Question

If the value of the definite integral $\underset{\pi /6}{\overset{\pi /4}{\int }}\frac{1+\cot x}{e^x\sin x}dx$, is equal to $a{e}^{-\pi /6}+b{e}^{-\pi /4}$ then $(a+b)$ equals

• A. $2-\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $2\sqrt{2}-2$
• D. $2\sqrt{3}-\sqrt{2}$

Answer 1

Answer: (A)

$\underset{\pi /6}{\overset{\pi /4}{\int }}{e}^{-x}(\mathrm{cosec}x+\cot x\mathrm{cosec}x)dx;\text{ put }-x=t;dx=-dt$
$-\underset{-\pi /6}{\overset{-\pi /4}{\int }}{e}^t(-\mathrm{cosec}(t)+\cot (t)\cdot \mathrm{cosec}(t))dt=\underset{-\pi /6}{\overset{-\pi /4}{\int }}{e}^t(\mathrm{cosec}(t)-\cot (t)\cdot \mathrm{cosec}(t))dt$
$=e^t\mathrm{cosec}(t{)}_{-\pi /6}^{-\pi /4}=-\sqrt{2}{e}^{-\frac{\pi }{4}}+2e^{-\frac{\pi }{6}}\Rightarrow 2e^{-\frac{\pi }{6}}-\sqrt{2}{e}^{-\frac{\pi }{4}}\Rightarrow a+b=2-\sqrt{2}$