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Question
Prove that `int_0^af(x)dx=int_0^af(a-x) dx`
hence evaluate `int_0^(pi/2)sinx/(sinx+cosx) dx`
Solution
Let I = `int_0^af(x)dx`
Put x = a – t
∴ dx = – dt
When x = 0, t = a - 0 = a
When x = a, t = a - a = 0
`I=int_0^af(x)dx=int_a^0f(a-t)(-dt)`
`=-int_a^0f(a-t)dt` ......[`∵int_a^bf(x)dx=-int_b^af(x)dx`]
`=int_0^af(a-x)dx` ......[`∵int_a^bf(x)dx=-int_b^af(t)dx`]
`therefore int_0^af(x)dx=int_0^af(a-x)dx`
Let I=`int_0^(pi/2)sinx/(sinx+cosx)` ........(i)
`I=int_0^(pi/2)sin(pi/2-x)/(sin(pi/2-x)+cos(pi/2-x))` ......[`∵int_0^af(x)dx=-int_0^af(a-x)dx`]
`=int_0^(pi/2)cosx/(cosx+sinx)dx` ............(ii)
Adding (i) and (ii), we get
`2I=int_0^(pi/2)(sinx+cosx)/(sinx+cosx)dx`
`=int_0^(pi/2)1 dx`
`=[x]_0^(pi/2)`
`=pi/2-0`
`2I=pi/2`
`I=pi/4`
`therefore int_0^(pi/2)sinx/(sinx+cosx)dx=pi/4`
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