Let f x = ∫ 0 x g t log e 1 - t 1 + t dt , where g is a continuous odd function. If ∫ - π 2 π 2 f x + x 2 cosx 1 + e x dx = π α 2 - α , then α is equal to _____.

Step-by-step solution

Given: f x = ∫ 0 x g t log 1 - t 1 + t dt ⇒ f - x = ∫ 0 - x g t log 1 - t 1 + t dt ⇒ f - x = - ∫ 0 x g - y log 1 + y 1 - y dy = - ∫ 0 x g ( y ) ln 1 - y 1 + y dy ( g is odd) f ( - x ) = - f ( x ) ⇒ f is also odd. Now, I = ∫ - π 2 π 2 f ( x ) + x 2 cosx 1 + e x dx . . . i ⇒ I = ∫ - π 2 π 2 f ( - x ) + x 2 e x cosx 1 + e x dx . . . ii ⇒ 2 I = ∫ - π 2 π 2 x 2 cosxdx ⇒ 2 I = 2 ∫ 0 π 2 x 2 cosxdx ⇒ I = ∫ 0 π 2 x 2 cosxdx ⇒ I = x 2 sinx 0 π 2 - ∫ 0 π 2 2 xsinxdx ⇒ I = π 2 4 - 2 - xcosx + sinx 0 π 2 ⇒ I = π 2 4 - 2 ( 0 +