Mathematics

$$\int_{\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{sin x +cos x}{\sqrt{sin 2x}}dx$$


SOLUTION

We have,

$$ \int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx} $$

$$ =\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\sin x+\cos x}{\sqrt{1-1+\sin 2x}}dx} $$

$$ =\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx} $$

$$ =\int_{-\dfrac{\pi }{6}}^{\dfrac{\pi }{6}}{\dfrac{\dfrac{d}{dx}\left( \sin x-\cos x \right)}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx}\,\,\,\,\,\,\therefore \dfrac{1}{\sqrt{1-{{x}^{2}}}}dx={{\sin }^{-1}}x $$

$$ ={{\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin x-\cos x \right) \right]}_{-\dfrac{\pi }{6}}}^{\dfrac{\pi }{6}} $$

$$ =\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin \left( -\dfrac{\pi }{6} \right)-\cos \left( -\dfrac{\pi }{6} \right) \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right) \right) \right] $$

$$ =\left[ \sqrt{2}{{\sin }^{-1}}\left( -\sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right) \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{6} \right)-\cos \left( \dfrac{\pi }{6} \right) \right) \right] $$

$$ =\left[ \sqrt{2}{{\sin }^{-1}}\left( -\dfrac{1}{2}-\dfrac{\sqrt{3}}{2} \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \dfrac{1}{2}-\dfrac{\sqrt{3}}{2} \right) \right] $$

$$ =\left[ \sqrt{2}{{\sin }^{-1}}\left( \dfrac{-1-\sqrt{3}}{2} \right) \right]-\left[ \sqrt{2}{{\sin }^{-1}}\left( \dfrac{1-\sqrt{3}}{2} \right) \right] $$

$$ =\sqrt{2}\left[ {{\sin }^{-1}}\left( \dfrac{-1-\sqrt{3}}{2} \right) \right]-\left[ {{\sin }^{-1}}\left( \dfrac{1-\sqrt{3}}{2} \right) \right] $$

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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