$\int_{0}^{\pi/2} \sqrt{1-\sin 2x}\,dx$ is equal to (A) $2\sqrt{2}$ (B) $2(\sqrt{2}+1)$ (C) $2$ (D) $2(\sqrt{2}-1)$

∫₀π/2 √(1-sin 2x)dx is equal to (A) 2√(2) (B) 2(√(2)+1) (C) 2 (D) 2(√(2)-1)

NCERT Exemplar · Q63

\int_{0}^{\pi/2} \sqrt{1-\sin 2x}\,dx

Punjab PsebMCQ· 1mImportance★★★★★

Appeared in past exams:MHT-CET 2024· Set pcm-2024-05-10-M· 2mreworded

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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 is to rewrite $1-\sin 2x$ as $(\sin x - \cos x)^2$ , then handle the absolute value that arises from the square root. The integral splits at $x = \pi/4$ , and the final value is $2(\sqrt{2} - 1)$ , which matches option (D).

We start with the integral

I = \int_{0}^{\pi/2} \sqrt{1-\sin 2x}\,dx.

The expression $1 - \sin 2x$ looks like it might be a perfect square. Recall the identity $\sin 2x = 2\sin x \cos x$ . Also, $1 = \sin^2 x + \cos^2 x$ . So:

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

That’s neat — the integrand becomes $\sqrt{(\sin x - \cos x)^2}$ , which is $|\sin x - \cos x|$ .

Watch out

A common mistake is to drop the absolute value and write $\sin x - \cos x$ directly. But $\sqrt{u^2} = |u|$ , not $u$ . The sign of $\sin x - \cos x$ changes over $[0, \pi/2]$ , so we must split the interval.

Now, where is $\sin x - \cos x$ positive or negative?

$\sin x = \cos x$ at $x = \pi/4$ . For $x < \pi/4$ , $\cos x > \sin x$ , so $\sin x - \cos x < 0$ . For $x > \pi/4$ , $\sin x > \cos x$ , so $\sin x - \cos x > 0$ .

Thus:

|\sin x - \cos x| = \begin{cases} \cos x - \sin x, & 0 \le x \le \pi/4, \\ \sin x - \cos x, & \pi/4 \le x \le \pi/2. \end{cases}

We break the integral accordingly:

I = \int_{0}^{\pi/4} (\cos x - \sin x)\,dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx.

Now integrate each piece.

First integral: $\int (\cos x - \sin x)\,dx = \sin x + \cos x$ (since derivative of $\sin x$ is $\cos x$ , derivative of $\cos x$ is $-\sin x$ , so the antiderivative of $\cos x - \sin x$ is $\sin x + \cos x$ ). Evaluate from $0$ to $\pi/4$ :

\left[\sin x + \cos x\right]_{0}^{\pi/4} = \left(\sin\frac{\pi}{4} + \cos\frac{\pi}{4}\right) - (\sin 0 + \cos 0) = \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1.

Second integral: …

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Q. $\int_{0}^{\pi/2} \sqrt{1-\sin 2x}\,dx$ is equal to
(A) $2\sqrt{2}$
(B) $2(\sqrt{2}+1)$
(C) $2$
(D) $2(\sqrt{2}-1)$

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: Perfect square under a root implies an absolute value

Steps

Common Mistakes

Q.∫0π/21−sin⁡2x dx\int_{0}^{\pi/2} \sqrt{1-\sin 2x}\,dx∫0π/2​1−sin2x​dx is equal to (A) 222\sqrt{2}22​ (B) 2(2+1)2(\sqrt{2}+1)2(2​+1) (C) 222 (D) 2(2−1)2(\sqrt{2}-1)2(2​−1)

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Q. $\int_{0}^{\pi/2} \sqrt{1-\sin 2x}\,dx$ is equal to
(A) $2\sqrt{2}$
(B) $2(\sqrt{2}+1)$
(C) $2$
(D) $2(\sqrt{2}-1)$

Unlock everything free for 14 days