Question

Prove that `int_0^af(x)dx=int_0^af(a-x) dx`

hence evaluate `int_0^(pi/2)sinx/(sinx+cosx) dx`

Answer 1

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`