Evaluate: $\int \dfrac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\,dx$

> [!TLDR] > The integrand simplifies to a constant sign function ($\pm 1$) because $\sqrt{1+\sin 2x} = |\sin x + \cos x|$. The integral therefore reduces to $\int \text{sgn}(\sin x + \cos x) \, dx$, which is a piecewise linear function. The final result is $\boxed{\text{sgn}(\sin x + \cos x) \cdot x + C}$, where $\text{sgn}$ is the sign function. The key insight here is that the denominator is not just a square root — it’s the **square root of a perfect square**. That changes everything. --- ### 1. Recognize the perfect square under the root We know $\sin 2x = 2\sin x \cos x$. So: $$ 1 + \sin 2x = 1 + 2\sin x \cos x = \sin^2 x + \cos^2 x + 2\sin x \cos x $$ That’s exactly $(\sin x + \cos x)^2$. Check: $$ (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x $$ So the integrand becomes: $$ \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^2}} $$ > [!WARNING] > A common mistake is to cancel $\sqrt{(\sin x + \cos x)^2}$ with $\sin x + \cos x$ and get $1$. But $\sqrt{t^2} = |t|$, not $t$. So cancellation is **not** automatic — it depends on the sign of $\sin x + \cos x$. --- ### 2. Write the integrand using absolute value We have: $$ \frac{\sin x + \cos x}{|\sin x + \cos x|} $$ This ratio is simply the **sign** (or signum) of $\sin x + \cos x$: $$ \frac{\sin x + \cos x}{|\sin x + \cos x|} = \begin{cases} 1, & \sin x + \cos x > 0 \\ -1, & \sin x + \cos x [!TIP] > The expression $\sin x + \cos x$ can be rewritten as $\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$. This makes it easy to see where it changes sign: it’s positive when $\sin\left(x + \frac{\pi}{4}\right) > 0$, i.e., when $x + \frac{\pi}{4} \in (2n\pi, (2n+1)\pi)$, and negative otherwise. --- ### 3. Integrate piecewise On any interval where $\sin x + \cos x$ has a constant sign, the integrand i

Evaluate: ∫ dfracsin x+cos x√(1+sin 2x)dx

NCERT Exemplar · Q8

\int \dfrac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\,dx

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Concept understanding — Trigonometric Simplification

Trigonometric Simplification

You know $\sin^2 x + \cos^2 x = 1$ — but the skill of turning a messy trig expression into that kind of clean form is trigonometric simplification. Because sines, cosines and their relatives are all tied together by identities from the unit circle, a tangled combination can almost always be rewritten as something shorter: a single term, a constant, or an easier combination.

Your core toolkit

\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta

How the process feels

Simplify $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}$ . Over a common denominator the numerator is $\sin^2 x+(1+\cos x)^2 = \sin^2 x + 1 + 2\cos x + \cos^2 x$ . The Pythagorean identity turns $\sin^2 x+\cos^2 x$ into $1$ , giving $2+2\cos x = 2(1+\cos x)$ , so

\frac{2(1+\cos x)}{\sin x\,(1+\cos x)} = \frac{2}{\sin x} = 2\csc x.

A two-term sum collapses to one term.

Strategies that usually work

Convert everything to sines and cosines — cancellations then appear.
Spot Pythagorean pairs and replace them with $1$ (or $\sec^2$ , $\csc^2$ ).
Factor and cancel as you would with ordinary algebra.
Multiply by a conjugate — e.g. multiply $\frac{1}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ to unlock a Pythagorean identity. …

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The integrand simplifies to a constant sign function ( $\pm 1$ ) because $\sqrt{1+\sin 2x} = |\sin x + \cos x|$ . The integral therefore reduces to $\int \text{sgn}(\sin x + \cos x) \, dx$ , which is a piecewise linear function. The final result is $\boxed{\text{sgn}(\sin x + \cos x) \cdot x + C}$ , where $\text{sgn}$ is the sign function.

The key insight here is that the denominator is not just a square root — it’s the square root of a perfect square. That changes everything.

1. Recognize the perfect square under the root

We know $\sin 2x = 2\sin x \cos x$ . So:

1 + \sin 2x = 1 + 2\sin x \cos x = \sin^2 x + \cos^2 x + 2\sin x \cos x

That’s exactly $(\sin x + \cos x)^2$ . Check:

(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x

So the integrand becomes:

\frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^2}}

Watch out

A common mistake is to cancel $\sqrt{(\sin x + \cos x)^2}$ with $\sin x + \cos x$ and get $1$ . But $\sqrt{t^2} = |t|$ , not $t$ . So cancellation is not automatic — it depends on the sign of $\sin x + \cos x$ .

2. Write the integrand using absolute value

We have:

\frac{\sin x + \cos x}{|\sin x + \cos x|}

This ratio is simply the sign (or signum) of $\sin x + \cos x$ :

\frac{\sin x + \cos x}{|\sin x + \cos x|} = \begin{cases} 1, & \sin x + \cos x > 0 \\ -1, & \sin x + \cos x < 0 \end{cases}

At points where $\sin x + \cos x = 0$ , the integrand is undefined (division by zero), so the integral is defined piecewise on intervals where the sign is constant.

Tip

The expression $\sin x + \cos x$ can be rewritten as $\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)$ . This makes it easy to see where it changes sign: it’s positive when $\sin\left(x + \frac{\pi}{4}\right) > 0$ , i.e., when $x + \frac{\pi}{4} \in (2n\pi, (2n+1)\pi)$ , and negative otherwise.

3. Integrate piecewise

On any interval where $\sin x + \cos x$ has a constant sign, the integrand is just $\pm 1$ . So:

If $\sin x + \cos x > 0$ , the integral is $\int 1 \, dx = x + C$ .
If $\sin x + \cos x < 0$ , the integral is $\int (-1) \, dx = -x + C$ . …

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Q.Evaluate: $\int \dfrac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\,dx$

Concept understanding — Trigonometric Simplification

Trigonometric Simplification

Your core toolkit

How the process feels

Strategies that usually work

1. Recognize the perfect square under the root

2. Write the integrand using absolute value

3. Integrate piecewise

Method: Perfect-square-under-a-root simplification

Steps

Common Mistakes

Q.Evaluate: ∫sin⁡x+cos⁡x1+sin⁡2x dx\int \dfrac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\,dx∫1+sin2x​sinx+cosx​dx

1. Recognize the perfect square under the root

2. Write the integrand using absolute value

3. Integrate piecewise

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Q.Evaluate: $\int \dfrac{\sin x+\cos x}{\sqrt{1+\sin 2x}}\,dx$

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