Question

The value of the definite integral $\int\limits_0^{\pi / 2}(\sin x+\cos x) \cdot \sqrt{\frac{e^x}{\sin x}} d x$ equals

• A. $2 \sqrt{ e ^{\pi / 2}}$
• B. $\sqrt{ e ^{\pi / 2}}$
• C. $2 \sqrt{ e ^{\pi / 2}} \cdot \cos 1$
• D. $\frac{1}{2} e ^{\pi / 4}$

Answer 1

Answer: (A)

$ I=\int\limits_0^{\pi / 2} \frac{e^x(\sin x+\cos x)}{\sqrt{e^x \sin x}} d x$
$\text { put } e^x \sin x=t^2 \Rightarrow e^x(\sin x+\cos x) d x=2 t d t $
$\therefore \left.I=2 \int\limits_0^{\pi / 4} d t=2 t\right]_0^{e^{\pi / 4}}=2 e^{\pi / 4}=2 \sqrt{e^{\pi / 2}} $