pi-4-pi-1-sin2x-dx-

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) = \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