JAMB 2013 · UTME · Q39Calculus (JAMB bridge)Evaluate ∫0π/2sinx dx\displaystyle\int_0^{\pi/2} \sin x\,dx∫0π/2sinxdx.A−2-2−2B2C1D−1-1−1Worked solution (try it first)sinx\sin xsinx integrates to −cosx-\cos x−cosx.[−cosx]0π/2=−cosπ2−(−cos0)\left[-\cos x\right]_0^{\pi/2} = -\cos\frac\pi2 - (-\cos0)[−cosx]0π/2=−cos2π−(−cos0)=0+1= 0 + 1=0+1=1= 1=1, option C.Report a problem with this question