Question

निम्नलिखित का मान निकालिए-

`int_"0"^pi  (x"d"x)/(1 + sin x)`

Answer 1

मान लीजिए I = `int_"0"^pi  (x"d"x)/(1 + sin x)`  .....(i)

= `int_0^pi (pi - x)/(1 + sin(pi - x)) "d"x`  ......`["उपयोग"  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)

(i) और (ii), को जोड़ने पर हमें प्राप्त होता है

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`

अत:, I = `pi`