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Question
Evaluate the following:
`int_"0"^pi (x"d"x)/(1 + sin x)`
Solution
Let I = `int_"0"^pi (x"d"x)/(1 + sin x)` .....(i)
= `int_0^pi (pi - x)/(1 + sin(pi - x)) "d"x` ......`["Using" int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x)"d"x]`
= `int_0^pi (pi - x)/(1 + sinx) "d"x` ......(ii)
Adding (i) and (ii), we get
2I = `int_0^pi (x/(1 + sinx) + (pi - x)/(1 + sinx)) "d"x`
= `int_0^pi ((x + pi - x)/(1 + sinx))"d"x`
= `int_0^pi pi/(1 + sin x) "d"x`
= `pi int_0^pi 1/(1 + sinx) "d"x`
= `pi int_0^pi (1.(1 - sinx))/((1 + sinx)(1 - sinx)) "d"x`
= `pi int_0^pi (1 - sinx)/(1 - sin^2x) "d"x`
= `pi int_0^pi (1 - sinx)/(cos^x) "d"x`
= `pi int_0^pi (1/(cos^2x) - sinx/(cos^2x))"d"x`
= `pi int_0^pi (sec^2x - secx tanx)"d"x`
= `pi[tanx - sec]_0^pi`
= `pi[tan pi - tan 0) - (sec pi - sec 0)]`
2I = `pi[0 - (-1 - 1)`
= `pi`(2)
∴ I = `pi`
Hence, I = `pi`
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