Evaluate: $\int \sqrt{1+\sin x}\,dx$

> [!TLDR] > The key idea is to rewrite $1+\sin x$ as a perfect square using the identity $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}$. This gives $\sqrt{1+\sin x} = \left|\sin\frac{x}{2}+\cos\frac{x}{2}\right|$, and the integral becomes $2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)\cdot\text{sgn}\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)+C$, or more commonly, $\pm 2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)+C$ depending on the interval. ## Why This Approach Works When you see $\sqrt{1+\sin x}$, your first instinct might be to try a direct substitution — but that leads nowhere because the expression under the square root isn't a simple derivative. The trick is to notice that $1+\sin x$ resembles the expansion of $(\sin\frac{x}{2}+\cos\frac{x}{2})^2$. This is a classic trigonometric identity play: using the double-angle formulas in reverse. Recall: - $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ - $1 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}$ So $1+\sin x = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2} = (\sin\frac{x}{2}+\cos\frac{x}{2})^2$. The square root then becomes $|\sin\frac{x}{2}+\cos\frac{x}{2}|$, and the integral reduces to something we can handle with a simple substitution. > [!WARNING] > A common mistake is to forget the absolute value when taking the square root of a square. $\sqrt{(\sin\frac{x}{2}+\cos\frac{x}{2})^2} = |\sin\frac{x}{2}+\cos\frac{x}{2}|$, not $\sin\frac{x}{2}+\cos\frac{x}{2}$. The sign matters, and the final answer must account for intervals where the expression is negative. ## Step-by-Step Solution 1. **Rewrite the integrand using the identity.** Start with $1+\sin x$. Write $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}$. Then: $$1+\sin x = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2.$$ 2. **Take the square root.** $$\sqrt{1+\sin x} = \sqrt

Evaluate: ∫ √(1+sin x)dx

NCERT Exemplar · Q9

\int \sqrt{1+\sin x}\,dx

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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 key idea is to rewrite $1+\sin x$ as a perfect square using the identity $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}$ . This gives $\sqrt{1+\sin x} = \left|\sin\frac{x}{2}+\cos\frac{x}{2}\right|$ , and the integral becomes $2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)\cdot\text{sgn}\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)+C$ , or more commonly, $\pm 2\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)+C$ depending on the interval.

Why This Approach Works

When you see $\sqrt{1+\sin x}$ , your first instinct might be to try a direct substitution — but that leads nowhere because the expression under the square root isn't a simple derivative. The trick is to notice that $1+\sin x$ resembles the expansion of $(\sin\frac{x}{2}+\cos\frac{x}{2})^2$ . This is a classic trigonometric identity play: using the double-angle formulas in reverse.

Recall:

$\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$
$1 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}$

So $1+\sin x = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2} = (\sin\frac{x}{2}+\cos\frac{x}{2})^2$ .

The square root then becomes $|\sin\frac{x}{2}+\cos\frac{x}{2}|$ , and the integral reduces to something we can handle with a simple substitution.

Watch out

A common mistake is to forget the absolute value when taking the square root of a square. $\sqrt{(\sin\frac{x}{2}+\cos\frac{x}{2})^2} = |\sin\frac{x}{2}+\cos\frac{x}{2}|$ , not $\sin\frac{x}{2}+\cos\frac{x}{2}$ . The sign matters, and the final answer must account for intervals where the expression is negative.

Step-by-Step Solution

Rewrite the integrand using the identity. Start with $1+\sin x$ . Write $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1 = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}$ . Then:

1+\sin x = \sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2.

Take the square root.

\sqrt{1+\sin x} = \sqrt{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2} = \left|\sin\frac{x}{2}+\cos\frac{x}{2}\right|.

Simplify the absolute value (optional but helpful).

Notice that $\sin\frac{x}{2}+\cos\frac{x}{2} = \sqrt{2}\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)$ . This form makes it easier to see where the expression is positive or negative. The absolute value becomes $\sqrt{2}\left|\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)\right|$ .
Set up the integral.

\int \sqrt{1+\sin x}\,dx = \int \left|\sin\frac{x}{2}+\cos\frac{x}{2}\right|\,dx.

Substitute to simplify. Let $u = \frac{x}{2}$ , so $dx = 2\,du$ . Then:

\int \left|\sin u + \cos u\right| \cdot 2\,du = 2\int \left|\sin u + \cos u\right|\,du.

Integrate the absolute value. The integral of $|\sin u + \cos u|$ depends on the interval. But we can find an antiderivative by considering the sign. Notice that $\frac{d}{du}(\sin u - \cos u) = \cos u + \sin u$ . So: …

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Q.Evaluate: $\int \sqrt{1+\sin x}\,dx$

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?

Why This Approach Works

Step-by-Step Solution

Method: Rewriting $\sqrt{1\pm\sin x}$ via half-angle squares

Steps

Common Mistakes

Q.Evaluate: ∫1+sin⁡x dx\int \sqrt{1+\sin x}\,dx∫1+sinx​dx

Why This Approach Works

Step-by-Step Solution

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Q.Evaluate: $\int \sqrt{1+\sin x}\,dx$

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