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Definite Integration

Section: Mathematics

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Question : 9 of 54

Marks: +1, -0

\int\limits_{5\pi}^{25\pi} \left| \sin 2x + \cos 2x \right| \, dx =

[AP EAPCET 22 May 2025 S2]

$20\sqrt{2}$
$10\sqrt{2}$
$40\sqrt{2}$
$80\sqrt{2}$

Solution:

\int\limits_{5\pi}^{25\pi} \sin 2x + \cos 2x \, \vert \, dx

\Rightarrow \left| \sin 2x + \cos 2x \right| = \sqrt{2} \left| \sin\left(2x + \frac{\pi}{4}\right) \right|

\Rightarrow \int\limits_{5\pi}^{25\pi} \sin 2x + \cos 2x \, \vert \, dx

= \int\limits_{5\pi}^{25\pi} \sqrt{2} \left| \sin\left(2x + \frac{\pi}{4}\right) \right| \, dx

Let

u = 2x + \frac{\pi}{4}

du = 2 \, dx

dx = \frac{1}{2} \, du

x = 5\pi, \quad u = 10\pi + \frac{\pi}{4}

x = 25\pi, \quad u = 50\pi + \frac{\pi}{4}

= \sqrt{2} \int_{10\pi + \frac{\pi}{4}}^{50\pi + \frac{\pi}{4}} \left| \sin u \right| \cdot \frac{1}{2} \, du

= \frac{\sqrt{2}}{2} \int_{10\pi + \frac{\pi}{4}}^{50\pi + \frac{\pi}{4}} \left| \sin u \right| \, du

The period of

\left| \sin u \right|

is

\pi

.

The length of the interval of integration is

= \left(50\pi + \frac{\pi}{4}\right) - \left(10\pi + \frac{\pi}{4}\right) = 40\pi

= \frac{\sqrt{2}}{2} \cdot 40 \int\limits_{0}^{\pi} \left| \sin u \right| \, du

= \frac{\sqrt{2}}{2} \times 40 \times 2

= 40\sqrt{2}

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