Question

$\underset{0}{\stackrel{\pi /2}{\int }}(\cos x-\sin x)e^{x}dx$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $-1$

$\int (\cos x-\sin x)e^{x}dx$

$=(\cos x-\sin x)e^x+\int (\cos x+\sin x)e^{x}dx$ ..... integration by parts

$=(\cos x-\sin x)e^x+(\cos x+\sin x)e^x-\int (\cos x-\sin x)e^{x}dx$ ..... integration by parts

$\Rightarrow 2\int (\cos x-\sin x)e^{x}dx=2e^x\cos x$

$\Rightarrow \int (\cos x-\sin x)e^{x}dx=e^x\cos x$

${\int }_0^{\frac{\pi }{2}}(\cos x-\sin x)e^{x}dx$

$={\left[e^x\cos x\right]}_0^{\frac{\pi }{2}}$

$=[0-1]$

$=-1$