∫_(π/4) ^π (√(1−sin2x)) dx


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Question Number 106125 by Ar Brandon last updated on 02/Aug/20

$\int_{\frac{π}{4}}^{π} \sqrt{1 - \sin 2 x} d x$
	Answered by Dwaipayan Shikari last updated on 02/Aug/20

	$\int_{\frac{π}{4}}^{π} sinx - cosx dx or \int_{\frac{π}{4}}^{π} cosx - sinx$ $- {[(sinx + cosx)]}_{\frac{π}{4}}^{π} = 1 + \sqrt{2}$ $or \int_{\frac{π}{4}}^{π} cosx - sinx = {[sinx + cosx]}_{\frac{π}{4}}^{π} = - (1 + \sqrt{2})$
	Commented by Ar Brandon last updated on 02/Aug/20

	$OK . Below is my view point . I^{'} ll like to have your$ $opinion .$
	Answered by Ar Brandon last updated on 02/Aug/20
	$f(x)=(√(1−sin2x))=(√((cosx−sinx)^2 )) =∣cosx−sinx∣=(√2)∣cos(x+(π/4))∣ f(x)= { (((√2)cos(x+(π/4)) for −(3/4)≤x≤(π/4))),((−(√2)cos(x+(π/4)) for (π/4)<x<((5π)/4))) :} ⇒∫_(π/4) ^π f(x)dx=−(√2)∫_(π/4) ^π cos(x+(π/4))dx =−(√2)[sin(x+(π/4))]_(π/4) ^π =−(√2)(−(1/(√2))−1) =(√2)+1$
	$f (x) = \sqrt{1 - \sin 2 x} = \sqrt{{(cosx - sinx)}^{2}}$ $=∣ cosx - sinx ∣= \sqrt{2} ∣ \cos (x + \frac{π}{4}) ∣$ $f (x) = {\begin{cases} \sqrt{2} \cos (x + \frac{π}{4}) for - \frac{3}{4} ⩽ x ⩽ \frac{π}{4} \\ - \sqrt{2} \cos (x + \frac{π}{4}) for \frac{π}{4} < x < \frac{5 π}{4} \end{cases}$ $\Rightarrow \int_{\frac{π}{4}}^{π} f (x) dx = - \sqrt{2} \int_{\frac{π}{4}}^{π} \cos (x + \frac{π}{4}) dx$ $= - \sqrt{2} {[\sin (x + \frac{π}{4})]}_{\frac{π}{4}}^{π} = - \sqrt{2} (- \frac{1}{\sqrt{2}} - 1)$ $= \sqrt{2} + 1$
	Commented by 1549442205PVT last updated on 03/Aug/20

	$By the hypothesis the interval to take$ $integration be [\frac{π}{4}; π], so \frac{π}{2} ⩽ \frac{π}{4} + x ⩽ \frac{5 π}{4}$ $\Rightarrow \cos (\frac{π}{4} + x) < 0$
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