Find  \int \sqrt{1-\sin2x}\: dx,\: \: \frac{\pi }{4}< x< \frac{\pi }{2}

 

 

 

 

 
 
 
 
 

Answers (1)

I=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sqrt{1-\sin 2x}\: dx

=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sqrt{\sin^2x+\cos^2 x-2\sin x \cos x}\: dx

=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}[(\cos x-\sin x)^2]^\frac{1}{2}\: dx

=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}|\cos x-\sin x|\: dx     

\left [ \cos x < \sin x \Rightarrow \cos x - \sin x <0 \right ]

  =\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}-(\cos x-\sin x)\: dx

=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}-\cos x\: dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin x\: dx

=-[\sin x + \cos x]_\frac{\pi}{4}^\frac{\pi}{2}

=-\left \{ \left ( \sin \frac{\pi}{2}+\cos \frac{\pi }{2} \right ) - \left ( \sin \frac{\pi}{4}+\cos \frac{\pi}{4} \right ) \right \}

=-\left \{ (1+0)-(2\cdot \frac{1}{\sqrt{2}}) \right \}

=\sqrt{2}-1

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