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Evaluate : int0^(pi/2)(sinx+cosx)dx...

Evaluate : `int_0^(pi/2)(sinx+cosx)dx`

Text Solution

Verified by Experts

`I = int_0^(pi/2) (sinx+cosx)dx`
`=>I = [-cosx+sinx]_0^(pi/2)`
`=>I = [-cos(pi/2)+sin(pi/2) + cos (0) - sin (0)]`
`=> I = [0+1+1 - 0]`
`=> I = 2`
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