MCQ
$\int_0^{\pi /2} {{e^x}\sin x\,dx = } $
- A$\frac{1}{2}({e^{\pi /2}} - 1)$
- ✓$\frac{1}{2}({e^{\pi /2}} + 1)$
- C$\frac{1}{2}(1 - {e^{\pi /2}})$
- D$2({e^{\pi /2}} + 1)$
$= - [{e^x}\cos x]_0^{\pi /2} + \int_0^{\pi /2} {{e^x}\cos x\,dx} $
$ = - [{e^x}\cos x]_0^{\pi /2} + [{e^x}\sin x]_0^{\pi /2} - \int_0^{\pi /2} {{e^x}\sin x\,dx} $
$\therefore $$2I = [{e^x}(\sin x - \cos x)]_0^{\pi /2} = ({e^{\pi /2}} + 1)$
Hence $\int_0^{\pi /2} {{e^x}\sin xdx = \frac{1}{2}({e^{\pi /2}} + 1)} $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.