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Evaluate :`int_0^(pi/2)dx/(1+cotx)`
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Solution
Let `I=int_0^(pi/2)dx/(1+cotx)`
` =int_0^(pi/2)dx/(1+(cosx)/(sinx))`
`=int_0^(pi/2)dx/((sinx+cosx)/sinx)`
`=int_0^(pi/2)sinx/(sinx+cosx)dx..................(i)`
We know
`int_0^af(x)dx=int_0^af(a-x)dx`
`I=int_0^(pi/2)(sin(pi/2-x))/(sin(pi/2-x)+cos(pi/2-x))`
`I=int_0^(pi/2)cosx/(cosx+sinx)dx.................(ii)`
Adding (i) and (ii)
`I+I=int_0^(pi/2)sinx/(sinx+cosx)dx+int_0^(pi/2)cosx/(cosx+sinx)dx`
`2I=int_0^(pi/2)(sinx+cosx)/(sinx+cosx)dx`
`2I=int_0^(pi/2)1dx=[x]_0^(pi/2)`
`I=1/2[x]_0^(pi/2)=1/2[pi/2-0]`
`I=pi/4`
Concept: Indefinite Integration - Integration by Parts
Is there an error in this question or solution?