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  4. The value of the integral ∫ limits(π/6)(π/2) cos ((1/ sin ((1) sin x))) ⋅ cos ((1/ sin x)) ⋅ cos x ⋅ (1/ sin 2 x ⋅ sin 2((1) sin x)) d x equals

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Q. The value of the integral $\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos \left(\frac{1}{\sin \left(\frac{1}{\sin x}\right)}\right) \cdot \cos \left(\frac{1}{\sin x}\right) \cdot \cos x \cdot \frac{1}{\sin ^2 x \cdot \sin ^2\left(\frac{1}{\sin x}\right)} d x$ equals

Integrals

Solution:

$\text { Consider } \frac{d}{d x} \sin \cdot \frac{1}{\sin \left(\frac{1}{\sin x}\right)} $
$\cos \frac{1}{\sin \left(\frac{1}{\sin x}\right)} \cdot(-) \frac{1}{\sin ^2\left(\frac{1}{\sin x}\right)} \cdot \cos \left(\frac{1}{\sin x}\right) \cdot(-1) \frac{1}{\sin ^2 x} \cdot \cos x $
$=\cos \frac{1}{\sin \left(\frac{1}{\sin x}\right)} \cdot \cos \left(\frac{1}{\sin x}\right) \cdot \cos x \cdot \frac{1}{\sin ^2 x \cdot \sin ^2\left(\frac{1}{\sin x}\right)}$
hence the given integrand is $\frac{ d }{ dx } \sin \frac{1}{\sin \left(\frac{1}{\sin x }\right)}$
$\therefore I=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{d}{d x}\left(\sin \frac{1}{\sin \left(\frac{1}{\sin x}\right)}\right)\right) d x=\left(\sin \frac{1}{\sin \left(\frac{1}{\sin x}\right)}\right)_{\pi / 6}^{\pi / 2}=\sin (\operatorname{cosec} 1)-\sin (\operatorname{cosec} 2)$