole point of the substitution. 3. **Integrate.** $$ \int 2 \cos u \, du = 2 \sin u + C. $$ 4. **Substitute back.** Since $u = \sqrt{x}$, we get: …

Integrate the following function: fraccos √(x)√(x)

Exercise 7.2 · Q26

\frac{\cos \sqrt{x}}{\sqrt{x}}

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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The integral $\int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx$ is solved by substituting $u = \sqrt{x}$ , which simplifies the integrand to $2 \cos u$ . The final answer is $2 \sin \sqrt{x} + C$ .

The key to this problem is noticing that the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$ , and we have a $\frac{1}{\sqrt{x}}$ factor sitting right next to $\cos \sqrt{x}$ . That’s a dead giveaway for U Substitution — we let the inner function be $u$ , and the rest of the integrand becomes its derivative (up to a constant factor).

Let’s walk through it.

Set up the substitution.

Let $u = \sqrt{x}$ . Then $x = u^2$ , so $dx = 2u \, du$ .

Why this choice? Because $\sqrt{x}$ appears inside the cosine, and its derivative $\frac{1}{2\sqrt{x}}$ is almost exactly the $\frac{1}{\sqrt{x}}$ we have. The substitution will collapse the whole expression into something clean.
Rewrite the integral in terms of $u$ .

The original integral is

\int \frac{\cos \sqrt{x}}{\sqrt{x}} \, dx.

Replace $\sqrt{x}$ with $u$ , and $dx$ with $2u \, du$ :

\int \frac{\cos u}{\sqrt{x}} \cdot 2u \, du.

But $\sqrt{x} = u$ , so $\frac{1}{\sqrt{x}} = \frac{1}{u}$ . That gives:

\int \frac{\cos u}{u} \cdot 2u \, du = \int 2 \cos u \, du.

The $u$ cancels beautifully — that’s the whole point of the substitution.

Integrate.

\int 2 \cos u \, du = 2 \sin u + C.

Substitute back. Since $u = \sqrt{x}$ , we get: …

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Q.Integrate the following function: $\frac{\cos \sqrt{x}}{\sqrt{x}}$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

Method: Substitute $u=\sqrt{x}$ when $\frac{1}{\sqrt{x}}$ accompanies it

Steps

Common Mistakes

Q.Integrate the following function: cos⁡xx\frac{\cos \sqrt{x}}{\sqrt{x}}x​cosx​​

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Q.Integrate the following function: $\frac{\cos \sqrt{x}}{\sqrt{x}}$

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