Question

${\int }_0^{\pi /4}cosx{e}^{sinx}dx$ is equal to
(a) e + 1 (b) e - 1 (c) e (d) -e

Answer 1

Let I = $I={\int }_0^{\pi /2}cosx{e}^{sinx}dxPutsinx=t\Rightarrow cosxdx=dtAsx\to 0,thent\to 0andx\to \pi /2,thent\to 1\therefore I={\int }_0^1{e}^{t}dt=[e^1{]}_0^1=e^1-e^0=e-1$