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Q.
If the value of the definite integral $\int\limits_{\pi / 6}^{\pi / 4} \frac{1+\cot x}{e^x \sin x} d x$, is equal to $a e^{-\pi / 6}+b e^{-\pi / 4}$ then $(a+b)$ equals
Integrals
Solution:
$\int\limits_{\pi / 6}^{\pi / 4} e^{-x}(\operatorname{cosec} x+\cot x \operatorname{cosec} x) d x ; \text { put }-x=t ; d x=-d t $
$-\int\limits_{-\pi / 6}^{-\pi / 4} e^t(-\operatorname{cosec}(t)+\cot (t) \cdot \operatorname{cosec}(t)) d t=\int\limits_{-\pi / 6}^{-\pi / 4} e^t(\operatorname{cosec}(t)-\cot (t) \cdot \operatorname{cosec}(t)) d t$
$=e^t \operatorname{cosec}(t)_{-\pi / 6}^{-\pi / 4}=-\sqrt{2} e^{-\frac{\pi}{4}}+2 e^{-\frac{\pi}{6}} \Rightarrow 2 e^{-\frac{\pi}{6}}-\sqrt{2} e^{-\frac{\pi}{4}} \Rightarrow a+b=2-\sqrt{2} $