Integrate the following function: $ \frac{\cos x}{\sqrt{1 + \sin x}} $

Integrate the following function: fraccos x√(1 + sin x)

Exercise 7.2 · Q28

\frac{\cos x}{\sqrt{1 + \sin x}}

Rajasthan RbseTextbookSubjective· 2mImportance★★★★★

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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 simplifies by substituting $u = 1 + \sin x$ , turning the integrand into $\frac{1}{\sqrt{u}}$ , which integrates to $2\sqrt{1 + \sin x} + C$ .

The key to this problem is noticing that the numerator, $\cos x$ , is exactly the derivative of $\sin x$ . That’s a dead giveaway for a u-substitution: whenever you see a function and its derivative multiplied together (or nearly so), substitution is your friend. Here, the denominator contains $1 + \sin x$ , and its derivative is $\cos x$ — which sits right in the numerator. That’s not a coincidence; it’s the signal to set $u = 1 + \sin x$ .

Let’s walk through it cleanly.

Choose the substitution. Let $u = 1 + \sin x$ . Then differentiate:

\frac{du}{dx} = \cos x \quad \Rightarrow \quad du = \cos x \, dx.

This is perfect because the numerator $\cos x \, dx$ is exactly $du$ .

Rewrite the integral in terms of $u$ . The original integral is

\int \frac{\cos x}{\sqrt{1 + \sin x}} \, dx = \int \frac{1}{\sqrt{u}} \, du.

Notice how the messy trigonometric expression collapses into a simple power of $u$ .

Integrate with respect to $u$ . Recall that $\frac{1}{\sqrt{u}} = u^{-1/2}$ . So $\int u^{-1/2} \, du = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C.$ …

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Q.Integrate the following function: $\frac{\cos x}{\sqrt{1 + \sin x}}$

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 the expression under the root; its derivative is the numerator

Steps

Common Mistakes

Q.Integrate the following function: cos⁡x1+sin⁡x\frac{\cos x}{\sqrt{1 + \sin x}}1+sinx​cosx​

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Q.Integrate the following function: $\frac{\cos x}{\sqrt{1 + \sin x}}$

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